You have a normal distribution with mean of 0 and variance of 1. Keeping the same probabilities and focusing only on half of the distribution (other half has it's original probabilities but x values of 0) what is the expected value of this?

Im trying to teach myself expected outcomes with weird constraints. I got 0.3989, hopefully this is right.

  • 2
    $\begingroup$ this may help en.wikipedia.org/wiki/Half-normal_distribution $\endgroup$
    – JYY
    Commented Apr 24, 2019 at 1:36
  • $\begingroup$ Welcome to CV, vt_og. Which half? $\endgroup$
    – Alexis
    Commented Apr 24, 2019 at 1:43
  • $\begingroup$ well the positive half. I would assume the negative half woudl just be -0.3989 $\endgroup$
    – confused
    Commented Apr 24, 2019 at 2:34
  • $\begingroup$ There are two parts to answering this question: (1) finding a formula for the distribution and (2) computing the integral. With which part do you need help? $\endgroup$
    – whuber
    Commented Apr 24, 2019 at 3:26

1 Answer 1


If we let $Z \sim \text{N}(0,1)$ denote a standard normal random variable, the random variable $|Z|$ follows a standard half-normal distribution (the positive half of the standard normal distribution) and has $\mathbb{E}(|Z|) = \sqrt{2/\pi}$. This means that:

$$\mathbb{E}(\max(Z,0)) = \frac{1}{2} \cdot \mathbb{E}(|Z|) = \frac{1}{2} \cdot \sqrt{\frac{2}{\pi}} = \frac{1}{\sqrt{2 \pi}} \approx 0.3989423,$$

$$\mathbb{E}(\min(-Z,0)) = \frac{1}{2} \cdot \mathbb{E}(-|Z|) = -\frac{1}{2} \cdot \sqrt{\frac{2}{\pi}} = -\frac{1}{\sqrt{2 \pi}} \approx -0.3989423.$$


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