# What is the purpose of $\frac{1}{\sigma \sqrt{2 \pi}}$ in $\frac{1}{\sigma \sqrt{2 \pi}}e^{\frac{(-(x - \mu ))^2}{2\sigma ^2}}$?

I have been studying the probability density function...

$$\frac{1}{\sigma \sqrt{2 \pi}}e^{\frac{(-(x - \mu ))^2}{2\sigma ^2}}$$

For now I remove the constant, and using the following proof, I prove that...

$$\int_{-\infty}^{\infty}e^{\frac{-x^2}{2}} = \sqrt{2 \pi }$$

The way I interpret this is that the area under the gaussian distribution is $$\sqrt{2 \pi }$$. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $$\sigma$$ as well. Why is this done?

• so the integral of the probability density function over the entire space is equal to one – J. W. Tanner Apr 10 '19 at 1:53

If you consider every possible outcome of some event you should expect the probability of it happening to be $$1$$, not $$\sqrt{2\pi}$$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
• (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is. – John Doe Apr 10 '19 at 2:02
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $$\sigma>1$$ you are decreasing your area by a factor $$\sigma$$ but you are increasing it by the same factor because you replace $$x$$ by $$x/\sigma$$ (the shift does not change the area of course)
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(x\mid\mu,\sigma^2)=\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac12\left(\frac{x-\mu}\sigma\right)^2\right)$$ where the parameter space is $$\mathit\Theta=\{(\mu,\sigma^2)\in\Bbb R^2:\sigma^2>0\}$$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.
Now consider the simple case where $$\mu=0$$ and $$\sigma^2=1$$. Then the standard normal distribution has p.d.f. $$f(x)=\frac1{\sqrt{2\pi}}\exp\left(-\frac12x^2\right).$$ If we integrate this in the interval $$(-\infty,\infty)$$, we will get $$1$$. This is by definition always the case as for all $$x\in\mathit X$$ (in this instance $$\mathit X=\Bbb R$$), $$\int_{\mathit X}f(x)\,dx=1.$$ That is, the sum of all the probabilities of $$x$$ being in each region in $$\mathit X$$ is $$1$$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.
A further example is the Beta distribution, with p.d.f. $$f(x\mid\alpha,\beta)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{\text B(\alpha,\beta)}$$ where $$1/\text B(\alpha,\beta)$$ is the normalising constant.