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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#### Best Answer

Write $displaystyle P{Z leq a} = Pleft{frac{X-mu}{sigma} leq aright} = P{Xleq mu+asigma}=int_{-infty}^{mu+asigma}f_X(x),mathrm dx$ 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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