I have a sample that is pretty normally distributed with a mean greater 0. Is there any way to get a closed form estimate for the mean of all values that are less than 0? In other words, given a mean, variance and a variable x, can I estimate the average of all values that are smaller than x?

  • 1
    $\begingroup$ Not in the usual sense of closed form; if you admit the use of the cdf of a standard normal, then, the answer is yes. $\endgroup$
    – cardinal
    Apr 20, 2012 at 1:14
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    $\begingroup$ See truncated normal distribution. $\endgroup$
    – Cyan
    Apr 20, 2012 at 1:17

1 Answer 1


Cyan offered a link that answers the question but since the question remains without an actual answer, I'll put one in.

While not strictly "closed form" by the usual definitions, I expect people doing statistical work will mostly want to admit the normal cdf (or the error function, which would serve the same purpose) into what they call "closed form".

Given $X\sim N(\mu,\sigma^2)$,

$$E(X\mid X<b)=\mu -\sigma \frac {\phi (\frac{b-\mu}{\sigma})}{\Phi (\frac{b-\mu}{\sigma})}$$

where $\phi$ is the standard normal density and $\Phi$ is the standard normal cdf.

For the present problem, $b=0$, giving us:

$$E(X\mid X<0)=\mu -\sigma \frac {\phi (\frac{-\mu}{\sigma})}{\Phi (\frac{-\mu}{\sigma})}$$

(An expression for the variance can be found at the above link.)


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