Q.7.4 - A First Course in Probability

Question

7.4. If X and Y have joint density function $f_{X,Y}(x,y)=\{\begin{array}{ll}1/y,& \text{if }0<y<1,0<x<y\\ 0,& \text{otherwise}\end{array}$ find

(a) E[X Y]

(b) E[X]

(c) E[Y]

Step-by-Step Solution

Answer

a) $E\left[XY\right]=\frac{1}{6}$

b) $E\left[X\right] = \frac{1}{4}$

c) $E\left[Y\right] = \frac{1}{2}$

Part(a) - Step 1: To find

Expectation of $XY$

Part (a) - Step 2: Explanation

Given :$f_{X,Y}(x,y)=\{\begin{array}{ll}1/y,& \text{if }0<y<1,0<x<y\\ 0,& \text{otherwise}\end{array}$

Formula to be used:

 $E\left[XY\right]={\int }_y{\int }_{x}xy\times f(x,y)dxdy$

Calculation: 
$\begin{array}{r}E\left[XY\right]={\int }_0^1 {\int }_0^y xy\times \frac{1}{y}dxdy\\ ={\int }_0^1 {\int }_0^y xdxdy\\ ={\left.{\int }_0^1 \left[\frac{x^2}{2}\right]\right|}_0^{y}dy\\ ={\int }_0^1 \left(\frac{y^2}{2}-0\right)dy\end{array}$

Now, integrating w.r.t $Y$

$\begin{array}{r}={\left.\left[\frac{y^3}{2\times 3}\right]\right|}_0^1\\ =\frac{1^3}{6}-0\\ =\frac{1}{6}\end{array}$

Hence $E\left[XY\right]=\frac{1}{6}$

Part (b) - Step 3: To find

Expectation of X

Part (b) - Step 4: Explanation

To find : $E\left[X\right]$

Formula to be used: 

$E\left[XY\right]={\int }_y{\int }_{x}xy\times f(x,y)dxdy$

Calculation:

$\begin{array}{r}E\left[X\right]={\int }_0^1 {\int }_0^y x\times \frac{1}{y}dxdy\\ ={\int }_0^1 \left({\left.\left[\frac{1}{y}\times \frac{x^2}{2}\right]\right|}_0^y\right)dxdy\\ ={\int }_0^1 \left(\frac{y}{2}-0\right)dy\\ ={\left.\left[\frac{y^2}{2\times 2}\right]\right|}_0^1=\frac{1^2}{4}-0\end{array}$

Therefore $E\left[X\right]=\frac{1}{4}$

Part (c) : To find

Expectation of $X$

Part(c) : Step 6: Explanation

To find: $E\left[Y\right]$

Formula to be used:$E\left[XY\right]={\int }_y{\int }_{x}xy\times f(x,y)dxdy$

Calculation:

$\begin{array}{r}E\left[X\right]={\int }_0^1 {\int }_0^y y\times \frac{1}{y}dxdy\\ ={\int }_0^1 \left({\left.\left[x\right]\right|}_0^y\right)dxdy\\ ={\int }_0^1 (y-0)dy\\ ={\left.\left[\frac{y^2}{2}\right]\right|}_0^1=\frac{1^2}{2}-0\end{array}$

Therefore $E\left[Y\right]=\frac{1}{2}$