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?
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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 '12 at 1:14
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4$\begingroup$ See truncated normal distribution. $\endgroup$ – Cyan Apr 20 '12 at 1:17
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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.)
answered Sep 21 '16 at 4:55
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