Question

The joint density of $X$and $Y$is given by

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

Compute $E\left[X^3\mid Y=y\right]$.

Short answer

$\mathbb{E}\left[X^3\mid Y=y\right]=\frac{y^3}{4}$

Step-by-step solution

Given information

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

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

Explanation

The joint density of $X$and $Y$is given by $f(x,y)=\frac{e^{-y}}{y}$on the district $0<x<y$and $0<y<\infty$. For simpler arrangement we'll introduce a diagram on the subsequent page. The goal is to work out the normal worth of$X^3$given $Y=y$. Initially note that:

$f(x\mid y)=\frac{f(x,y)}{f_Y\left(y\right)}=\frac{e^{-y}}{y}·\frac{1}{e^{-y}}=\frac{1}{y}$

Now we simply calculate:

$\mathbb{E}\left[X^3\mid Y=y\right]={\int }_0^y{x}^3·\frac{1}{y}dx={\left.\frac{1}{y}·\frac{x^4}{4}\right|}_0^y=\frac{y^3}{4}$

Graph

We integrate on the following region:

Final answer

$\mathbb{E}\left[X^3\mid Y=y\right]=\frac{y^3}{4}$