next up previous index
Next: Expectation and Variance 11/6 Up: Normal Random Variables 11/4 Previous: Normal Random Variable

How was the constant $\frac{1}{\sqrt{2\pi}}$ found?

We consider points distributed on the plane, with two independent standard Normal marginals, and we ask, what is the distribution of the distance to the origin?

We suppose we don't know the constant of integration in the Normal density, call it c. As the coordinates are independent the joint density distribution is the product of the marginal densities:

\begin{displaymath}f(x,y)=c^2e^{-\frac{1}{2}(x^2+y^2)}
\end{displaymath}

First we note that this is the same value, whatever the point (x,y), as long as it is at the same distance from the origin, call this distance $r=\sqrt{x^2+y^2}$.

What is the density distribution of the distance to the center, for a random point taken from this joint density? Here is its picture in 3 d:


 

\begin{displaymath}P(R\in dr)\approx f(r) dr \approx 2\pi r dr c^2e^{-\frac{1}{2}r^2}\end{displaymath}

From which we get :

\begin{displaymath}f(r)= 2\pi r c^2e^{-\frac{1}{2}r^2}\end{displaymath}

this has to integrate to one:

\begin{displaymath}1=\int_0^{+\infty} 2\pi r c^2 e^{-\frac{1}{2}r^2} dr =-2\pi c...
...\infty}_0 =2\pi c^2=1\qquad
\mbox{ so } c=\frac{1}{\sqrt{2\pi}}\end{displaymath}



Susan Holmes
1998-12-07