Problem
Let $X, Y \sim N(0,1)$ indepdendent and let $Z = 3X+4Y$. Find $E(X \, | \, Z)$.
Intuitive solution $$ E(X \, | \, Z = z) = E(X \, | \, 3X + 4Y = z) = E(X \, | \, X = (z - 4Y)/3) = E(z/3 - 4Y/3) = z/3. $$ This would imply that $E(X \, | \, Z) = Z/3$. I think there is something wrong with this solution because I get another result computing the conditional densities, but I don't know what the problem is.
Solution
Here I just find the conditional density and take expectations. We have $X \sim N(0,1)$, $Z \, | \, X = x \sim N(3x, 16)$, and $Z \sim N(0,25)$.
Integrating, we get $$ E(X \, | \, Z= z)= \int_{-\infty}^\infty x \frac{f_X(x) f_{Z \, | \, X=x}(z) }{f_Z(z)} \, \mathrm{d} x = \frac{3}{25}z, $$ which implies $E(X \, | \, Z) = 3Z/25$.
I would like to know what's wrong with the intuitive solution. Also, is there is a way to solve this problem without doing the integration?
Thanks!
The intuitive solution is based on the wrong belief that if $X$ and $U$ are independent then $E[X\mid X=U]=E[U]$, used for $U=\frac13(z-4Y)$. This is not true in general.
As a counterexample, consider $X$ uniform on $\{1,2,3,4\}$ and $U$ uniform on $\{4,5,6,7,8\}$, then $[X=U]\subseteq[X=4]$ hence $E[X\mid X=U]=4$, while $E[U]=6$ is not even in the support of $X$.
As an alternative to the other solution (which is correct), consider $T=4X-3Y$ and note that $T$ is centered, that $Z$ and $T$ are independent and that $25X=3Z+4T$. Hence, $$ 25E[X\mid Z]=3E[Z\mid Z]+4E[T\mid Z]=3Z+4E[T]=3Z.$$