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Given $X \sim N\left(\mu, \sigma^2 \right)$, I want to find the distribution of $$Y = \Phi \left( X \right),$$ with $\Phi \left(x\right) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-\frac{t^2}{2}}dt$, i.e., CDF of a Standard Normal Distribution.

More specifically, I want to get an expression for $SD \left[Y \right]$, the standard deviation of $Y$.

For the specific case with $\mu = 0, \sigma=1$, I would easily have $Y \sim U\left(0,1 \right)$, but what will be the case for any normal distribution?

Any pointers will be very helpful.

Zhanxiong
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augustine
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    If by $\Phi(X)$ you mean $Y=F(X)$ where $F(x)=\Pr(X \le x)$, your $Y\sim U(0,1)$ applies to all continuous distributions – Henry Jun 27 '23 at 10:05
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    @utobi $\Phi \left( x \right) $ is CDF of Standard Normal Distribution. I have updated my original post with this information – augustine Jun 27 '23 at 10:13
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    Derive $\mathbb P(\Phi(X)\le y)$ then. – Xi'an Jun 27 '23 at 10:21
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    There may not be a simple solution with your $\Phi$ being the standard normal CDF. Clearly $Y$ is constrained to be on $[0,1]$ and has a symmetric distribution about $0.5$ when $\mu=0$. The upper bound on the standard deviation is $0.5$, approached when $\mu$ is constant and $\sigma$ increases without bound (bimodal in this case). The lower bound on the standard deviation is $0$, approached when $\mu$ is constant and $\sigma$ decreases towards $0$. – Henry Jun 27 '23 at 10:22
  • @Henry This is insightful. How can I calculate the upper bound of standard deviation as 0.50? – augustine Jun 27 '23 at 11:31
  • @augustine Because the support of $Y$ is $[0,1]$ since a CDF can never be less than $0$ or more than $1$. The variance (and so standard deviation) is maximised when the probability is concentrated at the end points and when symmetric between these – Henry Jun 27 '23 at 11:43
  • But where does the value 0.50 come from? – augustine Jun 27 '23 at 11:53
  • Closely related: https://stats.stackexchange.com/questions/61080. – whuber Jun 27 '23 at 17:31
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    The 0.5 is explained at https://stats.stackexchange.com/questions/45588/variance-of-a-bounded-random-variable. – whuber Jun 27 '23 at 17:32

1 Answers1

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There is no closed form expression for the standard deviation but it may be useful to express it in terms of the bivariate normal cdf as follows. First, note that \begin{align} E(Y)&=E(\Phi(X)) \\&=P(Z\le X) && \text{where $Z\sim N(0,1)$} \\&=P(Z-X\le 0) \\&=\Phi\Big(\frac{\mu}{\sqrt{1+\sigma^2}}\Big). \end{align} Similarly, \begin{align} E(Y^2)&=E(\Phi(X)^2) \\&=P(Z_1 \le X \cap Z_2\le X) &&\text{where $Z_1,Z_2$ are iid $N(0,1)$} \\&=P(Z_1-X\le 0 \cap Z_2-X\le 0) \\&=\Phi_2\Big(\frac{\mu}{\sqrt{1+\sigma^2}},\frac{\mu}{\sqrt{1+\sigma^2}}\Big) \end{align} where $\Phi_2$ is the cdf of a bivariate standardnormal distribution with correlation $$ \rho=\frac{\sigma^2}{1+\sigma^2} $$ available as the function mvtnorm::pmvnorm in R. So from this you can compute the variance and standard deviation of $Y$.

Jarle Tufto
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