Modify problem 4 p.98/142 from this PDF:
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$f_{X,Y}(x,y)=\begin{cases} cxy,& \mbox{if } 0\leq x\leq y\leq 1,\\ 0,& \mbox{otherwise,}\end{cases}$
If we know $Y = 0.5$, is my conditional calculation of $f_{X|Y}(x\, |\, 0.5)$ below correct?
The joint PDF now becomes: $f_{X,Y}(x,y) = cxy, \textrm{for}~0 \leq x \leq y \leq 0.5 $
$\displaystyle \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f_{X,Y}(x,y) = c\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} xy\,dy\,dx = c\int\limits_{0}^{0.5}\int\limits_{x}^{0.5} xy\,dy\,dx = c\int\limits_{0}^{0.5} x\frac{y^2}{2}\Big |_{x}^{0.5}\,dx = \frac{c}{2}\int\limits_{0}^{0.5} x\cdot y^2\Big |_{x}^{0.5}\,dx$
$\displaystyle = \frac{c}{2}\int\limits_{0}^{0.5} x\cdot \left(\frac{1}{4} - x^2\right)\,dx = \frac{c}{2}\int\limits_{0}^{0.5} \left(\frac{x}{4} - x^3\right)\,dx = \frac{c}{2}\left(\frac{x^2}{8} - \frac{x^4}{4}\right)\Bigg |_{0}^{0.5} = \frac{c}{2}\left(\frac{1}{32} - \frac{1}{64}\right) = \frac{c}{128} = 1$
Thus, $c = 128$.
$f_{X,Y}(x,y) = 128\cdot xy, \textrm{for}~0 \leq x \leq y \leq 0.5 $
$\displaystyle f_{Y}(y) = 128\cdot \int\limits_{0}^{0.5}xy\,dx = 128y\cdot \int\limits_{0}^{0.5}x\,dx = 128y\cdot \frac{x^2}{2}\Bigg |_{0}^{0.5} = 64y\cdot x^2\Bigg |_{0}^{0.5} = 16y$
Therefore:
$\displaystyle f_{X|Y}(x\mid Y = 0.5) = \frac{f_{X,Y}(x,0.5)}{f_{Y}(y)} = \frac{128xy}{16y} = 8x $
Did I get it right?
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From Maxim's suggestion:
$f_{X,Y}(x,y) = cxy, \textrm{for}~0 \leq x \leq y \leq 1 $
$\displaystyle \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f_{X,Y}(x,y) = c\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} xy\,dy\,dx = c\int\limits_{0}^{1}\int\limits_{x}^{1} xy\,dy\,dx = c\int\limits_{0}^{1} x\frac{y^2}{2}\Big |_{x}^{1}\,dx = \frac{c}{2}\int\limits_{0}^{1} x\cdot y^2\Big |_{x}^{1}\,dx$
$\displaystyle = \frac{c}{2}\int\limits_{0}^{1} x\cdot \left(1 - x^2\right)\,dx = \frac{c}{2}\int\limits_{0}^{1} \left(x - x^3\right)\,dx = \frac{c}{2}\left(\frac{x^2}{2} - \frac{x^4}{4}\right)\Bigg |_{0}^{1} = \frac{c}{2}\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{c}{8} = 1$
Thus, $c = 8$.
$f_{X,Y}(x,y) = 8\cdot xy, \textrm{for}~0 \leq x \leq y \leq 1 $
$\displaystyle f_{Y}(y) = 8\cdot \int\limits_{0}^{y}xy\,dx = 8y\cdot \int\limits_{0}^{y}x\,dx = 8y\cdot \frac{x^2}{2}\Bigg |_{0}^{y} = 4y\cdot x^2\Bigg |_{0}^{y} = 4y^3,~y\in [0,1]$
Therefore:
$\displaystyle f_{X|Y}(x\mid Y = 0.5) = \frac{f_{X,Y}(x,0.5)}{f_{Y}(y)} = \frac{8xy}{4y^3} = \frac{2x}{y^2} = \frac{2x}{0.5^2} = 8x,~x\in [0,0.5]$
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Maxim's beautiful method:
$\displaystyle f_{X \mid Y = y}(x \mid Y = y) = \frac{f_{X,Y}(x,y)}{f_{Y}(y)} = \frac {cxy \, [0 \leq x \leq y]} {\int_0^y cxy \, dx} = \frac {cxy \, [0 \leq x \leq y]} {cy\int_0^y x \, dx} = \frac {cxy\, [0 \leq x \leq y]} {cy\frac{x^2}{2}\big|_0^y}$
$\displaystyle = \frac {cxy}{cy\frac{y^2}{2}}~[0 \leq x \leq y] = \frac {2 x} {y^2}~[0 \leq x \leq y]$
Thus, $\displaystyle f_{X \mid Y}(x \mid y) = \frac {2 x} {y^2},~\textrm{for}~0 \leq x \leq y \leq 1$