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Let $f(x,y)=c(x^2+y^3)I_{[0,1]^2}(x,y)$ be the density of a random vector. Compute $\Bbb E [Z|X]$, where $Z=\frac{Y+1}{X+1}$.

My approach: $f_x=\int_0^1 f(x,y)dy=cx^2+\frac{c}{4}$

$E[\frac{Y+1}{X+1}|X]=\frac{1}{x+1}E[Y+1|X]=\frac{1}{x+1}(E[Y|X]+1)=\frac{1}{x+1}(\int_{0}^1yf_{Y|X}(y|x)dy+1)=\frac{1}{x+1}(\int_{0}^1y\frac{c(x^2+y^3)}{cx^2+\frac{c}{4}}dy+1)=...$

Is this correct?

Max
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1 Answers1

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Hint: You have that $$\begin{align*}E[Z|X]&=\int_{X}E[Z|X=x]f_X(x)dx=\int_{X}E\left[\dfrac{Y+1}{X+1}| X=x\right]f_{X}(x)dx=\\&=\int_{X}E[Y+1|X=x]\dfrac{1}{x+1}f_X(x)dx=\\&=\int_{X}\left(\int_{Y}(y+1)f_{Y|X}(y|x)dy\right)\dfrac{1}{x+1}f_X(x)dx\end{align*}$$ where $$f_{Y|X}(y|x)=\dfrac{f_{XY}(x,y)}{f_X(x)}$$ for appropriate $x$ such that $f_X(x)>0$.

Jimmy R.
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