Conditional expectation of uniformly distributed variable

Question

I am given two independently distributed variables, X and Y. Both are uniform on the interval (-1,1).

What is $\mathrm{E}(X|Z)$ when $Z=\alpha + \beta X + Y$? I am a bit clueless on how to approach the problem. Is there something similar to the projection theorem (Projection theorem for conditional probability) for Uniform Distributions?

Many thanks in advance.

Answer 1

Hint: Let $W=Z-\alpha=\beta X+Y$. Since $\alpha$ is a constant, $\mathbb{E}\left[X\mid Z\right]=\mathbb{E}\left[X\mid W\right]$.

1. Compute the CDF of $X$ and $W$: $$ F_{X,W}(x,w) =\mathbb{P}(X\leq x,W\leq w) $$
2. Compute the joint PDF of $X$ and $W$ by taking derivatives: $$ f_{X,W} =\frac{\partial^{2}}{\partial x\partial w}\left[F_{X,W}\right] $$
3. Compute the density of $X$ conditional on $W$: $$ f_{X\mid W}(x\mid w)=\frac{f_{X,W}(x,w)}{f_{X}(x)} $$
4. Use the conditional density to resolve the expectation: $$ \mathbb{E}\left[X\mid W\right] =\int x f_{X\mid W}(x\mid w)dx $$

Answer 2

For all interested, I think the answer is along the following lines:

Suppose $X \sim U[0,1],Y \sim U[0,k], k \geq 1$ and $X$ and $Y$ are independent and $Z=X+Y$. Then, we have: $$f_z(z) = \int_{\Omega(z)}f_x(u)f_y(z-u)$$ Depending on $z$, we have three different intervals over which to integrate: $$ f_z(z)= \begin{cases} \int_0^z \frac{1}{k}du = \frac{z}{k}, & \text{for } 0\leq z\leq 1\\ \int_0^1 \frac{1}{k}du = \frac{1}{k}, & \text{for } 1\leq z\leq k\\ \int_{z-k}^1 \frac{1}{k}du = \frac{1}{k} - \frac{z-k}{k} , & \text{for } k\leq z\leq k+1 \end{cases} $$

The conditional pdf of $X$ given $Z$ is then given by: $$ f_{x|z}(x|z) = \frac{f_{z|x}(z|x) f_x(x)}{f_z(z)} $$

Thus, for the three intervals of $z$ we have:

$$ f_{x|z}(x|z)= \begin{cases} \frac{1}{z}, & \text{for } 0\leq z\leq 1 \text{ and } 0\leq x \leq z\\ 1, & \text{for } 1\leq z\leq k \text{ and } 0\leq x \leq 1\\ \frac{1}{k+1-z}, & \text{for } k\leq z\leq k+1 \text{ and } z-k\leq x \leq 1 \end{cases} $$

The expected value $\mathbb{E}[X|Z]$ follows immediately and is given by

$$ \mathbb{E}(x|z)= \begin{cases} \frac{\hat{z}}{2}, & \text{for } 0\leq z\leq 1\\ \frac{1}{2}, & \text{for } 1\leq z\leq k\\ \frac{1}{2} (1-k+\hat{z}), & \text{for } k\leq z\leq k+1 \end{cases} $$

The extension for the general case follows if $X = a+(b-a) \hat{X}$, $Y = c+(b-a) \hat{Y}$, whereby the variables with a hat are the ones that I used previously.