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Suppose $X \sim N(\mu, 1)$ is a standard normal with shifted mean. If I sample $X$ once and calculate $\Phi(X)$ (where $\Phi$ is the CDF of the standard normal), what is its expected value and variance?

If $\mu = 0$ then $X$ is just standard normal, so we can deduce that $E[\Phi(X)] = 1/2$, but how do you calculate its variance? I suppose I can go straight to evaluating the integral, but was wondering if there's a "Statistical" approach to it that cleverly uses the definition of $X$ or properties of $\Phi$

If $\mu > 0$, then the expected value is also not $1/2$ but again, I can evaluate the integral directly or I feel like there is a more clever way to do this.

Hendrata
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  • In case it helps, I found this discussion of [variance of a function of a random variable](https://math.stackexchange.com/questions/3285581/variance-of-function-of-random-variable#3285587). – TMBailey Oct 18 '21 at 18:48
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    Solve the problem for general continuous distribution functions $F$ rather than $\Phi:$ that will remove a lot of distracting, irrelevant information and enable you to focus on the essence of the problem. After you write the integral for any of the moments, that will leave you with only one possible thing to do: apply the change-of-variables formula. That takes care of the case $\mu=0.$ For other $\mu,$ you will need formulas like the one at https://stats.stackexchange.com/questions/61080. – whuber Oct 18 '21 at 19:25
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    You can see from [here](https://stats.stackexchange.com/q/61080/119261) that the expectation is $\Phi\left(\frac{\mu}{\sqrt 2}\right)$ and for the variance, see https://stats.stackexchange.com/q/212421/119261. If $\mu=0$, we can of course directly say that mean is $1/2$ and variance is $1/12$ because $\Phi(X)\sim \text{Uniform}(0,1)$. – StubbornAtom Oct 18 '21 at 19:39

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