91Ó°ÊÓ

Let$f(x,y)=24xy0≤x≤1,0≤y≤1,0≤x+y≤1$and let it equal 0 otherwise.

The formula for a joint distribution said to be density function ,

${∫}_y{∫}_{x}f(x,y)dxdy=1$

Applying the formula to showis a joint probability density function.

$$\begin{gathered} {∫}_y{∫}_{x}f(x,y)dxdy={∫}_0^1{∫}_0^{1-x}24xydxdy \\ ={∫}_0^{1}24x{\left[\frac{y^2}{2}\right]}_0^{1-x}dx \\ ={∫}_0^{1}24x\left[\frac{1+x^2-2x}{2}\right]dx \\ ={∫}_0^{1}12\left(x+x^3-2x^2\right)dx \end{gathered}$$

Further integrating with respect to $dx$

$$\begin{gathered} ={12\left(\frac{x^2}{2}+{\frac{x}{4}}^4-\frac{2x^3}{3}\right)}_0^1 \\ =12\left(\frac{1}{2}+\frac{1}{4}-\frac{2}{3}\right) \\ =12\left(\frac{6+3-8}{12}\right) \\ =12\left(\frac{1}{12}\right) \\ =1 \end{gathered}$$

Conclusion: $f(x,y)$is a joint probability density function$$\begin{gathered} {∫}_y{∫}_{x}f(x,y)dxdy=12{\left(\frac{x^2}{2}+\frac{x^4}{4}-\frac{2x^3}{3}\right)}^{1}0 \\ =12\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{3}\right) \\ =12\left(\frac{2+3-4}{4×3}\right) \\ =1 \end{gathered}$$

To find$E\left(X\right)$, first write the expression for $f_X\left(x\right)$,

$f_X\left(x\right)={∫}_y{f}_{X,Y}(x,y)dxdy={∫}_0^{1-x}24xydy=12x{(1-x)}^2$when $xâ‰\yen 0$and zero otherwise.

$$\begin{gathered} ∴E\left(X\right)={∫}_0^{1}x{f}_X\left(x\right)dx=12{∫}_0^1{x}^2{(1-x)}^{2}dx=12{∫}_0^1(x^2-2x^3+x^4)dx \\ =12\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)=\frac{2}{5} \end{gathered}$$

By symmetry, $f_Y\left(y\right)=12y{(1-y)}^2$when $yâ‰\yen 0$and zero otherwise

Since the Probability density functions are same, therefore by symmetry, the expectations will be equal.