# Conditional Expectation of pdf

Wish to identify what I'm doing wrong when finding the $$\operatorname E(X\mid Y=5)$$ of the following: $$f(x, y)=\begin{cases} 1/6 & \text{if } 0

My working: \begin{align}\operatorname E(X\mid Y=5)~&=~\int_0^2 x \frac{\frac{1}{6}}{f(y)} \, \mathsf d x\end{align} where \begin{align}\mathsf f(y)~&=~\int_0^{\frac{-y+6}{3}} \frac{1}{6} \,\mathsf d x = \frac{6-y}{18}\end{align}

Therefore, \begin{align}\mathsf E(X\mid Y{=}5)~&=~\int_0^2 x \frac{\frac{1}{6}}{\frac{6-y}{18}}\,\mathsf d x \mathsf ~&=~\int_0^2 x \frac{\frac{1}{6}}{\frac{6-5}{18}}\,\mathsf d x~&=~\int_0^2 3x \,\mathsf d x \end{align} = 6

But then when I compute $$\operatorname{Var}(X\mid Y=5),$$ the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.

My working for Variance: $$\operatorname{Var}(X\mid Y=5) = \operatorname E(X^2\mid Y=5) - (\operatorname E(X\mid Y=5) )^2 = 8 - 6^2 = -28$$

I appreciate any help.

• you forgot the constraint on the support of $X$ given $Y=y.$ – Xi'an Nov 1 '18 at 21:23

Think about this geometrically: we've got $$(X,Y)$$ uniformly distributed over the right triangle with vertices at $$(0,0)$$, $$(2,0)$$, and $$(0,6)$$.
If we imagine sampling over and over from this region uniformly, we can picture $$E(X\mid Y=5)$$ as the average $$x$$ coordinate of the points that end up on the horizontal line $$y=5$$. All of these points will necessarily have $$0 < x < 1/3$$, since if $$x \geq 1/3$$ then $$y = 5$$ can't happen, and since the points are uniform over this line, this image suggests $$E(X\mid Y=5) \stackrel {\text{?}}= 1/6$$ (this is the midpoint of the range of $$x$$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $$x$$ also depend on $$y$$. If $$y = 5$$ is observed then the biggest that $$x$$ can be is $$1/3$$, so really $$E(X\mid Y=5) = \int_0^{1/3}x \frac{3}{6-y}\,\text dx = 3 \cdot \frac 12 x^2\bigg|_{x=0}^{x=1/3} = \frac 16.$$
Another quick sanity check is that $$E(X\mid Y)$$ has to be within the range of $$X$$, so $$E(X\mid Y=5)\neq 6$$ can be recognized as incorrect before the negative variance shows up.