91Ó°ÊÓ

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

$1=âˆ\neg R^{2}fdxdy={∫}_0^{∞}\left({∫}_{-y}^{y}c\right(y^2-x^2\left)e^{-y}dx\right)dy$

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

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

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,

${∫}_0^{∞}{y}^3{e}^{-y}dy=-y^3{e}^{-y}+{∫}_0^{∞}3y^2{e}^{-y}dy=3{∫}_0^{∞}{y}^2{e}^{-y}dy$

Similarly we have,

${∫}_0^{∞}3y^2{e}^{-y}dy=-3y^2{e}^{-y}+3{∫}_0^{∞}2y{e}^{-y}dy=6{∫}_0^{∞}y{e}^{-y}dy$

and

$6{∫}_0^{∞}y{e}^{-y}dy=-6y{e}^{-y}+6{∫}_0^{∞}{e}^{-y}dy=6$

Therefore,

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

we have

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

The first integrant is equal to

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

The second integral is

${∫}_x^{∞}{x}^2{e}^{-y}dy=x^2{e}^{-x}$

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.

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

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

Using the definition of exception, we have that

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

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