Question

The joint density of $X$ and $Y$ is given by

$f(x,y)=\frac{1}{\sqrt{2\pi }}{e}^{-y}{e}^{-(x-y{)}^2/2}0<y<\infty$,

$-\infty <x<\infty$

(a) Compute the joint moment generating function of $X$ and $Y$.

(b) Compute the individual moment generating functions.

Short answer

a). The joint moment generating function of $X$and $Y$are $M_{(X,Y)}\left(t_1,t_2\right)=\exp \left(\frac{t_1^2}{2}\right)\frac{1}{\left(1-t_2-t_1\right)}$.

b). The individual moment generating functions are$M_X\left(t_1\right)=\exp \left(\frac{t_1^2}{2}\right){\left(1-t_1\right)}^{-1}$and$M_Y\left(t_2\right)={\left(1-t_2\right)}^{-1}$.

Step-by-step solution

Given Information (Part a)

$\begin{array}{r}f(x,y)=\frac{1}{\sqrt{2\pi }}{e}^{-y}{e}^{-(x-y{)}^2/2}0<y<\infty \\ -\infty <x<\infty \end{array}$

Explanation (Part a)

$M_{(X,Y)}\left(t_1,t_2\right)=E\left[\exp \left(t_{1}x+t_{2}y\right)\right]$

$=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }{\int }_{-\infty }^{\infty }\exp \left(t_{1}x+t_{2}y\right)\exp (-y)\exp \left(\frac{-(x-y{)}^2}{2}\right)dxdy$

$=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }{\int }_{-\infty }^{\infty }\exp \left(t_{1}x+t_{2}y-y-\frac{x^2}{2}-\frac{y^2}{2}+xy\right)dxdy$

$=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }\exp \left(t_{2}y-y-\frac{y^2}{2}\right)\left({\int }_{-\infty }^{\infty }\exp \left(\frac{-1\left(x^2-2bx\right)}{2}\right)dx\right)dy$

$=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }\exp \left(t_{2}y-y-\frac{y^2}{2}\right)\left({\int }_{-\infty }^{\infty }\exp \left(\frac{-(x-b{)}^2}{2}\right)\exp \left(\frac{b^2}{2}\right)dx\right)dy$

Explanation (Part a) 

$=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }\exp \left(\left(t_2+t_1-1\right)y+\frac{t_1^2}{2}\right)\sqrt{2\pi }dy$

$={\int }_0^{\infty }\exp \left(\frac{t_1^2}{2}\right)\exp \left(-y\left(1-t_2-t_1\right)\right)dy$

$=\exp \left(\frac{t_1^2}{2}\right){\int }_0^{\infty }\exp \left(-y\left(1-t_2-t_1\right)\right)dy$

$={\left.\exp \left(\frac{t_1^2}{2}\right)\frac{\exp \left(-y\left(1-t_2-t_1\right)\right)}{\left(-\left(1-t_2-t_1\right)\right.}\right|}_0^{\infty }$

Therefore,

$M_{(X,Y)}\left(t_1,t_2\right)=\exp \left(\frac{t_1^2}{2}\right)\frac{1}{\left(1-t_2-t_1\right)}$

Final Answer (Part a)

The joint moment generating functions of$X$and$Y$are

$M_{(X,Y)}\left(t_1,t_2\right)=\exp \left(\frac{t_1^2}{2}\right)\frac{1}{\left(1-t_2-t_1\right)}$.

Given Information (Part b)

$f(x,y)=\frac{1}{\sqrt{2\pi }}{e}^{-y}{e}^{-(x-y{)}^2/2}0<y<\infty$

$-\infty <x<\infty$

Explanation (Part b)

If $t_2=0$we have

$M_x\left(t_1\right)=\exp \left(\frac{t_1^2}{2}\right)\frac{1}{\left(1-0-t_1\right)}$

$=\exp \left(\frac{t_1^2}{2}\right){\left(1-t_1\right)}^{-1}$

If $t_1=0$we have

$M_Y\left(t_2\right)=\exp \left(\frac{0}{2}\right)\frac{1}{\left(1-t_2-0\right)}$

$={\left(1-t_2\right)}^{-1}$

Final Answer (Part b)

Individual moment generating functions,

If $t_2=0$, $M_x\left(t_1\right)=\exp \left(\frac{t_1^2}{2}\right)\frac{1}{\left(1-0-t_1\right)}$

If$t_1=0$,$M_Y\left(t_2\right)={\left(1-t_2\right)}^{-1}$.