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Let X and Y be continuous random variables having the joint pdf $$f(x,y) = 8xy , 0\leq{y}\leq{x}\leq{1}$$ Find $g(x|y=\frac{1}{2})$ the conditional pdf of $X$ given $Y = \frac{1}{2}$

I found that the marginal pdf of Y is $f_2(y) = 4y - 4y^3$.

And $g(x|y=\frac{1}{2}) = \frac{f(x,\frac{1}{2})}{f_2(\frac{1}{2})} = \frac{8x}{3}$. I thought that the support of this function would be $y ≤ x ≤ 1$, $0 ≤ y ≤ 1$. But when I evaluate the integral from $y$ to $1$, I don't get $1$. Does this mean I've done something wrong? Or is it that the support (domains) are incorrect?

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  • $\begingroup$ When you say "the support of this function would be $y ≤ x ≤ 1$, $0 ≤ y ≤ 1$", are you referring to $g(x|y=1/2)$? That isn't a function of $y$ though. The support is just $1/2\le x\le 1$. $\endgroup$
    – snarfblaat
    May 5, 2016 at 16:02

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Your calculation is correct. At a specific value of $y\in[0,1]$, the conditional density of $X$ given $Y=y$ has support $[y,1]$. So when $y=\frac12$ the support of the conditional density is $\frac12\le x\le1$. When you integrate the conditional density from $x=\frac12$ to $x=1$, the result is $1$.

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  • $\begingroup$ Ohhhh, thanks heaps $\endgroup$
    – Mathew
    May 6, 2016 at 0:35

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