(ProQuest: ... denotes non-US-ASCII text omitted.)
Academic Editor:G. M. N'Guérékata
Instituto Universitario de Matemática Multidisciplinar, Universitat Politècnica de València, Camino de Vera s/n, 46022 Valencia, Spain
Received 20 June 2013; Accepted 14 October 2013; 21 January 2014
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and Motivation
Over the last few decades, random differential equations (RDEs) have been demonstrated to be powerful tools to model numerous problems appearing in many different areas such as physics, engineering, economics, epidemiology, and hydrology. The consideration of randomness into their formulation through initial/boundary conditions, source terms, and/or coefficients adapts better than their deterministic counterpart to model the uncertainty associate to the experimental measurement required to set the above inputs as well as the inherent complexity involved in many real modelling problems. This approach leads to face new and exciting questions different from the corresponding ones appearing in the deterministic scenario. Indeed, instead of obtaining just the solution stochastic process (SP) of RDEs, the theory is also concerned with its probabilistic properties, mainly the computations of the expectation and variance functions. The computation of the first probability density function (1-PDF) of the solution SP, say x ( t ) , is much more desirable since, from it, one can compute the previous statistical functions as simple particular cases and, in addition, it provides a comprehensive probabilistic description of the solution SP for each time instant t . However, the computation of the 1-PDF constitutes a major challenge that can been achieved in only a few cases.
The aim of this paper is to determine the 1-PDF, f 1 ( z , t ) , of the solution SP Z ( t ) to the linear random initial value problem (IVP): [figure omitted; refer to PDF] where the data Z 0 , B and A are assumed to be continuous random variables (continuous RVs) defined on a common probability space ( Ω , ... , ... ) , whose domains are assumed to be [figure omitted; refer to PDF] respectively. Hereinafter, in order to avoid cumbersome notation, we will hide the ω sample dependence when writing domains of continuous RVs. In this way, for instance, the domain D Z 0 will be written as D Z 0 = { z 0 : z 0,1 ...4; z 0 ...4; z 0,2 } rather than the first expression in (2). The same can be said for D B , D A , and the domain of any other RV throughout this paper. We allow the left (right) endpoint of each interval of the domain takes the value - ∞ ( + ∞ ); that is, we also consider unbounded continuous RVs. Throughout the paper, we will denote by f Z 0 ( z 0 ) , f B ( b ) , and f A ( a ) the PDFs of the continuous RVs Z 0 , B , and A , respectively. The case where Z 0 , B and A are pairwise dependent continuous RVs will also be treated. In such case, f Z 0 , B ( z 0 , b ) , f Z 0 , A ( z 0 , a ) , and f B , A ( b , a ) will denote the joint PDFs of the random vectors: ( Z 0 , B ) , ( Z 0 , A ) , and ( B , A ) , respectively. Finally, we will also deal with the case where Z 0 , B , and A are dependent continuous RVs, then f Z 0 , B , A ( z 0 , b , a ) will represent their joint PDF. Notice that the domains of these two- and three-dimensional PDFs often can be written directly as products of the sets D Z 0 , D B , and D A given by (2).
In order to compute the 1-PDF f 1 ( z , t ) , random variable transformation (RVT) method will be applied. RVT is a probability technique that allows us to calculate the PDF f Y ( y ) of a RV Y resulting after the algebraic transformation of another RV, say X , whose PDF, f X ( x ) , is known. In its simplest scalar formulation, the method reads as follows: if X is a continuous RV lying on the domain or support D X = { x : x 1 ...4; x ...4; x 2 } , whose PDF is f X ( x ) > 0 and Y = r ( X ) being r : D X ⊆ ... [arrow right] ... a monotone mapping on D X , then [figure omitted; refer to PDF] where s ( y ) = x is the inverse function of r on D X , which is assumed to have a continuous derivative on D Y and | d s ( y ) / d y | denotes the modulus of the derivative of s ( y ) . In the particular case that r increases (decreases) on D X , the domain D Y of Y = r ( X ) is determined by D Y = { y : y 1 = r ( x 1 ) ...4; y ...4; r ( x 2 ) = y 2 } ( D Y = { y : y 1 = r ( x 2 ) ...4; y ...4; r ( x 1 ) = y 2 } ).
Notice that we are interested in computing the 1-PDF f 1 ( z , t ) to the solution Z ( t ) of (1) which is a SP rather than a RV, whereas RVT technique is mainly designed to handle (transformations of) continuous RVs. In order to take advantage of RVT, we first will fix t ^ : t ^ > t 0 and then we will apply RVT to the (transformed) RV Z ( t ^ ) . Therefore, we can say that RVT technique provides a time-transversal description of the 1-PDF f 1 ( z , t ) .
Some of the earliest applications of RVT method to RDEs can be found in [1, ch.6] where this technique is applied to study a linear oscillator assuming randomness just in the two initial conditions related to the position and velocity. Most of the subsequent contributions have focused on the study of particular equations assuming specific probabilistic distributions for the involved uncertainty which facilitates the analysis. Here, we point out some recent contributions that illustrate quite well the current trends of RVT method in dealing with RDEs. In [2], authors solve the radiative transfer equation in a semi-infinite continuous stochastic medium with Rayleigh scattering. RVT method is applied to obtain the 1-PDF of the solution when the optical depth space variable is assumed to be a RV belonging to the following particular distributions: exponential and Gaussian. Higher order statistical moments of the solution stochastic process are also computed. An analogous study on the stochastic transport equation of neutral particles with anisotropic scattering can be found in [3]. In [4, 5], authors apply RVT technique to develop a stochastic finite element method for solving some stochastic problems with random excitation.
The application of RVT technique to the exact determination of the 1-PDF of the solution SP of RDEs requires the previous computation of the exact solution of the RDE under study. However, in the outstanding contribution [6], author takes advantage of RVT method together with classical numerical techniques to illustrate, through a wide range of examples, the potentiality of this method to approximate the 1-PDF for the solution SP of some RDEs.
As it has been announced previously, in this paper, we will compute the 1-PDF of the IVP (1) whose exact solution is available. For the sake of clarity, in the presentation, we will divide the study in the three main IVPs (I)-(III) listed in Table 1. In Case (I), we consider the homogeneous (H) problem whereas Cases (II) and (III) deal with the nonhomogeneous (NH) cases. Within each case, we distinguish, in a systematic manner, the different possibilities regarding the randomness of each of the involved parameters Z 0 , B , and A . These casuistries include the situations where the parameters are statistically dependent. In this context, and as it has been pointed out previously, if for example Z 0 and B are statistically dependent, then f Z 0 , B ( z 0 , b ) will denote the joint PDF of the random vector ( Z 0 , B ) , and the same can be said for the rest of the possibilities. The IVPs (I) and (II) can be seen as particular cases of the IVP (III) when ... [ { ω ∈ Ω : B ( ω ) = 0 } ] = 1 and ... [ { ω ∈ Ω : A ( ω ) = 0 } ] = 1 , respectively. When uncertainty can only be attributted to Z 0 and B , and the parameter A can be set in a deterministic way, it is more realistic and convenient to assume that the joint PDF f Z 0 , B ( z 0 , b ) is known rather than f Z 0 , B , A ( z 0 , b , a ) . Notice that the construction of the joint PDF of only two continuous RVs from measured data can become a difficult problem which accuracy can deteriorate severely if one includes a new and inappropriate RV into its formulation [7]. The most accuracy of the PDF of the random input parameter, the best approximation of the 1-PDF of the solution SP of the IVP (1). Therefore, the consideration of all the thirteen separate cases listed in Table 1 turns out more recommendable from a practical point of view.
Table 1: List of the thirteen different cases considered to perform the full study depending on whether the IVP is homogeneous (H) or non-homogeneous (NH) and the form that uncertainty is considered into the problem: just through one RV: Z 0 , B , or A (Cases I-II-III.1, I-II-III.2, and III.3), two RVs: ( Z 0 , B ), ( Z 0 , A ), or ( B , A ) (Cases I-II.3 and III.4-5-6), and the three involved RVs: ( Z 0 , B , A ) (Case III.7).
Type | IVP | Case | |
H | Z ( t ) = A Z ( t ) Z ( t 0 ) = Z 0 (I) | Case I.1 | Z 0 is a random variable |
Case I.2 | A is a random variable | ||
Case I.3 | ( Z 0 , A ) is a random vector | ||
| |||
NH | Z ( t ) = B Z ( t 0 ) = Z 0 (II) | Case II.1 | Z 0 is a random variable |
Case II.2 | B is a random variable | ||
Case II.3 | ( Z 0 , B ) is a random vector | ||
Z ( t ) = A Z ( t ) + B Z ( t 0 ) = Z 0 (III) | Case III.1 | Z 0 is a random variable | |
Case III.2 | B is a random variable | ||
Case III.3 | A is a random variable | ||
Case III.4 | ( Z 0 , B ) is a random vector | ||
Case III.5 | ( Z 0 , A ) is a random vector | ||
Case III.6 | ( B , A ) is a random vector | ||
Case III.7 | ( Z 0 , B , A ) is a random vector |
For the sake of clarity in the development of all the cases listed in Table 1, throughout this paper, the input parameters, which are assumed to be continuous RVs, will be denoted by upper cases, while deterministic magnitudes will be written by lower cases. More precisely, for instance, in Case III.4, we will denote by Z 0 and B the random inputs, while the multiplicative coefficient in the RDE will be denoted by a rather than A .
The paper is organized as follows. In Section 2 we summarize the main results concerned with RVT that will be applied throughout the paper. We particularly state different versions of this useful technique including its application to different transformations that will facilitate the presentation of the results. Sections 3, 4, and 5 provide a detailed study where the 1-PDF of the solution SP, Z ( t ) , corresponding to the IVPs (I), (II), and (III) listed in Table 1, respectively, is computed. For each one of the thirteen casuistries, the study shows, at least, an illustrative example. The choice of the PDFs considered in the examples has been made to show the ability of the method to deal with both standard and non-standard dependent probability distributions. In Section 6 we include some considerations related to the application and better understanding of RVT method that we found particularly useful. Conclusions are drawn in the closing section.
2. Preliminaries
Below, we state several versions of the RVT technique as well as some related results emerging from its applications that will play a relevant role in our subsequent developments. Most of these results can be found in [8, 9] or they are a direct consequence of them.
Theorem 1 (RVT technique: scalar version).
Let X be a continuous RV with PDF f X ( x ) and domain D X = { x : f X ( x ) > 0 } . Let Y = r ( X ) be a new RV generated by the map r : ... [arrow right] ... which is assumed to be continuously differentiable on D X and such that r [variant prime] ( x ) ...0; 0 except at a finite number of points. Let one suppose that, for each y ∈ ... , there exist m ( y ) ...5; 1 points: x 1 ( y ) , x 2 ( y ) , ... , x m ( y ) ( y ) ∈ D X such that [figure omitted; refer to PDF] Then [figure omitted; refer to PDF]
Although Theorem 1 unifies the treatment of the different cases that one can present to compute the PDF, f Y ( y ) , in practice, it is easier to determine f Y ( y ) by dividing the domain of RV X into subintervals where the mapping r is monotone and then applying formula (3) on each subinterval. The process to compute f Y ( y ) on the whole domain of Y is completed by adding the corresponding expressions calculated previously for each subinterval.
In the simplest but significant case where the map r is linear, Theorem 1 reads as follows.
Proposition 2 (RVT technique: linear transformation).
Let X be a continuous RV with domain D X = { x : x 1 ...4; x ...4; x 2 } and PDF f X ( x ) . Then, the PDF f Y ( y ) of the linear transformation Y = α X + β , α ...0; 0 is given by [figure omitted; refer to PDF]
If α = 0 , then Y = β w.p. 1 stands for with probability 1 and [figure omitted; refer to PDF] where δ ( · ) denotes the Dirac delta distribution.
The following result is a direct application of Theorem 1 in the case that r ( x ) = α exp ... ( β x ) + γ .
Proposition 3 (RVT technique: exponential transformation).
Let X be a continuous RV with domain D X = { x : x 1 ...4; x ...4; x 2 } and PDF f X ( x ) . Then the PDF f Y ( y ) of the exponential transformation Y = α exp ... ( β X ) + γ , with α β ...0; 0 , is given by [figure omitted; refer to PDF]
If α = 0 or β = 0 , then Y = α + γ with probability 1 and [figure omitted; refer to PDF]
The computation of the joint PDF of two or more continuous RVs using the RVT method can also be performed by using the following generalization of formula (3).
Theorem 4 (RVT technique: multidimensional version).
Let X = ( X 1 , ... , X n ) be a random vector of dimension n with joint PDF f X ( x ) . Let r : ... n [arrow right] ... n be a one-to-one deterministic map which is assumed to be continuous with respect to each one of its arguments and with continuous partial derivatives. Then, the joint PDF f Y ( y ) of the random vector Y = r ( X ) is given by [figure omitted; refer to PDF] where s ( y ) is the inverse transformation of r ( x ) : x = r - 1 ( y ) = s ( y ) and J n is the Jacobian of the transformation; that is, [figure omitted; refer to PDF] which is assumed to be different from zero.
As we will see later, the analysis of Cases I-3, II-3, and III-3-6 requires the computation of the PDF of the sum and product of two continuous RVs which turns out by the application of Theorem 4 in its two-dimensional version. Thus, for the sake of clarity in the exposition, we specialize Theorem 4 in this significant case.
Theorem 5 (RVT technique: two-dimensional version).
Let X = ( X 1 , X 2 ) be a two-dimensional RV with joint PDF f X 1 , X 2 ( x 1 , x 2 ) . Let [figure omitted; refer to PDF] be a one-to-one deterministic map from ... 2 to ... 2 ; that is, there exists its inverse transformation: [figure omitted; refer to PDF] on the range of the map (12). Let one assume that both maps (12) and (13) are continuous. Let further assume that the following partial derivatives [figure omitted; refer to PDF] exist and are continuous and the Jacobian J 2 of the inverse map satisfies [figure omitted; refer to PDF] on the range of the transformation (12). Then, the joint PDF f Y 1 , Y 2 ( y 1 , y 2 ) of the two-dimensional RV Y = ( Y 1 , Y 2 ) = ( r 1 ( X 1 , X 2 ) , r 2 ( X 1 , X 2 ) ) is given by [figure omitted; refer to PDF]
Next, we apply Theorem 5 in the particular case that transformation r 1 only depends on variable x 1 and r 2 only depends on variable x 2 . As it will be seen later, this result will be crucial in further applications.
Proposition 6.
Let X = ( X 1 , X 2 ) be a two-dimensional RV with joint PDF f X 1 , X 2 ( x 1 , x 2 ) . Let [figure omitted; refer to PDF] be a one-to-one deterministic map from ... 2 to ... 2 ; that is, there exists its inverse transformation: [figure omitted; refer to PDF] on the range of the map (17). Let one assume that both maps (17) and (18) are continuous and the two following derivatives that exist are continuous and satisfy [figure omitted; refer to PDF] on the range of the transformation (17). Then, the joint PDF f Y 1 , Y 2 ( y 1 , y 2 ) of the two-dimensional RV Y = ( Y 1 , Y 2 ) = ( r 1 ( X 1 ) , r 2 ( X 2 ) ) is given by [figure omitted; refer to PDF]
On the other hand, applying Theorem 5 to r 1 ( x 1 , x 2 ) = x 1 + x 2 and r 2 ( x 1 , x 2 ) = x 1 (or r 2 ( x 1 , x 2 ) = x 2 ) and r 1 ( x 1 , x 2 ) = x 1 x 2 and r 2 ( x 1 , x 2 ) = x 1 (or r 2 ( x 1 , x 2 ) = x 2 ), we obtain the PDF of the sum and product of two continuous RVs, respectively. We state both results in the two following propositions.
Proposition 7 (RVT technique: sum of two continuous RVs).
Let ( X 1 , X 2 ) be a continuous random vector with joint PDF f X 1 , X 2 ( x 1 , x 2 ) and respective domains D X 1 = { x 1 : x 1,1 ...4; x 1 ...4; x 1,2 } and D X 2 = { x 2 : x 2,1 ...4; x 2 ...4; x 2,2 } . Then the PDF f Y 1 ( y 1 ) of their sum Y 1 = X 1 + X 2 is given by [figure omitted; refer to PDF] or, equivalent, by [figure omitted; refer to PDF] If X 1 and X 2 are independent continuous RVs, since f X 1 , X 2 ( x 1 , x 2 ) = f X 1 ( x 1 ) f X 2 ( x 2 ) , where f X i ( x i ) denotes the PDF of X i , i = 1,2 , (21) and (22) inform one that the PDF of the sum of two independent continuous RVs is just the convolution of their respective PDFs: [figure omitted; refer to PDF]
Proposition 8 (RVT technique: product of two continuous RVs).
Let ( X 1 , X 2 ) be a continuous random vector with joint PDF f X 1 , X 2 ( x 1 , x 2 ) with respective domains D X 1 = { x 1 ...0; 0 : x 1,1 ...4; x 1 ...4; x 1,2 } and D X 2 = { x 2 : x 2,1 ...4; x 2 ...4; x 2,2 } . Then the PDF f Y 1 ( y 1 ) of their product Y 1 = X 1 X 2 is given by [figure omitted; refer to PDF] Equivalently, if D X 1 = { x 1 : x 1,1 ...4; x 1 ...4; x 1,2 } and D X 2 = { x 2 ...0; 0 : x 2,1 ...4; x 2 ...4; x 2,2 } , then [figure omitted; refer to PDF] If X 1 and X 2 are independent continuous RVs with PDF's f X 1 ( x 1 ) and f X 2 ( x 2 ) , respectively, then (24) and (25) become [figure omitted; refer to PDF] respectively.
As usual we have not specified the domain of variation of y 1 in (24) and (25) since it is cumbersome. However, later we will detail it in some illustrative cases where it appears (see for instance Case III.6).
We close this section by extending Proposition 7 for the case of three terms since it will be required to deal with Case III.7. This result comes directly from the application of Theorem 4.
Proposition 9 (RVT technique: sum of three continuous RVs).
Let ( X 1 , X 2 , X 3 ) be a continuous random vector with joint PDF f X 1 , X 2 , X 3 ( x 1 , x 2 , x 3 ) and respective domains D X 1 = { x 1 : x 1,1 ...4; x 1 ...4; x 1,2 } , D X 2 = { x 2 : x 2,1 ...4; x 2 ...4; x 2,2 } , and D X 3 = { x 3 : x 3,1 ...4; x 3 ...4; x 3,2 } . Then the PDF f Y 1 ( y 1 ) of their sum Y 1 = X 1 + X 2 + X 3 is given by [figure omitted; refer to PDF] or, equivalent, by [figure omitted; refer to PDF] or [figure omitted; refer to PDF]
3. Case Study: Initial Value Problem (I)
This section is devoted to obtain the 1-PDF f 1 ( z , t ) of the solution SP Z ( t ) to the IVP (I) in each of the three cases listed in Table 1. Notice that Z ( t ) has the following expression: [figure omitted; refer to PDF]
3.1. Case I.1: Z 0 Is a Random Variable
As we pointed out previously for the sake of clarity in the presentation, we rewrite (30) by distinguishing the deterministic character of parameter a (which is written with a lower case letter): [figure omitted; refer to PDF] Next, we first fix t : t ...5; t 0 and denote Z = Z ( t ) . Then we apply Proposition 2 to [figure omitted; refer to PDF] Then, taking into account that the domain of RV Z 0 is given by (2), one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF] We illustrate the previous development in the following example where Z 0 is assumed to be a standard continuous RV, although further distributions, not necessarily standard, could be considered.
Example 10.
Let us assume that Z 0 has a Gaussian distribution, Z 0 ~ N ( μ ; σ 2 ) , μ ∈ ... , and σ 2 > 0 . Therefore, according to (33)-(34), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] For each t ...5; t 0 , the domain of z has been determined taking into account in (34) that in this case z 0,1 = - ∞ and z 0,2 = + ∞ . It can be checked that f 1 ( z , t ) is a PDF for each t ...5; t 0 . Figure 1 shows f 1 ( z , t ) at different values of t in the particular case that t 0 = 0 , a = - 1 and Z 0 ~ N ( 0 ; 1 ) , so μ = 0 and σ = 1 .
Figure 1: Plot of f 1 ( z , t ) given by (35) in Example 10 at different values of t = { 0,0.25,0.5,0.75 , ... , 2 } (corresponding to the solid lines) in the case that Z 0 ~ N ( μ = 0 ; σ 2 = 1 ) , t 0 = 0 , and a = - 1 .
[figure omitted; refer to PDF]
3.2. Case I.2: A Is a Random Variable
First, we rewrite (30) by highlighting the deterministic character of the initial condition z 0 : [figure omitted; refer to PDF] Next, we fix t : t > t 0 and denote Z = Z ( t ) . In the first analysis, we will assume that z 0 ...0; 0 . Then we apply Proposition 3 to [figure omitted; refer to PDF] Then, taking into account that z / z 0 = e a ( t - t 0 ) > 0 and the domain of RV A is given by (2), one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF] According to (2), if z 0 = 0 , then [figure omitted; refer to PDF] For t = t 0 , from (36) Z ( t ) = Z ( t 0 ) = z 0 , which is a deterministic initial condition. Then its 1-PDF can be written through the Dirac delta function as follows: [figure omitted; refer to PDF]
In the example below, we illustrate the previous development in the case where z 0 > 0 .
Example 11.
Let us assume that A has a beta distribution, A ~ Be ( α ; β ) , α , β > 0 , and z 0 > 0 . Therefore, according to (38)-(39), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where B ( α , β ) denotes the beta deterministic special function. Since z = z 0 e a ( t - t 0 ) and 0 ...4; a ...4; 1 , it is guaranteed that 0 ...4; ( 1 / ( t - t 0 ) ) ln ... ( z / z 0 ) ...4; 1 . As a consequence, f 1 ( z , t ) given by (42) is well defined. For each t > t 0 , the domain of z has been determined taking into account in (39) that in this case a 1 = 0 and a 2 = 1 . It can be checked that f 1 ( z , t ) is a PDF for each t > t 0 . Figure 2 shows f 1 ( z , t ) at different values of t in the particular case that A ~ Be ( 2 ; 3 ) , t 0 = 0 , and z 0 = 1 . For t = 0 , according to (41), f 1 ( z , 0 ) = δ ( z - 1 ) , - ∞ < z < ∞ .
Figure 2: Plot of f 1 ( z , t ) given by (42) in Example 11 at different values of t = { 0.1,0.2,0.3 , ... , 1 } (corresponding to the solid lines) in the case that A ~ Be ( 2 ; 3 ) , t 0 = 0 , and z 0 = 1 .
[figure omitted; refer to PDF]
3.3. Case I.3: ( Z 0 , A ) Is a Random Vector
Let us denote by f Z 0 , A ( z 0 , a ) the joint PDF of the random vector ( Z 0 , A ) . Now, we rewrite (30) in the following equivalent form: [figure omitted; refer to PDF] In order to apply RVT, we fix t : t > t 0 and denote Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 Z 2 = Z ( t ) . To compute the PDF of Z , first we will determine the joint PDF of Z 1 and Z 2 by applying Proposition 6 to [figure omitted; refer to PDF] Then taking into account [figure omitted; refer to PDF] one gets [figure omitted; refer to PDF] Once the joint PDF of ( Z 1 , Z 2 ) has been determined, the computation of the PDF of Z = Z 1 Z 2 follows directly by applying Proposition 8. Indeed, as Z 2 = e A ( t - t 0 ) ...0; 0 , we will apply formula (25) to X 1 = Z 1 = Z 0 , X 2 = Z 2 = e A ( t - t 0 ) > 0 and Y 1 = Z = Z 1 Z 2 . This yields [figure omitted; refer to PDF] or equivalently by using (46): [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Notice that if t = t 0 , then Z ( t ) = Z ( t 0 ) = Z 0 and f 1 ( z , t ) is just the PDF of RV Z 0 , which is easily obtained from the datum f Z 0 , A ( z 0 , a ) as the following marginal distribution: [figure omitted; refer to PDF]
Example 12.
Let ( Z 0 , A ) be a two-dimensional RV whose PDF is defined by [figure omitted; refer to PDF] Notice that we are implicitly assuming independence between Z 0 and A since the joint PDF f Z 0 , A ( z 0 , a ) factorizes as the product of the marginal PDFs f Z 0 ( z 0 ) = 2 z 0 , 0 < z 0 < 1 , and f A ( a ) = 2 a , 0 < a < 1 . Then, taking into account (48)-(50) and (52), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] For the sake of clarity in the graphical representation of f 1 ( z , t ) , Figure 3 shows two equivalent plots of f 1 ( z , t ) in the case that t 0 = 0 . Notice that the plot on the left side is the resultant surface where some 1-PDF f 1 ( z , t ) have been highlighted for different fixed times, whereas these 1-PDFs have been represented in these fixed times on the right side. As the two integrals appearing into the expression (53) can be computed explicitly, an equivalent expression to f 1 ( z , t ) with t 0 = 0 is given by [figure omitted; refer to PDF] For t = 0 , [figure omitted; refer to PDF]
Plots of f 1 ( z , t ) given by (54) in Example 12 at different values of t = { 0.1,0.2,0.3 , ... , 1.5 } (corresponding to the solid lines in both plots) in the case that ( Z 0 , A ) is a two-dimensional RV whose PDF is given by (52) and t 0 = 0 .
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
4. Case Study: Initial Value Problem (II)
This section is addressed to determine the 1-PDF f 1 ( z , t ) of the solution SP Z ( t ) to the IVP (II) in each of the three cases listed in Table 1. Notice that Z ( t ) has the following expression: [figure omitted; refer to PDF]
4.1. Case II.1: Z 0 Is a Random Variable
As we did in Case I.1, for the sake of clarity in the presentation, we rewrite (56) by emphasizing the deterministic character of parameter b (which is written with a lower case letter): [figure omitted; refer to PDF] Next, we fix t : t ...5; t 0 and denote Z = Z ( t ) . Then we apply Proposition 2 to [figure omitted; refer to PDF] Taking into account that the domain of RV Z 0 is given by (2), one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
Example 13.
Let us assume that Z 0 has a gamma distribution, Z 0 ~ Ga ( α ; β ) , α , β > 0 . Therefore, according to (59)-(60), the 1-PDF of Z ( t ) is given by: [figure omitted; refer to PDF] where Γ ( α ) denotes the deterministic gamma special function. For each t ...5; t 0 , the domain of z has been determined taking into account in (60) that, in this case, z 0,1 = 0 and z 0,2 = + ∞ . It can be checked that f 1 ( z , t ) is a PDF for each t ...5; t 0 . Figure 4 shows f 1 ( z , t ) at different values of t in the particular case that Z 0 ~ Ga ( 2 ; 1 ) , b = 1 , and t 0 = 0 .
Figure 4: Plot of f 1 ( z , t ) given by (61) in Example 13 at different values of t = { 0,0.25,0.5 , ... , 2 } (corresponding to the solid lines) in the case that Ga ( 2 ; 1 ) , b = 1 , and t 0 = 0 .
[figure omitted; refer to PDF]
4.2. Case II.2: B Is a Random Variable
First, we rewrite (56) by highlighting the deterministic character of the initial condition z 0 : [figure omitted; refer to PDF] Next, we fix t : t > t 0 and denote Z = Z ( t ) . Then we apply Proposition 2 to [figure omitted; refer to PDF] This yields [figure omitted; refer to PDF] where [figure omitted; refer to PDF] For t = t 0 , as it also occurred in Case I.2, Z ( t ) = Z ( t 0 ) = z 0 and therefore [figure omitted; refer to PDF]
Example 14.
Let us assume that B has a χ 2 -distribution with ν degrees of freedom, B ~ χ 2 ( ν ) , and ν > 0 . Therefore, according to (64)-(65), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] For each t : t > t 0 , the domain of z has been determined taking into account in (65) that in this case b 1 = 0 and b 2 = ∞ . It can be checked that f 1 ( z , t ) is a PDF for each t > t 0 . Figure 5 shows f 1 ( z , t ) at different values of t in the particular case that B ~ χ 2 (3), t 0 = 0 , and z 0 = 1 . For t = 0 , according to (66), f 1 ( z , 0 ) = δ ( z - 1 ) , - ∞ < z < ∞ .
Figure 5: Plot of f 1 ( z , t ) given by (67) in Example 14 at different values of t = { 0.1,0.2,0.3 , ... , 1 } (corresponding to the solid lines) in the case that B ~ χ 2 ( 3 ) , t 0 = 0 , and z 0 = 1 .
[figure omitted; refer to PDF]
4.3. Case II.3: ( Z 0 , B ) Is a Random Vector
We will denote by f Z 0 , B ( z 0 , b ) the joint PDF of continuous RVs Z 0 and B . Let us rewrite (56) in the following equivalent form: [figure omitted; refer to PDF] In order to compute the 1-PDF of Z ( t ) , f 1 ( z , t ) , we first fix t : t > t 0 and consider the continuous RVs Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 + Z 2 = Z ( t ) . Then, we will apply Propositions 6 and 7. Indeed, in a first step we compute the joint PDF f Z 1 , Z 2 ( z 1 , z 2 ) by applying Proposition 6 to [figure omitted; refer to PDF] Taking into account that [figure omitted; refer to PDF] one gets [figure omitted; refer to PDF] where z 1 and z 2 lie on [figure omitted; refer to PDF] Finally, we apply Proposition 7 using the following identification: X 1 = Z 1 , X 2 = Z 2 , and Y 1 = Z . This yields [figure omitted; refer to PDF] or more explicitly, using that Z 1 = Z 0 and (71), [figure omitted; refer to PDF] where [figure omitted; refer to PDF] If t = t 0 , then Z ( t ) = Z ( t 0 ) = Z 0 and, similar to Case I.3, f 1 ( z , t ) is just the PDF of RV Z 0 , which is easily obtained from the datum f Z 0 , B ( z 0 , b ) as the following marginal distribution: [figure omitted; refer to PDF]
Example 15.
Let ( Z 0 , B ) be a two-dimensional vector whose joint PDF is given by [figure omitted; refer to PDF] Then, carrying out the involved computations according to (74)-(75), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] Taking into account that t 0 = 0 , the value of this integral is [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
As we did in the example of Case I . 3 (see Figure 3), for the sake of clarity, Figure 6 shows two equivalent plots of f 1 ( z , t ) given by (79)-(80).
Plot of f 1 ( z , t ) given by (79)-(80) in Example 15 at different values of t = { 0,0.1,0.2 , ... , 1.5 } (corresponding to the solid lines in both plots) in the case that ( Z 0 , B ) is a two-dimensional RV whose PDF is given by (77) and t 0 = 0 .
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
5. Case Study: Initial Value Problem (III)
This section deals with the computation of the 1-PDF f 1 ( z , t ) of the solution SP Z ( t ) to the IVP (III) in each of the seven cases listed in Table 1. Notice that Z ( t ) has the following expression: [figure omitted; refer to PDF] Throughout the sequent analysis depending on whether A is considered to be a RV (Cases III.3, III.5, III.6, and III.7) or a deterministic constant (Cases III.1, III.2, and III.4), see Table 1; we will assume that ... [ { ω ∈ Ω : A ( ω ) ...0; 0 } ] = 1 or a ...0; 0 , respectively.
5.1. Case III.1: Z 0 Is a Random Variable
In order to take advantage of the RVT method and to study this case, it is convenient to rewrite (81) in the equivalent form: [figure omitted; refer to PDF] Next, we first fix t : t ...5; t 0 and denote Z = Z ( t ) . Then we apply Proposition 2 to [figure omitted; refer to PDF] Then, taking into account that the domain of RV Z 0 is given by (2), one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
Example 16.
Let Z 0 be an exponential RV, Z 0 ~ Exp ( λ ) , and let λ > 0 . Then, according to (84)-(85), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where the domain of z has been determined taking into account in (85) that in this case z 0,1 = 0 and z 0,2 = + ∞ . Again, it can be checked that f 1 ( z , t ) is a PDF for each t ...5; t 0 . Figure 7 shows f 1 ( z , t ) at different values of t in the case that λ = 1 , t 0 = 0 , a = - 1 , and b = 1 .
Figure 7: Plot of f 1 ( z , t ) given by (86) in Example 16 at different values of t = { 0,0.25,0.5 , ... , 2 } in the case that Z 0 ~ Exp ( 1 ) , t 0 = 0 , a = - 1 , and b = 1 .
[figure omitted; refer to PDF]
Before closing this case, we provide an example where the usefulness of the 1-PDF f 1 ( z , t ) to determine the statistical moments of the solution Z ( t ) and to compute the probability of sets of interest is shown. To illustrate these applications, we will choose the context of Example 16, although it could be applied to any example throughout this paper.
Example 17.
Let us consider the context of Example 16 where the 1-PDF of Z ( t ) , f 1 ( z , t ) , has been computed (see expression (86)). Then, the statistical moment of order n of Z ( t ) with respect to the origin can be computed directly in terms of f 1 ( z , t ) as follows: [figure omitted; refer to PDF] As a consequence, the following expressions for the mean and the variance of Z ( t ) are obtained: [figure omitted; refer to PDF] Figure 8 shows the expectation and variance of Z ( t ) for the same values of λ , t 0 , a , and b considered in Example 16.
Expectation (a) and variance (b) of Z ( t ) given by (88) in Example 17 in the case that λ = 1 , t 0 = 0 , a = - 1 , and b = 1 .
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
The computation of probabilities also can be performed directly through the 1-PDF. For instance, it may be of interest to compute the probability that the solution lies between two fixed values, say, v 1 = 2 and v 2 = 3 : [figure omitted; refer to PDF]
5.2. Case III.2: B Is a Random Variable
Now we assume that only B is a RV in the IVP (1) and f B ( b ) denotes its PDF. For convenience, in this case, we rewrite (81) as follows: [figure omitted; refer to PDF] First, we fixed t > t 0 and denote Z = Z ( t ) . Then we again apply Proposition 2 to determine the 1-PDF f 1 ( z , t ) of Z ( t ) , taking into account that [figure omitted; refer to PDF] Notice that α > 0 independently of the sign of the deterministic parameter a ...0; 0 . Then, [figure omitted; refer to PDF] where [figure omitted; refer to PDF] For t = t 0 , as it also occurred in Cases I.2 and II.2, Z ( t ) = Z ( t 0 ) = z 0 and therefore [figure omitted; refer to PDF]
Example 18.
Let us assume that B follows a gamma distribution of parameters α , β > 0 ; that is, B ~ Ga ( α ; β ) . Therefore, according to (92)-(93), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] For each t > t 0 , the domain of z has been determined taking into account in (93) that in this case b 1 = 0 and b 2 = ∞ . It can be checked that f 1 ( z , t ) is a PDF for each t > t 0 . Figure 9 shows f 1 ( z , t ) at different values of t in the particular case that B ~ Ga ( 2 ; 4 ) , t 0 = 0 , z 0 = 1 , and a = - 1 . For t = 0 , according to (94), f 1 ( z , 0 ) = δ ( z - 1 ) , - ∞ < z < ∞ .
Figure 9: Plot of f 1 ( z , t ) given by (95) in Example 18 at different values of t = { 0.05,0.1,0.2,0.3 , ... , 1 } in the case that B ~ Ga ( 2 ; 4 ) , t 0 = 0 , z 0 = 1 , and a = - 1 .
[figure omitted; refer to PDF]
5.3. Case III.3: A Is a Random Variable
So far we have observed that the application of RVT method to determine the PDF of a RV, say Y , generated by a transformation Y = r ( X ) , relies strongly on the feasibility of computing the inverse of the map y = r ( x ) . Fortunately, this has been done exactly in the previous cases, but, in general, it is not possible in the current case where just A is assumed to be a RV. In fact, once t : t > t 0 has been fixed, we must isolate A in the equation: [figure omitted; refer to PDF] which is not possible to perform in an exact manner. To circumvent this drawback, we will apply the Lagrange-Bürmann formula which gives the Taylor series expansion of the inverse of an analytic function.
Theorem 19 (Lagrange-Bürmann formula, see [10]).
Suppose that z is defined as a function of the variable a by an equation of the form: z = r ( a ) where r is analytic about the point a 0 and r [variant prime] ( a 0 ) ...0; 0 . Then, it is possible to invert (or to solve) the equation for a : a = s ( z ) on a neighbourhood ...A9; ( r ( a 0 ) ; δ ) , δ > 0 of r ( a 0 ) : [figure omitted; refer to PDF]
Although this result permits obtaining, from a theoretical stand point, the inverse function of r ( a ) , in practice, often this can only be achieved in an approximate manner since the infinite series (97) must be truncated to be kept computationally feasible. Moreover, the representation (97) of the inverse is only valid in a certain neighbourhood ...A9; ( r ( a 0 ) ; δ ) , δ > 0 of r ( a 0 ) , whose diameter δ > 0 must be determined carefully in each case study.
Let us apply Theorem 19 to determine the 1-PDF f 1 ( z , t ) of Z ( t ) . Fixing t : t > t 0 and denoting Z = Z ( t ) , we consider the map (96) which is analytic about any numerical value 0 ...0; a = A ( ω ) , ω ∈ Ω on the assumed domain to the RV A . As the map (96), in general, is not monotone, we have to apply Theorem 1. In a first step, we divide the domain of the map r (or equivalently, the domain of the RV A ) into k subintervals: ...9C; 1 , ...9C; 2 , ... , ...9C; k where r is monotone. Then, we will fix a such subinterval ...9C; j = ] a ini , j , a end , j [ , 1 ...4; j ...4; k where the contribution to the total 1-PDF f 1 ( z , t ) is going to be calculated. For this, we select a point a 0 , j ∈ ...9C; j where condition r [variant prime] ( a 0 , j ) ...0; 0 is met. By applying the Lagrange-Bürmann formula, we construct the inverse of the map r ( a ) = r j ( a ) on ...9C; j that, in the sequel, will be denoted by s j ( z ) : [figure omitted; refer to PDF] In practice, this infinite series could only converge on a subset of ] r j ( a ini , j ) , r j ( a end , j ) [ (if r j increases on ...9C; j ) or ] r j ( a end , j ) , r j ( a ini , j ) [ (if r j decreases on ...9C; j ) . In such case, the function s j ( z ) can be completed on the whole interval by taking m j another (or other, if necessary) appropriate point(s), say a 0 l , j ∈ ...9C; j , 1 ...4; l ...4; m j and then repeating the above process. In this manner, a piecewise inverse function s j ( z ) on ...9C; j will be constructed.
In accordance with RVT method, besides constructing the inverse s j ( z ) , one requires to compute its derivative. We again take advantage of Lagrange-Bürmann formula to complete this computation. In fact, notice that once s j ( z ) has been constructed, from (98), one gets [figure omitted; refer to PDF] Notice that if s j ( z ) has been defined by means of a piecewise function, then d s j ( z ) / d z will also be defined in the same way.
The process will be culminated by repeating again the previous argument on all the subintervals ...9C; 1 , ...9C; 2 , ... , ...9C; k . Following the previous development, the 1-PDF f 1 ( z , t ) of Z ( t ) can be computed as follows: [figure omitted; refer to PDF] where s j ( z ) and d s j ( z ) / d z are defined by (98) and (99), respectively.
Often, the infinite series (98) has to be truncated at the term N j to control computational burden. In this way, we obtain an approximation of the inverse: [figure omitted; refer to PDF] Thus, an approximation of its derivative is [figure omitted; refer to PDF] The infinite series (98) and (99) have the same convergence radius; however, often in practice, we need to handle their corresponding truncations (101) and (102), respectively. In this case, the selection of the appropriate a 0 l , j depends also on the quality of the approximations provided by both truncated series.
Repeating the foregoing process on each interval ...9C; j , 1 ...4; j ...4; k , one gets the corresponding approximation of f 1 ( z , t ) given by [figure omitted; refer to PDF]
In the following example, we illustrate the previous development.
Example 20.
Let us assume that A has a beta distribution of parameters α = 2 , β = 3 , A ~ Be ( 2 ; 3 ) , t 0 = 0 , z 0 = 1 and b = 1 . Figure 10 shows the approximation of f 1 ( z , t ) at different values of t . The approximation has been performed by (103), (101), and (102) with k = 1 being ...9C; 1 = [ 0,1 ] because of monotony of r ( A ) . In order to carry out the computations, ...9C; 1 has been split into 7 subintervals in accordance with the process described previously. In each subinterval, an approximation of degree N j = 2 has been used.
Plot of f 1 ( z , t ) given by (96) in Example 20 at different values of t = { 0,0.1,0.2 , ... , 1 } (a) and t = { 0.9,1 , 1.1 , ... , 2 } (b) in the case that A ~ Be ( 2 ; 3 ) , t 0 = 0 , z 0 = 1 , and b = 1 . In order to highlight better the shape of f 1 ( z , t ) , we have split its representation into two plots. Notice the different scales in the vertical axes for each plot.
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
5.4. Case III.4: ( Z 0 , B ) Is a Random Vector
Let us consider the IVP (1) and suppose that both Z 0 and B are continuous RVs with joint PDF f Z 0 , B ( z 0 , b ) . We rewrite (81) in the equivalent form: [figure omitted; refer to PDF] In order to compute the 1-PDF of Z ( t ) , f 1 ( z , t ) , we first fix t : t > t 0 and consider the continuous RVs Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 + Z 2 = Z ( t ) . Then, we will apply Propositions 6 and 7. Indeed, in a first step, we compute the joint PDF f Z 1 , Z 2 ( z 1 , z 2 ) by applying Proposition 6 to [figure omitted; refer to PDF] and taking into account that [figure omitted; refer to PDF] This leads to [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Notice that to determine the variation of z 2 , we have used that 1 / a ( e a ( t - t 0 ) - 1 ) > 0 for each a and t : t > t 0 . Finally, we apply Proposition 7 using the following identification: X 1 = Z 1 , X 2 = Z 2 , and Y 1 = Z . This yields [figure omitted; refer to PDF] or more explicitly, using (107), [figure omitted; refer to PDF] where z 1,1 , z 1,2 , z 2,1 , and z 2,2 are defined by (108).
If t = t 0 , then Z ( t ) = Z ( t 0 ) = Z 0 and, similar to Cases I.3 and II.3, f 1 ( z 0 , t 0 ) is the PDF of RV Z 0 . It is obtained as the following marginal distribution of the f Z 0 , B ( z 0 , b ) : [figure omitted; refer to PDF]
Example 21.
Let ( Z 0 , B ) be a two-dimensional Gaussian vector, η = ( Z 0 , B ) T ~ N ( μ η , Σ η ) , where [figure omitted; refer to PDF] and ρ Z 0 , B denotes the correlation coefficient between Z 0 and B . Then, according to (110) and (108), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Notice that to determine the integration limits in (113), we have considered in (108) that z 0,1 = - ∞ and z 0,2 = + ∞ . Figure 11 shows f 1 ( z , t ) at different values of t in the case that μ Z 0 = 1 , μ B = 0 , σ Z 0 = 0.1 , σ B = 0.1 , ρ Z 0 , B = 0.5 , t 0 = 0 , and a = - 1 .
For t = 0 , as the marginal distribution of a bivariate Gaussian distribution is also a Gaussian distribution with mean and variance of the corresponding component of the random vector, one gets [figure omitted; refer to PDF]
Figure 11: Plot of f 1 ( z , t ) given by (113)-(114) in Example 21 at different values of t = { 0,0.25,0.5 , ... , 2 } in the case that η = ( Z 0 , B ) T ~ N ( μ η , Σ η ) where μ η and Σ η are given by (112) with μ Z 0 = 1 , μ B = 0 , σ Z 0 = 0.1 , σ B = 0.1 , ρ Z 0 B = 0.5 , t 0 = 0 , and a = - 1 .
[figure omitted; refer to PDF]
5.5. Case III.5: ( Z 0 , A ) Is a Random Vector
Let us consider the IVP (1) and now we assume that both Z 0 and A are continuous RVs with joint PDF f Z 0 , A ( z 0 , a ) . We rewrite (81) in the following equivalent form: [figure omitted; refer to PDF] In order to apply RVT, we fix t : t ...5; t 0 and denote Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 + Z 2 = Z ( t ) . To compute the PDF of Z , first we will determine the joint PDF of Z 1 and Z 2 by applying Theorem 5 to [figure omitted; refer to PDF] Then taking into account ∂ s 2 ( z 2 ) / ∂ z 1 = 0 , the involved Jacobian simplifies to [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] where [figure omitted; refer to PDF] In the case that a 1 a 2 < 0 , the computation of the domain of variation of z 1 and z 2 in (119) is more complicated. To express it, we first introduce the numbers: [figure omitted; refer to PDF] Then, after elaborated computations, one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Notice that a 0 - and/or a 0 + can become 0 . In this case, the extremes of the intervals above that depend on a 0 - and/or a 0 + become ± ∞ , depending on the sign of parameter b .
Once the joint PDF of ( Z 1 , Z 2 ) has been determined, the computation of the PDF of Z = Z 1 + Z 2 follows directly by applying Proposition 7. Indeed, as Z 2 = - b / A ...0; 0 (since into Case III, b ...0; 0 ; otherwise we would be in Case I), we will apply formula (22) to X 1 = Z 1 = ( Z 0 + ( b / A ) ) e A ( t - t 0 ) , X 2 = Z 2 = - b / A ...0; 0 , and Y 1 = Z = Z 1 + Z 2 . This yields [figure omitted; refer to PDF] where D Z 2 denotes the domain of variation of Z 2 which, according to (120)-(124), depends on both the sign of the products a 1 a 2 and b . Using (119), it is equivalent to [figure omitted; refer to PDF] The range of variation of Z , denoted by D Z , can be straightforwardly computed taking into account Z = Z 1 + Z 2 and the domains of Z 1 and Z 2 which have been determined previously in (120)-(124). As its practical determination in specific examples is simple from the previous exposition, in order to avoid an unwieldy notation, we do not rewrite the final general expression.
Example 22.
Let ( Z 0 , A ) be a two-dimensional Gaussian vector, η = ( Z 0 , A ) T ~ N ( μ η , Σ η ) , where [figure omitted; refer to PDF] And let ρ Z 0 , A denotes the correlation coefficient between Z 0 and A . Then, according to (126), (122), and (124) to determine the domain of z 2 , the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Hereinafter, we will assume that b > 0 to facilitate our discussion through the example. We are in the case that z 0,1 = - ∞ , z 0,2 = + ∞ , and a 1 = - ∞ ; a 2 = + ∞ ; hence, a 1 a 2 < 0 , a 0 - = 0 - , and a 0 + = 0 + ; thus, z 2,1 = - ∞ and z 2,2 = + ∞ , which, according to (126), (122), and (124), determines the limits of integration of (128). Specifically, [figure omitted; refer to PDF]
Following an analogous reasoning, it is easy to check from (121) and (123) that I 1,1 + = ] - ∞ , + ∞ [ and I 1,2 + = ] - ∞ , + ∞ [ . Therefore, the domain of variation to Z 1 is ] - ∞ , + ∞ [ . As a consequence, in (128), Z lies in - ∞ ...4; z ...4; + ∞ .
Figure 12 shows f 1 ( z , t ) at different values of t in the case that μ Z 0 = 1 , μ A = 0 , σ Z 0 = 0.1 , σ A = 0.1 , ρ Z 0 , A = 0.5 , t 0 = 0 , and b = 1 .
Plot of f 1 ( z , t ) given by (128)-(129) in Example 22 at different values of t = { 0,0.1,0.2 , ... , 2 } in the case that η = ( Z 0 , A ) T ~ N ( μ η , Σ η ) where μ η and Σ η are given by (127) with μ Z 0 = 1 , μ A = 0 , σ Z 0 = 0.1 , σ A = 0.1 , ρ Z 0 , A = 0.5 , t 0 = 0 , and b = 1 .
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
5.6. Case III.6: ( B , A ) Is a Random Vector
Let us consider the IVP (1) and let us assume that B and A are continuous RVs with joint PDF f B , A ( b , a ) . For convenience, we rewrite (81) in the following equivalent form: [figure omitted; refer to PDF] In order to apply RVT method, we fix t : t > t 0 and denote Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 + Z 2 = Z ( t ) . We first consider the case where z 0 ...0; 0 . To compute the PDF of Z , first we will determine the joint PDF of Z 1 and Z 2 by applying Theorem 5 to [figure omitted; refer to PDF] As ∂ a / ∂ z 1 = 0 , in order to compute the Jacobian J 2 , it is enough to calculate the two following partial derivatives: [figure omitted; refer to PDF] which are well defined since t > t 0 , z 2 = z 0 e a ( t - t 0 ) ...0; 0 , and z 2 - z 0 = z 0 ( e a ( t - t 0 ) - 1 ) ...0; 0 due to by hypothesis z 0 ...0; 0 and ... [ { ω ∈ Ω : A ( ω ) ...0; 0 } ] = 1 . Also notice that ln ... | z 2 | - ln ... | z 0 | = a ( t - t 0 ) ...0; 0 . Moreover, taking into account that z 0 / z 2 = e - a ( t - t 0 ) > 0 , one gets [figure omitted; refer to PDF] The values z 1,1 , z 1,2 , z 2,1 , and z 2,2 that determine the domain of variation of Z 1 and Z 2 can be computed taking into account that the function h 1 ( a ) = ( 1 / a ) ( e a ( t - t 0 ) - 1 ) is positive and increasing on the whole real line in the current case where t - t 0 > 0 . This yields [figure omitted; refer to PDF] As it happened in the foregoing Case III.5, for the determination of the domains of Z 1 and Z 2 , we will use the notations a 0 - and a 0 + introduced in (121). After non difficult but elaborated computations, one gets [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
Finally, we apply Proposition 7 with the following identifications: X 1 = Z 1 , X 2 = Z 2 , and Y 1 = Z . As the variation of Z 2 given by (136), (138) and (139) is easily controlled in terms of the data than of Z 1 , in order to facilitate in practice the determination of the limits of integration of the integral which define the 1-PDF f 1 ( z , t ) , we will use formula (22) rather than (21). This yields [figure omitted; refer to PDF] where D Z 2 is defined by (136) and (138) or (139) depending on the signs of z 0 and the product a 1 a 2 . As in Case III.5, we do not explicit the range of variation of z , denoted by D Z , since its writing is cumbersome but not difficult from previous exposition.
Now, we deal with the case where z 0 = 0 and we keep the assumption t > t 0 . For convenience, we rewrite (81) in the following equivalent form: [figure omitted; refer to PDF] To apply RVT method, we fix t : t > t 0 and denote Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , and Z = Z 1 Z 2 = Z ( t ) . To compute the PDF of Z , first we will determine the joint PDF of Z 1 and Z 2 by applying Theorem 5 to [figure omitted; refer to PDF] As ∂ a / ∂ z 1 = 0 , in order to compute the Jacobian J 2 , it is enough to calculate the two following partial derivatives: [figure omitted; refer to PDF] which are well defined since t > t 0 and 1 + z 2 = e a ( t - t 0 ) ...0; 0,1 , due to by hypothesis ... [ { ω ∈ Ω : A ( ω ) ...0; 0 } ] = 1 . Then, one gets [figure omitted; refer to PDF] Again, the values z 1,1 , z 1,2 , z 2,1 , and z 2,2 determining the domain of variation of Z 1 and Z 2 can be computed taking into account that the function h 2 ( a ) = e a ( t - t 0 ) - 1 is increasing on the whole real line in the current case where t - t 0 > 0 . After non difficult but elaborated computations, this yields, [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
Next, we apply Proposition 8 with the following identifications: X 1 = Z 1 , X 2 = Z 2 , and Y 1 = Z . As the variation of Z 2 given by (145) and (147), which depends on the sign of a 1 , a 2 , and a 1 a 2 , is easily controlled in terms of the data than of Z 1 , in order to facilitate in practice the limits of integration of the integral defining the 1-PDF f 1 ( z , t ) , we will use formula (25) rather than (24). This yields [figure omitted; refer to PDF] where D Z 2 denotes the domain of variation of Z 2 . Again, we do not explicit the range of variation of Z , denoted by D Z , since its writing is cumbersome but simple from previous exposition.
Finally, we consider the case t = t 0 that implies Z ( t ) = Z ( t 0 ) = z 0 . Therefore, [figure omitted; refer to PDF]
Example 23.
With the aim of showing the generality of the obtained results to deal with different classes of continuous RVs, in this example, we will consider that the joint PDF of the input continuous RVs B and A is constructed by a copula function. Let B and A be two uniform continuous RVs on the interval ] 0,1 [ ; that is, B , A ~ Un ( ] 0,1 [ ) . Using the Farlie-Gumbel-Morgenstern copula [11], we construct the random vector ( B , A ) with joint PDF [figure omitted; refer to PDF] hence, B and A are statistically dependent. As a characteristic of copula functions, notice that the two marginal distributions of f B , A ( b , a ) are just the individual distributions of B and A . In the following, we will consider the case previously developed for z 0 = 0 and we will take t 0 = 0 . First, notice that according to (141) and taking into account that 0 < A , B < 1 , the domains of variation of Z 1 and Z 2 are z 1 ∈ ] 0 , ∞ [ and z 2 ∈ ] 0 , e t - 1 [ , respectively; thus, Z = Z 1 Z 2 lies on z ∈ ] 0 , ∞ [ . However, we must refine the domain of integration D Z 2 to Z 2 of (148), in such a way that the two arguments of the PDF f B , A ( · , · ) lie inside the interval ] 0,1 [ , where the input continuous RVs B and A are defined. Thus, fixing t > 0 and z > 0 , we must determine z 2 such as [figure omitted; refer to PDF] hold. By (141), notice that in our context z 2 = e a t - 1 , with 0 < a < 1 ; hence, second inequality in (151) is guaranteed. As t , z , z 2 , ln ... ( 1 + z 2 ) > 0 , the first inequality in (151) is equivalent to ln ... ( 1 + z 2 ) / z 2 < t / z . Therefore, D Z 2 in (148) is given by D Z 2 : ] min ... ( max ... ( 0 , z 2 * ) , e t - 1 ) , e t - 1 [ , where z 2 * is the solution of the equation ln ... ( 1 + z 2 ) / z 2 = t / z . At this point, notice that the function ln ... ( 1 + z 2 ) / z 2 is decreasing for z 2 > 0 and this justifies the consideration of the previous term: max ... ( 0 , z 2 * ) . To summarize, the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where z 2,1 = min ... ( max ... ( 0 , z 2 * ) , e t - 1 ) . In Figure 13, we have plotted f 1 ( z , t ) at different values of t .
Plot of f 1 ( z , t ) given by (152) in Example 23 at different values of t = { 0.1,0.2 , ... , 1 } (a) and t = { 1,1.1 , ... , 2 } (b) in the case that z 0 = 0 and t 0 = 0 and the joint PDF of ( B , A ) is given by (150). In order to better highlight the shape of f 1 ( z , t ) , we have split its representation into two plots. Notice the different scales in the vertical axes for each plot.
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
5.7. Case III.7: ( Z 0 , B , A ) Is a Random Vector
Finally, we consider the IVP (1) and assume that every input is randomized; that is, ( Z 0 , B , A ) is a random vector with joint PDF f Z 0 , B , A ( z 0 , b , a ) . For convenience, we rewrite (81) in the following equivalent form: [figure omitted; refer to PDF] In order to apply RVT method, we fix t : t > t 0 and denote Z 1 = Z 1 ( t ) , Z 2 = Z 2 ( t ) , Z 3 = Z 3 ( t ) , and Z = Z 1 + Z 2 + Z 3 = Z ( t ) . To compute the PDF of Z , first we will determine the joint PDF of Z 1 , Z 2 , and Z 3 by applying Theorem 4 to [figure omitted; refer to PDF] Notice that - z 2 / z 3 = e a ( t - t 0 ) > 0 so s 2 ( z 2 , z 3 ) and s 3 ( z 2 , z 3 ) are well defined as B and A are continuous RVs such as ... [ { ω ∈ Ω : B ( ω ) ...0; 0 } ] = 1 and ... [ { ω ∈ Ω : A ( ω ) ...0; 0 } ] = 1 , respectively. After some computations, it is easy to check that the absolute value of the Jacobian J 3 required to apply Theorem 4 is given by [figure omitted; refer to PDF] Notice that we have used z 3 ...0; 0 . Therefore, [figure omitted; refer to PDF] Now, we will apply Proposition 9 with the following identifications: X 1 = Z 1 , X 2 = Z 2 , X 3 = Z 3 , and Y 1 = Z . As the variation of Z 2 and Z 3 is easily controlled in terms of the data than of Z 1 , in order to facilitate in practice the specification of the limits of integration of the integral defining the 1-PDF f 1 ( z , t ) , we will use formula (27). This yields [figure omitted; refer to PDF] for z ∈ D Z and t > t 0 . As in the previous case, we do not detail the range of variation of z , D Z , since it is very involved.
Finally, we consider the case t = t 0 . From (1), one gets Z ( t ) = Z ( t 0 ) = Z 0 . Therefore, [figure omitted; refer to PDF]
Example 24.
Let ( Z 0 , B , A ) be a three-dimensional Gaussian vector, η = ( Z 0 , B , A ) T ~ N ( μ η , Σ η ) , where μ η = ( μ 1 , μ 2 , μ 3 ) T is a vector in ... 3 which represents the mean and Σ η is a symmetric positive definite real matrix of size 3 × 3 usually referred to as the variance-covariance matrix. Then, according to (157), the 1-PDF of Z ( t ) is given by [figure omitted; refer to PDF] where [figure omitted; refer to PDF] In the following, we will take t 0 = 0 : [figure omitted; refer to PDF] Now, we do not provide an explicit expression of f 1 ( z , t ) since it is very sophisticated. Figure 14 shows f 1 ( z , t ) at different values of t ( t = 0,0.1,1 , 2 ).
Plots of f 1 ( z , t ) given by (159) in Example 24 at different values of t with t 0 = 0 in the case that ( Z 0 , B , A ) is a three-dimensional Gaussian vector with mean and variance-covariance matrix given by (161).
(a) [figure omitted; refer to PDF]
(b) [figure omitted; refer to PDF]
(c) [figure omitted; refer to PDF]
(d) [figure omitted; refer to PDF]
6. Some Final Remarks
In this section, we will point out some considerations related to the practical application of RVT method in dealing with the computation of the 1-PDF.
Remark 25.
Throughout this paper, we have determined the 1-PDF of the solution SP to IVP (1) in all the cases listed in Table 1. It must be pointed out that it has depended heavily on doing an appropriate choice of the involved variables when applying the RVT method. To illustrate this statement, let us consider the foregoing Case III.5 when Z 0 and A are the only input continuous RVs. When we dealt with this case, we first decomposed the solution SP (81) in the form given by (116) and then we applied Theorem 5 as is shown in (117). This decomposition was carefully chosen to guarantee the successful application of Theorem 5. In fact, alternative decompositions of (81) are possible; however, they could not be adequate to achieve our goal. For example, if we rewrite (81) in the following form: [figure omitted; refer to PDF] then, the application of Theorem 5 to [figure omitted; refer to PDF] does not lead to fruitful results since we cannot isolate z 0 = s 1 ( z 1 , z 2 ) and a = s 2 ( z 1 , z 2 ) and this would ruin our goal. In other cases, we can find out that two or more choices do not have the previous drawback and then the best selection will be the one which facilitates the involved computations (the easiest expression for the Jacobian, the simplest determination of the domains, etc.).
Remark 26.
Although in this paper we have concentrated on the determination of the 1-PDF f 1 ( z , t ) , which describes the probabilistic behaviour of the solution SP Z ( t ) at each time instant t , it must be pointed out that the computation of higher PDFs is also possible and desirable. In fact, for instance, the 2 -PDF, say, f 2 ( z 1 , t 1 ; z 2 , t 2 ) provides a whole probabilistic information of X ( t ) in two time instants t 1 and t 2 . Therefore, from it, we can calculate relevant probabilistic properties such as the correlation function [figure omitted; refer to PDF] As it can be guested, from the previous development, the computation of f 2 ( z 1 , t 1 ; z 2 , t 2 ) follows in broad outline that f 1 ( z , t ) . In Example 27, we compute the 2 -PDF within the framework of foregoing Case II.3 including the correlation function.
Example 27.
Let us fix t 1 , t 2 such as t 2 > t 1 ...5; t 0 and denote Z 1 = Z ( t 1 ) and Z 2 = Z ( t 2 ) . We want to determine the joint PDF of continuous RVs Z 1 and Z 2 . For it, we will apply Theorem 5 to [figure omitted; refer to PDF] where the expression of the solution SP given by (56) has been considered. Now, taking into account that [figure omitted; refer to PDF] then, the involved Jacobian is given by [figure omitted; refer to PDF] This leads to [figure omitted; refer to PDF] where [figure omitted; refer to PDF]
In spite of having computed the 2-PDF in the previous example, it must be noticed that, in general, its determination to problem (1) becomes very difficult even prohibitive, particularly the specification of the associated domains.
Remark 28.
In some situations, the determination of the 1-PDF provides the full probabilistic information of the solution SP; hence, the computation of higher PDFs is not necessary. This is illustrated in the following example.
Example 29.
Let the IVP be considered in Case III.1 where the only random input is the initial condition Z 0 . We know that [figure omitted; refer to PDF] Its 1-PDF is given by (84). From it, we can compute any probabilistic information for every time instant, say t 1 . Let us consider another time instant, say t 2 . Notice that, in this case, the solution SP at t 2 can be represented as follows: [figure omitted; refer to PDF] From this expression, we see that the behaviour of the solution Z ( t ) at the time instant t 2 is deterministically given by (a linear transformation of) Z ( t 1 ) . Therefore, the computation of the 2-PDF is not required. This can be confirmed from another point of view. Let us assume without loss of generality that the expectation of the initial condition is zero: E [ Z 0 ] = 0 and its variance is σ Z 0 2 > 0 . From (170), it is easy to check that [figure omitted; refer to PDF] Then, the correlation coefficient function is given by [figure omitted; refer to PDF]
7. Conclusions
In this paper, we have determined the first probability density function (1-PDF) of the solution stochastic process of the linear random differential equation taking advantage of random variable transformation (RVT) method. The study has been made in a systematic way in order to facilitate and clarify the development and exposition of the results as well as to facilitate their practical use. The wide range of the exhibited examples, that include both standard and non-standard probabilistic distributions, show the ability of RVT technique to deal with the computation of the 1-PDF to models based on linear random differential equations. Notice that throughout the paper no independence among the random parameters has been assumed in order to achieve general results.
Acknowledgments
This work has been partially supported by the Ministerio de Economía y Competitividad under Grant no. DPI2010-20891-C02-01 and Universitat Politècnica de València under Grant no. PAID06-11-2070.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Abstract
Deterministic differential equations are useful tools for mathematical modelling. The consideration of uncertainty into their formulation leads to random differential equations. Solving a random differential equation means computing not only its solution stochastic process but also its main statistical functions such as the expectation and standard deviation. The determination of its first probability density function provides a more complete probabilistic description of the solution stochastic process in each time instant. In this paper, one presents a comprehensive study to determinate the first probability density function to the solution of linear random initial value problems taking advantage of the so-called random variable transformation method. For the sake of clarity, the study has been split into thirteen cases depending on the way that randomness enters into the linear model. In most cases, the analysis includes the specification of the domain of the first probability density function of the solution stochastic process whose determination is a delicate issue. A strong point of the study is the presentation of a wide range of examples, at least one of each of the thirteen casuistries, where both standard and nonstandard probabilistic distributions are considered.
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Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer