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Basically, I am told that

$\varepsilon$~$N(0,1)$, and

$\omega$~$IG(\frac{v}{2}$,$\frac{v}{2})$ where $IG$ is the inverted gamma distribution

Now, I am told that the distribution of:

$\varepsilon(\frac{v-2}{2} \omega )^\frac{1}{2}$

is Student-t distributed. But I can not figure out why.

When inputting the code into mathematica I get the following PDF:

$\frac{(1+\frac{x^2}{v-2})^{-\frac{v+1}{2}}.\Gamma(\frac{v+1}{2})}{\pi^{\frac{1}{2}}.(v-2)^{\frac{1}{2}}.\Gamma(\frac{v}{2})}$ for $v>2$

Of course it looks very similar to the PDF of the student-t distribution but sometimes the $v$ is replaced with $v-2$.

Can I nevertheless conclude that it is student-t distributed and if so, with how many degrees of freedom, $v$ of $v-2$?

JSM77
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1 Answers1

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Ignoring multiplicative constants and assuming you have not made any errors, the key term you have is $(1+\frac{x^2}{v-2})^{-\frac{v+1}{2}}$, while in a $t$-distribution it would be $(1+\frac{x^2}{\nu})^{-\frac{\nu+1}{2}}$. So this is not a $t$-distribution as such.

But it is a scaled $t$-distribution: let $y = \sqrt{\frac{v}{v-2}}x$ and you would have $(1+\frac{y^2}{v})^{-\frac{v+1}{2}}$ which is proportional to the pdf of a $t$-distribution with $v$ degrees of freedom

Henry
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