8
$\begingroup$

Given $X \sim N(0, \sigma^2)$ (that is, $X:\mathbb{R} \to \mathbb{R}$ is a normal random variable with mean $0$ and variance $\sigma^2$), I'm trying to calculate the expected value of $X$ given that $X>0$. I thought that integrating $$ \int_{0}^{\infty} x\cdot \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2\sigma^2}}dx $$ would do it, but the value, $\frac{\sigma}{\sqrt{2\pi}}$, seems to be off by a factor of 2 based on some other information I have; I think the answer should be $\sqrt{\frac{2}{\pi}}\sigma$.

Question: How should the expected value of $X$, given that $X>0$, be computed?

  • 7
    You computed $E(X\mathbf 1_A)$, where $A=[X\gt0]$, instead of $E(X\mid A)$.2011-12-03
  • 0
    Wolfram gives the same [answer](http://www.wolframalpha.com/input/?i=integrate+%281%2F%5Csqrt%282*pi*t%5E2%29%29+x+exp%28-x%5E2%2F2t%5E2%29+from+0+to+infinity) as what you get. Why do you think your answer is off by factor of 2?2011-12-03
  • 1
    @tards I know the computation of the integral is correct, I'm just don't think that the integral represents what I actually want, as Michael Lugo's answer confirms.2011-12-03
  • 0
    "Exact duplicate"? I think I posted the answer to this same question about two months or so ago.2011-12-03
  • 0
    Take a look at this: http://math.stackexchange.com/questions/71537/derivation-of-chi-squared-pdf-with-one-degree-of-freedom-from-normal-distributio/71594#715942011-12-03
  • 1
    @MichaelHardy I don't see how the post you liked is a duplicate. Please explain.2011-12-04

1 Answers 1

12

Let $f(x)$ be the density of $X$; let $F(x)$ be its CDF.

Then the density of $X$, conditional on it being positive, is $f(x)/P(X \ge 0)$ if $x \ge 0$, and $0$ otherwise.

Of course $P(X \ge 0) = 1/2$ by symmetry, so the density of $X$ conditional on $X \ge 0$ is $2f(x)$ (on $x \ge 0$).

So you need to do the integral $$ \int_0^\infty 2xf(x) \: dx $$ which is twice the integral you've written.