The Normal Density

Suppose $X \sim N(\mu, \sigma^2)$ , so $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ . Write down the density $f_X(x)$ . Where does the constant factor in front come from?

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#The density

f_X(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right). \tag{1}

#The constant is a normaliser

A density must integrate to $1$ , so the factor in front is whatever forces that, the reciprocal of the integral of the bare exponential shape. Call that shape integral

Z = \int_{-\infty}^{\infty} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) dx. \tag{2}

The substitution $t = (x-\mu)/\sigma$ gives $dx = \sigma\, dt$ and strips out the parameters,

Z = \sigma \int_{-\infty}^{\infty} e^{-t^2/2}\, dt. \tag{3}

#Evaluating the Gaussian integral

Let $I = \int_{-\infty}^{\infty} e^{-t^2/2}\, dt$ . Squaring turns one integral into a plane integral that polar coordinates collapse,

I^2 = \int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty} e^{-(u^2+v^2)/2}\, du\, dv = \int_0^{2\pi}\!\!\int_0^{\infty} e^{-r^2/2}\, r\, dr\, d\theta. \tag{4}

The radial piece is $\int_0^{\infty} e^{-r^2/2}\, r\, dr = 1$ , so $I^2 = 2\pi$ and $I = \sqrt{2\pi}$ .

#Normalising the Gaussian integral

Therefore $Z = \sigma\sqrt{2\pi}$ , and the normalising constant is its reciprocal,

\frac{1}{Z} = \frac{1}{\sigma\sqrt{2\pi}}, \tag{5}

the exact factor standing in front of $f_X$ .

The exponential alone fixes the bell shape, peaking at the mean and falling off over a width of one standard deviation. The leading constant scales it so the shaded area is exactly 1, which is what makes the curve a probability density.