Q.50 - A First Course in Probability

Question

The joint density of $X$  and $Y$ is given by $f(x,y)=\frac{e^{-x/y}{e}^{-y}}{y}$,$0<x<\infty , 0<y<\infty$ ,  Compute $E\left[X^2\mid Y=y\right]$.

Step-by-Step Solution

Answer

$\mathbb{E}\left[X^2\mid Y=y\right]=2y^2$

Step 2: Given information

Given in the question that, The joint density of $X$ and $Y$ is given by

$f(x,y)=\frac{e^{-x/y}{e}^{-y}}{y}, 0<x<\infty , 0<y<\infty$.

Step 2: Explanation

The joint density of $X$ and $Y$ is given by $f(x,y)=\frac{e^{-\frac{x}{y}}{e}^{-y}}{y}$where $0<x<\infty ,0<y<\infty$. We have to compute $\mathbb{E}\left[X^2\mid Y=y\right]$.

From the definition we have:

$\mathbb{E}[X\mid Y=y]={\int }_{-\infty }^{\infty }x{f}_{x\mid y}(x\mid y)dx$

Therefore it simply follows:

$\mathbb{E}\left[X^2\mid Y=y\right]={\int }_{-\infty }^{\infty }{x}^2·\frac{f(x,y)}{f_Y\left(y\right)}dx$

$={\int }_0^{\infty }{x}^2·\frac{\frac{e^{-\frac{x}{y}}{e}^{-y}}{y}}{{\int }_{-\infty }^{\infty }f(x,y)dx}dx$

$={\int }_0^{\infty }{x}^2·\frac{e^{-\frac{x}{y}}{e}^{-y}}{y{e}^{-y}}dx$

$=\frac{1}{y}{\int }_0^{\infty }{x}^2·e^{-\frac{x}{y}}dx$

$=\frac{1}{y}·2y^3=2y^2$

Step 3: Final answer

$\mathbb{E}\left[X^2\mid Y=y\right]=2y^2$