Is there a closed-form solution for the variance of $Y = \Phi\left(X\right)$, where $X \sim N \left(\mu, \sigma^2\right)$ and $\Phi$ is the standard normal CDF? I can find the variance for some specific values of $\mu$ and $\sigma$ kind of "trivially" ($\mu = 0$, $\sigma = 1$), but is there a general analytical solution? If not, is there at least one if I know the specific values of $\mu$ and $\sigma$? I tried the method described here, but when I compare the result to that from a simulation, they're not close.
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Because Chebyshev's inequality applies broadly (to any distribution having a variance) you cannot expect the Chebyshev _bound_ on a probability in a particular instance to be close to the probability. For example if $Z$ is standard normal, then $P(|Z| < 2) \approx 0.95$ but the Chebyshev bound guarantees $P(|Z| < 2) \le 3/4.$ – BruceET Sep 08 '20 at 19:17
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1@BruceET it's $\geq3/4$ – gunes Sep 08 '20 at 20:16
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Indeed yes. Typo. – BruceET Sep 08 '20 at 21:30
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See https://stats.stackexchange.com/questions/212421/variance-of-a-cumulative-distribution-function-of-normal-distribution – Kieran Sep 09 '20 at 07:55
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This is a duplicate as pointed out by @Kieran in the comments. The post below provides the answer:
Skumin
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