# Box-Muller transform

1. Feb 3, 2016

### rabbed

In derivation of the box-muller transform, the joint distribution p(x,y) = e^(-r^2/2)/(2*pi) is interpreted as the product of a uniform distribution 1/(2*pi) and an exponential distribution e^(-x/2), but isn't an exponential distribution defined as k*e^(-k*x)? What happened to the coefficient?

2. Feb 3, 2016

### mathman

You are missing a constant.

3. Feb 3, 2016

### mathman

derivation: Start with $\frac{1}{2\pi}e^{\frac{-x^2-y^2}{2}}dxdy$. Change to polar coordinates. $\frac{1}{2\pi}e^{\frac{-r^2}{2}} rdrd\theta$. For you picture $s=r^2,\ so\ ds=2rdr,\ or\ rdr=\frac{ds}{2}$. There's the coefficient.

4. Feb 4, 2016

### rabbed

Thanks mathman :)