The joint probability density function of X and Y is given by

f(x, y) = c(y² − x²)e^-y −y … x … y, 0 < y < q .

(a) Find c.

(b) Find the marginal densities of X and Y.

(c) Find E[X].

A capacity or function used to describe the likelihood dispersion of a consistent arbitrary vector is the joint likelihood thickness work. It's a multivariate speculation of the likelihood thickness work, which portrays a persistent arbitrary variable's dispersion.

The joint density function must satisfy

$1=\iint R^{2}fdxdy={\int }_0^{\infty }\left({\int }_{-y}^{y}c\right(y^2-x^2\left)e^{-y}dx\right)dy$

Thus, we have that the double integrals above is equal to

$c{\int }_0^{\infty }(y^2{e}^{-x}x-e^{-y}\frac{x^3}{3})dy=\frac{4c}{3}{\int }_0^{\infty }{y}^3{e}^{-y}dy$

In order to solve the remaining integral, use the integration by parts. Define $u=y^3$and role="math" localid="1647316497578" $dv=e^{-y}dy$, we have,

${\int }_0^{\infty }{y}^3{e}^{-y}dy=-y^3{e}^{-y}+{\int }_0^{\infty }3y^2{e}^{-y}dy=3{\int }_0^{\infty }{y}^2{e}^{-y}dy$

Similarly we have,

${\int }_0^{\infty }3y^2{e}^{-y}dy=-3y^2{e}^{-y}+3{\int }_0^{\infty }2y{e}^{-y}dy=6{\int }_0^{\infty }y{e}^{-y}dy$

and

$6{\int }_0^{\infty }y{e}^{-y}dy=-6y{e}^{-y}+6{\int }_0^{\infty }{e}^{-y}dy=6$

Therefore,

$$\begin{gathered} 1=\frac{4c}{3}\times 6 \\ c=\frac{1}{8} \end{gathered}$$

we have

$f_X\left(x\right)={\int }_{R}f(x,y)dy={\int }_x^{\infty }\frac{1}{8}(y^2-x^2)e^{-y}dy=\frac{1}{8}{\int }_x^{\infty }(y^2{e}^{-y}-x^2{e}^{-y})dy$

The first integrant is equal to

$$\begin{gathered} {\int }_x^{\infty }{y}^2{e}^{-y}dy=-y^2{e}^{-y}+2{\int }_x^{\infty }y{e}^{-y}dy=x^2{e}^{-x}+2(-y{e}^{-y}+{\int }_x^{\infty }{e}^{-y}dy) \\ =x^2{e}^{-x}+2x{e}^{-x}+2e^{-x} \end{gathered}$$

The second integral is

${\int }_x^{\infty }{x}^2{e}^{-y}dy=x^2{e}^{-x}$

Here we have that

$f_X\left(x\right)=\frac{1}{8}(x^2{e}^{-x}+2x{e}^{-x}+2e^{-x}-x^2{e}^{-x})=\frac{1}{4}{e}^{-x}(1+x)$

In order to calculate the marginal density of Y, observe that it is equal to zero.

$f_Y\left(y\right)={\int }_{-y}^y\frac{1}{8}(y^2-x^2)e^{-y}dx=\frac{1}{8}{\int }_{-y}^y{y}^2{e}^{-y}-x^2{e}^{-y}dx=\frac{1}{8}.\frac{4}{3}{y}^3{e}^{-y}$

$=\frac{1}{6}{y}^3{e}^{-y}$

Using the definition of exception, we have that

$E\left(X\right)={\int }_{R}xf\left(x\right)dx={\int }_{-\infty }^{\infty }x.\left(\frac{1}{4}{e}^{-x}\right(1+x\left)dx\right)$

But this integral is equal to zero since we integrate odd functions over symmetric region. Hence,$E\left(X\right)=0$