I am failing to see where the $\sigma$ is going in the below.

Given that normal distribution's pdf is:

$$p(x) = \frac{1}{\sigma \sqrt{2 \pi}}\exp \left( -\frac{(x-\mu)^2}{2 \sigma^2} \right)$$

and we set $\phi = \frac{x-\mu}{\sigma}$.

I obtain:

$$\Phi(\phi) = \frac{1}{\sigma \sqrt{2 \pi}}\exp\left( -\frac{1}{2}\phi^2 \right)$$

and yet, the correct result it appears, is:

$$\Phi(\phi) = \frac{1}{\sqrt{2 \pi}}\exp\left( -\frac{1}{2}\phi^2 \right).$$

Is there something there going on with the fact that $\sigma$ is $1$ for normal standard distribution, and they imply that fact in the equation?

And then a bonus question, how can you use just the $\phi$ to show that expected value and variance is 0 and 1, without using the distribution. I know that $\mathbb{E}[X]=\int_{\mathbb{R}}x f(x) dx$, but it is shown in my notes that mean of standard normal distribution is zero by simply using this: $\mathbb E\left[\frac{X-\mu}{\sigma}\right]$. I fail to understand this step.

  • 1
    $\begingroup$ Let $X \sim Unif(0, 1)$ with density $f_X(x) = 1$ for $x \in (0,1)$. Then let $Y = 2X \sim Unif(0, 1)$ with density $f_Y(y) = 1/2$ for $y \in (0,2).$ Are you shocked that the PDF is changed by a constant in the second distribution? $\endgroup$
    – BruceET
    Dec 22, 2015 at 1:43

1 Answer 1


I'm going to change your notation a bit. Let $X\sim N(\mu, \sigma^2)$, and let $$Y = \frac{X-\mu}{\sigma}.$$ Since this is one-to-one over the support, we can apply a one-to-one change of variable. Thus $X = \sigma Y+\mu$, and $$f_Y(y) = \frac{f_X(\sigma y+\mu)}{\left|\frac{dy}{dx}\right|} = \sigma\cdot\frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{1}{2}\left(\frac{\sigma y+\mu-\mu}{\sigma}\right)^2\right\} = \frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}y^2\right\}$$ which is what you are seeking. It is also possible to integrate $P(Y<y)$, but you can do that on your own.

Notice that this is the density of a a standard normal. Thus, we know that $$E[Y] = 0$$ and $$\text{Var}[Y] = 1.$$

In other words we have just shown that if $X\sim N(\mu, \sigma^2)$, then the transformation $$\frac{X-\mu}{\sigma}$$ yields a standard normal distribution $N(0,1)$. That is why $$E\left[\frac{X-\mu}{\sigma}\right] = 0,$$ immediately.

  • 1
    $\begingroup$ Perhaps it is helpful to give a little more explanation about the formula in the forth line: $f_Y(y)=f_X(\sigma y+\mu)/|dy/dx|$, especially about the origin of the factor $|dy/dx|$. In short, this factor comes from the normalization of $\Phi(\phi)$, that is, $\int \mathrm{d}\phi \Phi(\phi)=\int \mathrm{d}x p(x)=1$. $\endgroup$ Dec 21, 2015 at 16:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .