# What is the conditional $f_{X|Y}$ and expectation of $X|Y$ over a irregular area $f_{X,Y}(x,y)=.01$

I am confused as to how to integrate, would one split this up into sections ?

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## 1 Answer

Recall (or see) that $f_{X|Y}=f_{X,Y}/f_Y$, for a particular value of $Y$, is just the joint $f_{X,Y}$ density restricted to that $Y$ (like if we'd made a cut on the above graph, over an horizontal line) and properly normalized. Then, if the joint density is uniform over some convex region on ${\mathbb R}^2$, you should see that $f_{X|Y}$, for each given $Y$, is uniform over some interval that you should be able to draw; once you see that, you note that the expectation of a uniform variable lies in the midpoint of the interval.

In the example above (it's more easy to draw it than to explain it), you have for example that for $Y=1.5$, $f_{X|Y}$ is uniform in $[-1,1]$, and hence the mean is zero. For $Y=0.2$, $f_{X|Y}$ is uniform in $[-1,2]$, and hence the mean is $-1+(2-(-1))/2=0.5$

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I think I understood everything except the last part about the midpoint of the interval. so if I found the fx|y by splitting the problem over the intervals x (-1,1) and take that rectangle and then do the other rectangle? Would that work? – Olivia Irving Jan 25 '13 at 16:49
If a variable is uniform in the interval $[3,7]$ the mean lies in the midpoint ($E(X)=5$). I added some examples. – leonbloy Jan 25 '13 at 16:53
Oh, so this problem wants me to use the value of y some particular value I choose. Then just use that to calculate the probability also why then do they give me the joint pdf? – Olivia Irving Jan 25 '13 at 17:16