# If $X \sim N(\mu, \sigma ^2)$, show that $(X - \mu) / \sigma \sim N(0,1)$ [closed]

I don't know how to do this. Do I need to use converge in distribution? (I thought this can only been used if $n$ involves)

## closed as off-topic by saz, Did, user223391, Paramanand Singh, Claude LeiboviciJul 10 '15 at 7:47

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• What have you studied so far? Which topics/level etc ... the answer to these questions would help people here to provide you with an appropriate answer. – Math-fun Jul 9 '15 at 20:01
• @Math-fun, so far, random variable, joint distribution, expected values, now focus on distribution derived from the normal distribution. Very basic beginning to stats – ElleryL Jul 9 '15 at 20:06

Okay, so we know that $X\sim N(\mu,\sigma)$, looks like this: $$Pr(X<x) = F_X(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt$$ Next observe that $$Pr(\frac{X-\mu}{\sigma}<x)=Pr(X<x\sigma+\mu)=\int_{-\infty}^{x\sigma+\mu}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt$$ We then let $y = \frac{t-\mu}{\sigma}$, so $dy = \frac{dt}{\sigma}$ and $y(x\sigma+\mu) = \frac{x\sigma+\mu-\mu}{\sigma}=x$, while $y(-\infty)=-\infty$.

Hence $$Pr(\frac{X-\mu}{\sigma}<x)=Pr(X<x\sigma+\mu)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-y^2}dy$$ Which is the distribution of $N(0,1)$, so $\frac{X-\mu}{\sigma}\sim N(0,1)$

Note that this is a very simple solution, there are much more insightful proofs that you can do, I just feel that for your level this should be sufficient. As practice try and do the same with the pdf, rather than the cdf as I did. $\ddot\smile$

• thanks you ! It just meets my level... ;) – ElleryL Jul 10 '15 at 1:21

The transformations in $\frac {X -\mu}{\sigma}$ are just translation and scaling, i.e., this is a change of coordinate from $X$ to $\frac {X-\mu}{\sigma}$ given by the function $X'=\phi(X)=\frac {X-\mu}{\sigma}$. More formally, you can use the Jacobian to study the effect of the change of variables in the density function:

You are making a (linear) change of variable in $X$, given by $X':=\frac{X-\mu}{\sigma}$, or in the notation of the link:

$X':=\phi(X)= \frac {X-\mu}{\sigma}$ , so that $\phi^{-1}=\sigma X'+\mu$, and

then $p(X')= p(\phi^{-1}(X))d\phi^{-1}(X)$.

Just write out the explicit formula for the gaussian (https://en.wikipedia.org/wiki/Normal_distribution)

and then perform the transformation. The algebra shouldn't be difficult.

• Actually, I think this just shows that the curve is transformed the way you want it do. However, I am not sure about the "~" for the random variable X. Perhaps this answer is not rigorous, but atleast from the perspective of what the distribution looks like it should be fine. – Andy Yao Jul 9 '15 at 20:25

One straightforward argument is to calculate the moment generating function of $(X - \mu) / \sigma$,

\begin{align} M_{(X - \mu) / \sigma}(t) &= M_{X - \mu}(t / \sigma) \\ &= M_{X}(t / \sigma) M_{-\mu} (t / \sigma) \\ &= e^{t \mu / \sigma + t^2 / 2} e^{ - t \mu / \sigma} \\ &= e^{t^2 / 2} , \end{align}

which is the standard normal m.g.f.