# What distribution is this? [duplicate]

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 $$v-2$$?

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