# Kullback Leibler divergence between normal and Cauchy distributions? [duplicate]

Let $$\phi(x)$$ be the standard normal probability density function and $$f(x)$$ be the Cauchy probability density function.

How can I calculate the Kullback-Leibler divergences between $$\phi$$ and $$f$$. This is

$$KL1 = \int f(x) \log \frac{f(x)}{\phi(x)}dx,$$ and $$KL2 = \int \phi(x) \log \frac{\phi(x)}{f(x)}dx?$$

Numerically, I have found that $$KL2=0.259245$$ and $$KL1 = \infty$$. Is this correct? Is $$KL2<\infty$$ and $$KL1 = \infty$$?

• The limiting value of the integrand of $KL1$ as $x\to \pm \infty$ is nonzero, because up to a normalizing constant of $1/\pi$ it behaves like $x^{-2}\log(x^{-2}/\exp(-x^2/2))=x^{-2}(-2\log(x) + x^2/2)\approx 1/2.$ Thus its integral must diverge. The integrand of $KL2$ is a multiple of $\phi,$ which goes to zero so rapidly its integral must be small in size, and certainly finite.
– whuber
Feb 5, 2022 at 0:48
• This is answered by user Xi'an at stats.stackexchange.com/questions/351947/… and also used as an example at stats.stackexchange.com/questions/188903/… Feb 5, 2022 at 0:49