Standardising: the one curve behind them all
Every normal is a shifted, stretched copy of a single master curve. The
standard normal Z \sim N(0,1) has mean
0 and variance 1; its cumulative
distribution function is written \Phi(z) = \mathbb{P}(Z \le z) — the
area under the bell to the left of z. There is no elementary formula
for \Phi, which is exactly why it gets its own symbol (and why it
shows up inside the Black–Scholes formula).
To turn any X \sim N(\mu, \sigma^2) into a standard normal, subtract
the mean and divide by the standard deviation:
Z = \frac{X - \mu}{\sigma} \sim N(0,1).
This standardisation measures X in units of
\sigma away from the mean, so a single table of
\Phi answers questions about every normal. But why does
the recipe Z = (X-\mu)/\sigma land us back on the standard
bell, every single time? That is a claim about densities, and it deserves a real proof.
Standardisation, derived line by line
We will start from the density of X \sim N(\mu, \sigma^2),
introduce the standardised variable Z = (X-\mu)/\sigma, and chase
through the change of variables until the density of
Z emerges. The only tool is the change-of-variables rule for a
monotone transformation: if z = (x-\mu)/\sigma is the new variable,
then the density f_Z of Z relates to the
density f_X of X by
f_Z(z) = f_X(x)\,\left|\frac{dx}{dz}\right|,
where x is rewritten in terms of z. The
factor \lvert dx/dz\rvert is the Jacobian: it
rescales the density so that probability mass is conserved when we stretch the
x-axis into the z-axis. Let us assemble
each piece.
Step 1 — write the density of X. This is just the
definition we started with:
f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}\,\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).
Step 2 — invert the substitution. From
z = (x-\mu)/\sigma solve for x by
multiplying through by \sigma and adding
\mu:
x = \mu + \sigma z.
Step 3 — differentiate to get the Jacobian. Since
\mu and \sigma are constants,
differentiating x = \mu + \sigma z with respect to
z gives
\frac{dx}{dz} = \sigma, \qquad \text{so}\qquad \left|\frac{dx}{dz}\right| = \sigma
(the absolute value is harmless here because \sigma > 0). This is
the single \sigma that will do all the cancelling.
Step 4 — substitute x = \mu + \sigma z into the
exponent. The numerator (x-\mu)^2 becomes
(x-\mu)^2 = (\mu + \sigma z - \mu)^2 = (\sigma z)^2 = \sigma^2 z^2,
so the whole exponent simplifies, the \sigma^2 upstairs cancelling the \sigma^2 downstairs:
-\frac{(x-\mu)^2}{2\sigma^2} = -\frac{\sigma^2 z^2}{2\sigma^2} = -\frac{z^2}{2}.
Step 5 — assemble. Put the substituted density and the Jacobian into the
change-of-variables formula:
f_Z(z) = f_X(\mu + \sigma z)\,\left|\frac{dx}{dz}\right| = \frac{1}{\sigma\sqrt{2\pi}}\,\exp\!\left(-\frac{z^2}{2}\right)\cdot \sigma.
Step 6 — watch the \sigma cancel.
The \sigma from the Jacobian sits in the numerator; the
\sigma in 1/(\sigma\sqrt{2\pi}) sits in
the denominator. They divide out completely:
f_Z(z) = \frac{\sigma}{\sigma\sqrt{2\pi}}\,\exp\!\left(-\frac{z^2}{2}\right) = \frac{1}{\sqrt{2\pi}}\,\exp\!\left(-\frac{z^2}{2}\right).
That last expression is the standard normal density
\varphi(z) — no \mu, no
\sigma left anywhere. The shift by \mu and
the scale by \sigma have been completely absorbed, exactly as
promised. So Z \sim N(0,1), whatever
\mu and \sigma were.
If X \sim N(\mu, \sigma^2) and
Z = \dfrac{X-\mu}{\sigma}, then
Z \sim N(0,1); its density is the standard normal
\varphi(z) = \frac{1}{\sqrt{2\pi}}\,e^{-z^2/2}.
Consequently every normal probability reduces to one about Z:
\mathbb{P}(X \le x) = \Phi\!\left(\dfrac{x-\mu}{\sigma}\right).
For \varphi to be a genuine density it must integrate to
1, which forces the famous Gaussian integral
I = \int_{-\infty}^{\infty} e^{-x^2/2}\,dx = \sqrt{2\pi}.
There is no elementary antiderivative of e^{-x^2/2}, so we cannot
attack I head-on. The trick is to compute
I^2 instead, as a double integral, and switch to polar
coordinates. First write I^2 as a product of two identical
one-dimensional integrals, renaming the dummy in the second copy from
x to y:
I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2/2}\,dx\right)\!\left(\int_{-\infty}^{\infty} e^{-y^2/2}\,dy\right).
Because each factor is a constant with respect to the other's variable, the product of two
integrals is a single integral over the whole plane:
I^2 = \int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty} e^{-x^2/2}\,e^{-y^2/2}\,dx\,dy = \int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty} e^{-(x^2+y^2)/2}\,dx\,dy.
Now change to polar coordinates
x = r\cos\theta, y = r\sin\theta. Two
things happen: the exponent collapses because
x^2 + y^2 = r^2, and the area element picks up the polar Jacobian
dx\,dy = r\,dr\,d\theta. The whole plane is swept out by
r \in [0, \infty) and \theta \in [0, 2\pi):
I^2 = \int_{0}^{2\pi}\!\int_{0}^{\infty} e^{-r^2/2}\,r\,dr\,d\theta.
The extra factor of r is exactly what we needed — it makes the
radial integral elementary. Substitute u = r^2/2, so
du = r\,dr, turning the inner integral into
\int_{0}^{\infty} e^{-r^2/2}\,r\,dr = \int_{0}^{\infty} e^{-u}\,du = \left[-e^{-u}\right]_{0}^{\infty} = 0 - (-1) = 1.
The outer integral over \theta is then trivial:
I^2 = \int_{0}^{2\pi} 1\,d\theta = 2\pi.
Taking the positive square root (the integrand is positive, so
I > 0) gives
I = \sqrt{2\pi}. That is precisely the constant sitting under the
square root in the density: dividing by \sqrt{2\pi} is what makes
the total area equal to 1.
The empirical rule is not folklore — it is three values of
\Phi dressed up. Because the standard normal is symmetric about
0, the probability of landing within
k standard deviations of the mean is the area between
-k and +k:
\mathbb{P}(-k \le Z \le k) = \Phi(k) - \Phi(-k) = \Phi(k) - \bigl(1 - \Phi(k)\bigr) = 2\Phi(k) - 1,
where we used the reflection identity
\Phi(-k) = 1 - \Phi(k) (the area to the left of
-k equals the area to the right of +k).
Now read three values of \Phi off a table:
-
k=1: \Phi(1) \approx 0.8413, so
2(0.8413) - 1 = 0.6826 \approx 68\%.
-
k=2: \Phi(2) \approx 0.9772, so
2(0.9772) - 1 = 0.9544 \approx 95\%.
-
k=3: \Phi(3) \approx 0.9987, so
2(0.9987) - 1 = 0.9974 \approx 99.7\%.
By standardisation these percentages transfer to any
N(\mu,\sigma^2): "within k\sigma of
\mu" for X is the same event as "within
k of 0" for Z.