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The joint probability density function for random variables $X$, $Y$ is given by $$f(x, y)=\begin{cases} 2xy & \text{if } 0\leq x\leq 2y\leq 2 \\ 0 & \text{otherwise} & \end{cases}.$$ When the conditional expectation of $X$ is $E(Y | X=aY)=\frac{3}{10}$, what is the real number $a(>1/2)$?

My first solution is $$f(y \mid x=ay)=\frac{f(ay, y)}{f_{x}(ay)}=\frac{2ay^2}{\int_{0}^{\frac{1}{a}}2ay^2 dy}=3a^3y^2$$ $$\frac{3}{10}=E(Y\mid X=aY)=\int_{0}^{\frac{1}{a}} 3a^3y^3 dy=\frac{3}{4a} \Rightarrow a=\frac{5}{2}\,.$$

My second solution is

Let $Z=\frac{X}{Y}, \; W=Y$. Then, $X=YZ, \: \: Y=W, \;$, the Jacobian is $W$ and $g(z,w)=f(wz, w)w=2zw^3 \; (\frac{1}{2}\leq z\leq \frac{1}{w}, \; 0\leq w\leq 2)$ $$g(w \mid z=a)=\frac{g(a, w)}{g_{Z}(a)}=\frac{2aw^3}{\int_{0}^{\frac{1}{a}}2aw^3 dw}=4a^4w^3$$ $$\frac{3}{10}=E(W\mid Z=a)=\int_{0}^{\frac{1}{a}} 4a^4w^4 dw=\frac{4}{5a} \Rightarrow a=\frac{8}{3}\,.$$

Why do I get two different values of $a$?

Lorents
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purecj
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  • The question seeks $E(X | Y=aX)$ but your working is for $E(Y | X=ay)$ – Math Lover Jun 19 '21 at 21:22
  • @MathLover sorry! It was modified. – purecj Jun 20 '21 at 03:30
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    There's an answer to this related question which I think can help sort out some of the confusion here. – Lorents Mar 16 '23 at 09:17
  • Various issues: (A) you are not going to get $E(Y \mid X=aY) < \frac 12$ for any $a$, as the density increases with $y \in[0,1]$; (B) you are not going to have $a>2$ if $X \le 2Y$; (C) $X=aY$ is an event of probability $0$ so you run the risk of the Borel–Kolmogorov paradox - there is no reason to expect $\lim\limits_{\delta \to 0^+} E(Y\mid X-aY \in [-\delta,\delta])$ to be the same as $\lim\limits_{\epsilon \to 0^+} E(Y\mid \frac XY -a \in [-\epsilon,\epsilon])$ – Henry Jul 10 '25 at 19:00

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