Berger and Casella's Statistical Inference shows that given a random variable X ~ N($\mu, \sigma^2$), a random variable Z = (X - $\mu$) / $\sigma$ is distributed standard normal N(0, 1) by comparing expressions for the cumulative distribution functions. They say this fact is "easily established by writing":

$$ \begin{align} P(Z \le z) &= P \left(\frac{X-\mu}{\sigma} \le z\right) \\ &=P(X \le z\sigma + \mu) \\ &=\frac{1}{\sqrt{2\pi}\sigma}\int^{z\sigma+\mu}_{-\infty}e^{-(x-\mu)^2/2\sigma^2}dx \\ &=\frac{1}{\sqrt{2\pi}}\int^{z}_{-\infty}e^{-t^2/2}dx\qquad \left(substitute\, t = \frac{x - \mu}{\sigma}\right) \end{align} $$

Can somebody please expand on the last step here? The transition to $-t^2/2$ in the exponent is a clear consequence of substituting $\frac{x-\mu}{\sigma}$ for $t$, but how does $z\sigma+\mu$ tranform to $z$ in the upper limit, and where does the $\sigma$ in the constant go? I think I'm missing a straightforward piece of calculus, or something even simpler ...


1 Answer 1


how does zσ+μ tranform to z in the upper limit

If x=zσ+μ, what's t?

That's how substitution always works with limits; the limit is now in terms of the new variable.

where does the σ in the constant go? I think I'm missing a straightforward piece of calculus

You seem to be forgetting the Jacobian. What's "dx" in terms of "dt"?

It might pay to go back and revisit the basics of transformation in integrals.


  • $\begingroup$ Indeed, I was just forgetting my substitution rules! $\endgroup$
    – tef2128
    Jul 8, 2021 at 14:05

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