The joint probability density function for random variables $X$, $Y$ is given by $$f(x, y)=\begin{cases} 2(x+y) & \text{if } 0<x<y<1 \\ 0 & \text{otherwise} & \end{cases}.$$ When the conditional expectation of $X$ is $E(X | Y=aX)=\frac{2}{9}$, what is the real number $a$?
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Hint: For $a > 1$, the conditional density of $X$ given $Y = aX$ is $f_{X \mid Y=aX}(x) := \frac{f_{X,Y}(x, ax)}{\int_0^{1/a} f_{X,Y}(t, at) \, dt}$ for $x \in (0,1/a)$.
Continuing from above, $$f_{X \mid Y=aX}(x) := \frac{f_{X,Y}(x, ax)}{\int_0^{1/a} f_{X,Y}(t, at) \, dt}= \frac{x+ax}{\int_0^{1/a} (t+at) \, dt} = \frac{(1+a)x}{(1+a)/(2a^2)} = 2a^2 x.$$ Now that we have the conditional density, the desired conditional expectation is $$\int_0^{1/a} x (2 a^2 x) \, dx = \frac{2}{3a}.$$
By definition, $$\begin{align}\mathsf E(X\mid Y{=}aX) ~&=~\dfrac{\displaystyle\int_\Bbb R x\, f(x,ax)\,\mathsf d x}{\displaystyle\int_\Bbb R f(x,ax)\,\mathsf d x}\\[2ex]~&=~\dfrac{\displaystyle\int_0^{1/a} 2(1+a)x^2\,\mathsf d x}{\displaystyle\int_0^{1/a} 2(1+a)x\,\mathsf d x}&&\text{if }a\geq 1\end{align}$$