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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.

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

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