# Why should integrating the product of PDF and CDF have the same result given the same mean to standard deviation ratio?

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?

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

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 $$\begin{eqnarray} \frac{1}{\sigma}\int_{-\infty}^{+\infty}\phi\left(\frac{x-\mu}{\sigma}\right)\Phi^V\left(\frac{x-\mu}{\sigma}\right)\,dx =\int_{-\infty}^{+\infty}\phi\left(x-\frac{\mu}{\sigma}\right)\Phi^V\left(x-\frac{\mu}{\sigma}\right)\,dx. \end{eqnarray}$$