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Let $X$ be a standard normal random variable with cumulative distribtuion function $\Phi$. That is, $X\sim N(0,1)$ and the standard normal cdf is denoted by $\Phi(x)$. We know that $\Phi(X) \sim \operatorname{Uniform}(0,1)$. But can we say anything about the distribution of $\Phi(a+bX)$ where $a$ and $b>0$ are arbitrary real numbers? Can we express the cdf of $\Phi(a+bX)$ using other (known) functions?

Michael Hardy
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kbandesz
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1 Answers1

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Same basic way you show $\Phi(X)$ is $U(0,1)$ or more generally calculate the CDF of a monotonic function of a random variable whose CDF you know.

Let $Y = \Phi(a+bX)$ and calculate $$ P(Y\le y) \\= P(\Phi(a+ bX) \le y) \\= P(a+bX\le \Phi^{-1}(y) )\\=P\left(X\le \frac{\Phi^{-1}(y) -a}{b}\right) \\=\Phi\left(\frac{\Phi^{-1}(y) -a}{b}\right)$$ (assuming $b>0$). If $a=0$ and $b=1$, this gives $y$ which is the CDF of the uniform distribution.

spaceisdarkgreen
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  • Thanks a lot, I accepted the answer as it perfectly addresses the question the way I wrote it. But I wasn't clear enough, unfortunately. What I wanted to ask is whether the distribution of this transformed variable coincides with any specific "known" distribution. I know this is not well-defined, but I guess what i mean is if we can express either the cdf or pdf in closed form. – kbandesz Aug 14 '17 at 11:30
  • I'd consider the above cdf to be closed form, but I don't believe it corresponds to any standard distribution. – spaceisdarkgreen Aug 14 '17 at 13:46