Probability – How to Find Conditional Expected Value of Continuous Uniform RV Given Multiple Conditions

Question

Let $X_1, X_2, X_3$ be continuous, mutually independent, uniformly distributed random variables in $[0, 1]$.
Let $Y_1=X_1+X_2$ and $Y_2 = X_1 + X_3$. From linearity of the expected value, and from symmetry of the uniform distribution, it is easy to see that $\mathbb{E}[X_1|Y_1=a]=\mathbb{E}[X_2|Y_1=a] = \frac{a}{2}$ and that $\mathbb{E}[X_1|Y_2=b]=\mathbb{E}[X_3|Y_2=b] = \frac{b}{2}$, with $a,b\in[0,2]$. But how do you compute the expected conditional values $\mathbb{E}[X_1|Y_1=a, Y_2=b]$ and $\mathbb{E}[X_2|Y_1=a, Y_2=b]$? What if you also have other conditions on the range of $X_1,X_2,X_3$? For instance, what if you also know that $X_1 < c$, with $c\in[0,1]$?

Answer 1

If we know $Y_1 = a, Y_2 = b$ then we know \begin{align*} X_1 + X_2 &= a,\\ X_1 + X_3 &= b \end{align*} Therefore $$X_1 = b - X_3, \textrm{ and } X_2 = a - b + X_3.$$ We know that each of $X_1, X_2, X_3$ are in $[0, 1]$ so $$0 \leq b - X_3 \leq 1, \ 0 \leq a - b + X_3 \leq 1, \textrm{ and } 0 \leq X_3 \leq 1.$$ This places the constraints $$\max(b - 1, b - a, 0) \leq X_3 \leq \min(b, b - a + 1, 1)$$ on $X_3$. Since $X_3$ is uniform, the conditional distribution of $X_3$ is now uniform on $$[\max(b - 1, b - a, 0), \min(b, b - a + 1, 1)]$$ and so $$\mathbb{E}(X_3 \mid Y_1 = a, Y_2 = b) = \frac{\max(b - 1, b - a, 0) + \min(b, b - a + 1, 1)}{2}.$$ And therefore $$\mathbb{E}(X_1 \mid Y_1 = a, Y_2 = b) = b - \frac{\max(b - 1, b - a, 0) + \min(b, b - a + 1, 1)}{2}$$ and $$\mathbb{E}(X_2 \mid Y_1 = a, Y_2 = b) = a - b + \frac{\max(b - 1, b - a, 0) + \min(b, b - a + 1, 1)}{2}$$

If we have other constraints e.g. $X_1 < c$ then it would just add the constraint $$c < X_1 \iff b - X_3 < c \iff b - c < X_3$$ to the constraints on $X_3$, so the lower bound would be $$\max(b - 1, b - a, 0, b - c).$$