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If $X$ follows a normal distribution with parameters $\mu$ and $\sigma^2$ show that $Z = (X- \mu)/\sigma$ follows a standard normal distribution. This doesn't seem to intuitive to me. We shift $X$ so that the top of the bell curve is over $0$, that bit makes sense. But then we scale the curve by a scalar, wouldn't this change the area under the curve, making it no longer a pdf since if we integrate $Z$ over the real line we need to get $1$? I guess my intiution is failing because I'm trying to imagine area along an infinite line.

Now I would like to go about proving this using the definition of the normal distribution. The definition I'm working with is $$f(x,\mu,\sigma) = \frac{1}{\sigma \sqrt{2 \pi}}e^{\frac{-(x- \mu)^2}{2 \sigma^2}}$$

I thought the proof would be showing that $$f(\frac{x-\mu}{\sigma},\mu,\sigma) = \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}$$ But this didn't work, I got an ugly expression in the exponent and I couldn't see why the $\sigma$ in the denominator would disappear at the front of the expression.

How Should I go about proving this is a fundamental way? I think their is probably an easier proof using properties such as how adding and multiplying normal distribution's by scalars modifies it. But what I'm really after is a proof using the normal distribution definition

Thanks in advance.

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    $\begingroup$ The CDF for $Z$ is $P(Z \le z) = P(X \le z\sigma + \mu)$. Now, the density is the derivative of the CDF, so you are missing a factor of $\sigma$, (and you got the substitution the wrong way round :)). I.e. it is $(d/dz)P(X \le z\sigma + \mu) = \sigma f(z\sigma + \mu)$, which equals the standard normal density. BTW, I added the "self-study" tag, as this is a textbook problem. $\endgroup$ – P.Windridge Apr 13 '15 at 12:52
  • $\begingroup$ Thanks for the help @P.Windridge I think I've gotten to the last line, I am just not entirely sure why it equals the standard normal density! Yes it is, thanks for that $\endgroup$ – HBeel Apr 13 '15 at 13:24
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Write $\displaystyle P\{Z \leq a\} = P\left\{\frac{X-\mu}{\sigma} \leq a\right\} = P\{X\leq \mu+a\sigma\}=\int_{-\infty}^{\mu+a\sigma}f_X(x)\,\mathrm dx$ and then make a change of variable in the integral, setting $z=\frac{x-\mu}{\sigma}$, hopefully not forgetting to change the upper limit appropriately. DO NOT attempt to actually integrate: just do a change of variables, draw a box around the integral that you have obtained, and admire the contents of the box.

After the admiration is over, argue that the formula $\displaystyle P\{Z \leq a\} = F_Z(a) = \int_{-\infty}^a f_Z(z)\,\mathrm dz$ allows us to conclude something about the density function $f_Z(z)$ via the admirable contents of the box you just drew.

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  • $\begingroup$ So we have $\mathbb{P}(Z \leq a) = \int\limits_{-\infty}^{\mu + a \sigma} f_X(x) dx$. If we make the substitution you suggest I find $dx = \sigma dz$ and then the integral becomes $\sigma \int\limits_{-\infty}^{a} f_X(\sigma z + \mu) dz$. From this and the definition of $f_Z(z)$ we have that $\sigma \int\limits_{-\infty}^{a} f_X(\sigma z + \mu) dz = \int\limits_{-\infty}^a f_Z(z)dz$. Differentiation both sides I find $\sigma f_X(a \sigma + \mu) = f_Z(a)$. I feel like this isn't quite right, and that I'm going in circles :(! $\endgroup$ – HBeel Apr 13 '15 at 13:21
  • $\begingroup$ Oh I just use the definition of $f_X$ and $\sigma f_X(a \sigma + \mu)$ comes out as the definition of the standard normal distribution as posed in my question. Thanks for your help @Dilip! Also there is a small latex error in your answer that I'm not allowed to edit because the edit is so small $\endgroup$ – HBeel Apr 13 '15 at 13:27
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as I see you are thinking the area is divided by $$\sigma(standard deviation)$$ .....because Random Variable is divided by $\sigma$, but $\sigma$ is also multiplied by all Probability function so the area remains same.

Now, for you second query disappearance of $\sigma$ as you Put $$X=Z\sigma+\mu$$ you must put $$dx=\sigma*dz$$(because you changed the Random Variable)
so the $\sigma$ canceled out.

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