<s>[INST] What is $\mathbb{E}\!\left(X^2 \mid X + 2Y \right)$ if $X,Y \sim \mathcal{U}\left(\left[-1,1\right]\right)$ independently? $X$ and $Y$ are independent random variables of the uniform distribution on $\left[-1,1\right]$, $X,Y \sim \mathcal{U}\!\left(\left[-1,1\right]\right)$.
What is $\mathbb{E}\!\left(X^2 \mid  X + 2Y \right)$?
My attempt:
We need the joint pdf of $X$ and $X+2Y$ (as suggested). So we do a change of variables, let $W = X,~Z = X + 2Y$. This is a linear transformation, so the jacobian will be a constant. We need to find the image of the square $-1 \leq x, y \leq 1$ under this transformation. Since it is a linear transformation, the four edges of boundaries will be mapped to line segments. So we can find them by computing where the four corners of the square are mapped:
$$\begin{align}
\left(x,y\right)&|\left(w,z\right)\\
(-1,-1) &| (-1, -3)\\
(-1,+1) &| (-1,+1)\\
(+1,+1) &| (+1,+3)\\
(+1,-1) &| (+1,-1)
\end{align}$$
The image of the square is the parallelogram with vertices as above and in the picture below:

The joint density of $W$ and $Z$ will be uniform on this region, let A denote the interior of the parallelogram. Since it has area of 8, we can conclude
$f_{W,Z}\left(w,z\right) = \frac{1}{8}\unicode{x1D7D9}\left(\left(w,z\right) \in A\right) = \begin{cases}
\frac{1}{8},  & \text{if $-1\leq w \leq 1,~-2+w\leq z\leq2+w$} \\
0, & \text{otherwise}
\end{cases}.$
The computations for $\mathbb{E}\left(X|X+2Y\right)$ and $\mathbb{E}\left(X^2|X+2Y\right)$ are below.

(for $\mathbb{E}\left(X|X+2Y\right)$, $Z = X + 2Y$)
Now we can figure out what $f_{X|Z}\left(x|z\right)$ is. We must consider three cases:

*

*$-3 \leq z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+2+z}{2} = \frac{z}{2} + \frac{1}{2}$


*$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+1}{2} = 0$


*$1 \leq z \leq 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{1-2+z}{2} = \frac{z}{2} - \frac{1}{2}$
Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X|Z=z\right)$.

(for $\mathbb{E}\left(X^2|X+2Y\right)$, $Z = X + 2Y$)
We must consider three cases:

*

*$-3 < z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$ (so probability density function $f_{X|Z}(x|z) = \frac{1}{(z+2) - (-1)} = \frac{1}{z+3}$). So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{z+2}x^2 \cdot \frac{1}{z+3}dx = \frac{1}{3} \cdot \frac{(z+2)^3+1}{z+3}$


*$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$ (probability density function $f_{X|Z}(x|z) = \frac{1}{2}$).
So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{1}x^2 \cdot \frac{1}{2}dx = \frac{1}{3}$


*$1 \leq z < 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_1^{z-2}x^2\frac{1}{z-3}dx = \frac{1}{3}\cdot\frac{1}{z-3}\cdot\left((z-2)^3-1\right)$
Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X^2|Z=z\right)$.
 [/INST] Define $h:\mathbb{R} \to \mathbb{R}$ such that $h(x)=\frac 1 3 \chi_{[-3,3]}$.
The claim is that $\mathbb{E}\left(X^2 | X + 2Y \right)=h(X+2Y)$.

Suppose $B \in \mathscr{B}([-3,3])$.
Let us prove $\int \chi_{B}(x+2y)x^2dx dy=\int \chi_{B}(x+2y)h(x+2y)dxdy$.
If $g(x,y)=(x+2y,x)$ then $|\det g (x,y)|=2$ and thus we have to prove
$$\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx$$
thanks to the change of variable.
Now $\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\mu(B)\frac 1 3$ and $\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)\frac 1 3 dzdx=\mu(B)\frac 1 3$ which conclude the proof.
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