2

Suppose $X_1,X_2$ are independent and standard normal. I want to compute $E(X_1|X_1+X_2>0)$. How do I do it?

Here is my attempt:

\begin{align} E(X_1|X_1+X_2>0)&=\frac{\int_{-x_2}^{+\infty} x_1 f_{X_1}(x_1)dx_1}{P(X_1+X_2>0)}\\ &=\frac{\int_{-x_2}^{+\infty} x_1 \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x_1^2}dx_1}{0.5}\\ &=\frac{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x_2^2}}{0.5}\\ &=\frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}x_2^2} \end{align} But I suppose the answer should not have $x_2$ in it. What am I missing?

sreysus
  • 721
Accept Cookies
  • 109
  • 4
  • What variables should a conditional expectation depend on? – GEdgar Oct 19 '23 at 01:28
  • You can't use the marginal distribution of $X_1$ as you already counted even those $X_2$ making $X_1+X_2\leq 0$. Use the joint distribution $(X_1,X_2)$ and change variables to $(X_1,X_1+X_2)$. – user10354138 Oct 19 '23 at 01:40

1 Answers1

3

The key here is that, since $X_1$ and $X_2$ are iid, we have that $$ \mathbf{E}(X_1 | X_1 + X_2 > 0) = \mathbf{E}(X_2 | X_1 + X_2 > 0). $$ which we can use to compute \begin{align*} \mathbf{E}(X_1 | X_1 + X_2 > 0) &= \frac{1}{2}[\mathbf{E}(X_1 | X_1 + X_2 > 0) + \mathbf{E}(X_2 | X_1 + X_2 > 0)] \\ &= \frac{1}{2}\mathbf{E}(X_1 + X_2 | X_1 + X_2 > 0) \\ &= \frac{\sqrt{2}}{2}\mathbf{E}(Z|Z>0), \end{align*} where $Z = (X_1 + X_2) / \sqrt{2} \sim N(0, 1)$.

Now, it is a standard fact about Gaussian distributions (details, for example, here: Expected value of normal distribution given that distribution is positive) that $\mathbf{E}(Z | Z>0) = \sqrt{2/\pi}$, from which we have that $$ \mathbf{E}(X_1 | X_1 + X_2 > 0) = \frac{1}{\sqrt{\pi}}. $$

Damian Pavlyshyn
  • 3,722