Suppose, $X\sim N(\mu,\sigma^2)$.

Can anyone help in finding the following : $\text{Var }\Phi(X)$ ?

Here, $\Phi(x)$ is the "Cumulative Distribution Function" of the above-mentioned normal distribution.

Thanking you in advance.


1 Answer 1


Some of the following conditions can be left out, but let's keep it easy.

Let $F$ as CDF of random variable $X$ be continuous, strictly increasing and taking values in $(0,1)$.

Then $F:\mathbb R\to (0,1)$ is a bijection, so has an inverse. For $u\in(0,1)$ we find:$$F(X)\leq u\iff X\leq F^{-1}(u)$$ hence:$$P(F(X)\leq u)= P(X\leq F^{-1}(u))=F(F^{-1}(u))=u$$

So what is the distribution of $F(X)$?

If you know the distribution then you can also find the corresponding variance.

Apply this on $F=\Phi$.

  • $\begingroup$ First of all, I would like to thank you so much for providing me with the solution, @drhab !! I have one more query, sir, with regards to the above problem. Suppose, now we are to find $$ \text{Var} \, \big[\Phi \big(\frac{x + c}{d}\big) \big].$$ Here c,d are both positive. Then what changes take place in your above solution ?? Thanking you again in anticipation. $\endgroup$ May 13, 2016 at 15:43
  • $\begingroup$ Things will get complicated then, and this in such a way that I have no proper answer to that question. Integrals like $\int F(x)dF(x)$ or $\int F(x)^2dF(x)$ can easily be replaced and solved by $\int udu=\frac12$ and $\int u^2du=\frac13$ but that convenience gets lost if the integrands are replaced by $F(\frac{x+c}{d})$ and $F(\frac{x+c}{d})^2$. $\endgroup$
    – drhab
    May 13, 2016 at 18:37

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