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Using the following expression where $f(x)$ is the PDF of a normal distribution with a mean $\mu$ and standard deviation $\sigma$, and $F$ is the CDF of another normal distribution with mean of $0$, but the same standard deviation:

$$ \int_{-\infty}^{\infty}f(x|\mu,\sigma)(F(x|\mu=0, \sigma)^V)dx $$

$V$ is always an integer $\ge 1$. I've noticed from simulations that the results are always the same whenever the ratio $\mu/\sigma$ is the same. In other words:

$$ \int_{-\infty}^{\infty}f(x|\mu=\mu_a,\sigma=\sigma_a)(F(x|\mu=0, \sigma=\sigma_a)^V)dx = \int_{-\infty}^{\infty}f(x|\mu=\mu_b,\sigma=\sigma_b)(F(x|\mu=0, \sigma=\sigma_b)^V)dx $$

whenever $\frac{\mu_a}{\sigma_a} = \frac{\mu_b}{\sigma_b}$, for any $V\geq 1$.

It is not clear to me why this should be true, or how to show it. Is there a way to prove this relationship?

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  • $\begingroup$ Did you find that this holds for any $\sigma_a,\sigma_b$ when $\mu_a=\mu_b=0$? $\endgroup$
    – Annika
    Aug 5, 2021 at 1:15
  • $\begingroup$ @Bey Yes, still holds $\endgroup$
    – user144552
    Aug 5, 2021 at 13:18

1 Answer 1

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Let $\phi$ and $\Phi$ be the standard normal pdf and cdf. When the mean is $\mu$ and the standard deviation is $\sigma$ the pdf $f(x)$ and cdf $F(x)$ are $$ f(x)=\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right),\quad\quad F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right). $$ By a change of variable it is obvious that the integral \begin{eqnarray} \frac{1}{\sigma}\int_{-\infty}^{+\infty}\phi\left(\frac{x-\mu}{\sigma}\right)\Phi^V\left(\frac{x}{\sigma}\right)\,dx =\int_{-\infty}^{+\infty}\phi\left(x-\frac{\mu}{\sigma}\right)\Phi^V(x)\,dx,\quad\quad V\ge 1, \end{eqnarray} only depends on $\mu/\sigma$. The same holds for the integral \begin{eqnarray} \frac{1}{\sigma}\int_{-\infty}^{+\infty}\phi\left(\frac{x-\mu}{\sigma}\right)\Phi^V\left(\frac{x-\mu}{\sigma}\right)\,dx =\int_{-\infty}^{+\infty}\phi\left(x-\frac{\mu}{\sigma}\right)\Phi^V\left(x-\frac{\mu}{\sigma}\right)\,dx. \end{eqnarray}

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  • $\begingroup$ Very elegant answer +1 $\endgroup$
    – Annika
    Aug 5, 2021 at 13:22

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