# Derive cauchy distribution as a scale mixture of normal distributions

I doing Bayesian modelling these days. I found that cauchy distribution can be written as a scale mixture of normal based on following source. Link

So I started to derive this. Somehow, I am not getting what I want. I am trying to prove the third implementation in the above source.

It says :

$$Cauchy(x|0,1)=\int N(x|0,\tau^{-1/2})InvGamma(\tau|1/2,1/2)d\tau$$

Or in other words $$X=x_a*\sqrt{x_b}$$ follows a cauchy(0,1) if $$x_a \sim N(0,1)$$ and $$x_b \sim invGamma(1/2,1/2)$$

So here is my working :

$$N(x|0,\tau^{-1/2}) \propto \tau \times e^{x^2\tau/2}$$ and $$IG(\tau|1/2,1/2) \propto \tau^{1/2-1} \times e^{-2\tau}$$.

So the integral,

$$\int N(x|0,\tau^{-1/2})$$ $$IG(\tau|1/2,1/2)d\tau \propto \int\tau^{1/2} \times e^{-2\tau(x^2+1)} d\tau$$

$$\propto 1/(x^2+1)^{3/2}$$. (by creating the inverse gamma integral inside)

Ideally, I should get 1 as the exponent of the $$x^2$$+1 . What am I missing here ?

Can anyone help me to figure this out ?

Thank you.

• Shouldn't the leading term in your Normal be $t^{1/2}$ (reciprocal sd rather than reciprocal variance)? – Thomas Lumley Oct 4 '20 at 1:25
• thank you. I missed that. – student_R123 Oct 4 '20 at 1:36

If you just want to prove that it's a scale mixture of Normals, it's easier to work directly. The Cauchy distribution is the $$t_1$$ distribution.
The $$t_1$$ distribution is a Normal divided by the square root of an independent chi-squared. So, let $$Z\sim N(0,1)$$ and $$S^2\sim \sigma^2\chi^2_1$$. The distribution of $$Z/S$$ is Cauchy.
But that's equivalently an $$N(0,T^2)$$ conditional on $$T=1/S$$, so it's a scale mixture of Normals.