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In particular i have to find the variance of this random variable

$$U = \int_{-\infty}^{y} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-T)^2} dx$$

where T is a ranadom variable distributed with a normal distribution with mean $\mu$ and variance $\sigma^2$.

So i have to find the $Var(U)$. How can i find it?

Probably this is a cumulative function of a standard normal with some modification, but i don't know how i can approach it.

Tazz
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    At https://stats.stackexchange.com/questions/482345 I have recently posted an analysis of precisely such an integral (for a Weibull distribution). The trick is to evaluate the integral! – whuber Sep 01 '20 at 21:01
  • @whuber ah ok! so if i calculate the integral i have, with a sobsitution $u =x-T$ , i obtain that $U =\phi(y-T)$. But now, how can i calculate the variance of the cumulative function? because in the variance formula, i have to calculate $E[U^2]$, so how this expectation is? – Tazz Sep 02 '20 at 07:35
  • Apply any of the techniques illustrated at https://stats.stackexchange.com/questions/61080. – whuber Sep 02 '20 at 12:42

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