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Be $X\sim N(\mu,1)$ and $Y\sim Inverse-Gamma(\alpha,\beta)$. For the Inverse-Gamma, I usually use the parameterization which leads to the following probability distribution function for Y:
$f(y;\alpha,\beta)=\frac{\beta^{\alpha}}{\Gamma (\alpha)}(\frac{1}{x})^{\alpha+1}e^{-\frac{\beta}{x}}$

I need to find the distribution of $T=X\sqrt{Y}$.
According to my calculations, T is not a non-central Student's t-distribution , it is a non-standardized Student's t instead with $2\alpha$ degrees of freedom, location parameter $\mu$ and a scale parameter $\sqrt{\frac{\beta}{\alpha}}$.
Is it correct?
Thank you.

EDIT: These are my calculations: enter image description here enter image description here

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    $\begingroup$ As what appears to be fairly routine book-work, this question should carry the self-study tag (see its tag wiki for the special way such questions may be dealt with). Since you can only have 5 tags, I'd suggest replacing the distributions tag. Perhaps you could (i) explain why you expect it to be non-central-t (since you raise that possibility), and (ii) show your work (i.e. why you now think it isn't), so that we can give more specific guidance. $\endgroup$
    – Glen_b
    Commented Jan 13, 2015 at 17:15
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    $\begingroup$ We can't really give 'yes' or 'no' - or even one sentence - as answers, so it would also help in that way if you could give details we could respond to. But as far as that goes, I think it should be non-central t $\endgroup$
    – Glen_b
    Commented Jan 13, 2015 at 17:18
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    $\begingroup$ There are lots of competing definitions as to what an InverseGamma(a,b) distribution is or isn't ... so the question is not defined until you specify the functional form of the pdf you are assuming. $\endgroup$
    – wolfies
    Commented Jan 13, 2015 at 17:23
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    $\begingroup$ Check your calculations: @Glen_b is correct. $\endgroup$
    – whuber ♦
    Commented Jan 13, 2015 at 18:34
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    $\begingroup$ The distribution of your T will not generally be symmetrical. By contrast, the Student's t pdf is symmetrical. Therefore, your T is not distributed as Student's t. $\endgroup$
    – wolfies
    Commented Jan 13, 2015 at 18:53

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You have a term:

$$\frac{X}{\sqrt{\frac{\chi^2_{2\alpha}}{2\alpha}}}$$

which you then say is $\sim$ student-$t(2\alpha)$

But for that to be the case, $X$ would need to have mean 0.

When it has mean other than 0, you have non-central $t$.

Call the denominator $R$ and consider writing it as

$\qquad\frac{X-1+1}{R} = \frac{X-1}{R}+\frac{1}{R}$

The left hand term is a (central) t-distribution, while the other term is right skew (and the two terms are dependent because the variable on the denominator is shared).

Informally, it seems like the sum of the symmetric and the right-skew term might be right skewed. It turns out to be the case.

Note that having $X$ with mean other than zero does not result in the same distribution as a location-shifted $t$

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  • $\begingroup$ You are right. I didn't notice it. So $\mu$ is the non-centrality parameter. The problem is that now I can't lead back the formula to the general definition of a non-central Student's t-distribution as $T=\frac{Z+\mu}{\sqrt{\frac{V}{v}}}$ where $Z$ is a standard normal and $V$ is a Chis-squared with $v$ degrees of freedom, according to this definition. Where do I put $\sqrt{\frac{\beta}{\alpha}}$? Is it a generalization of the non-central student's t? Thank you. $\endgroup$
    – John M
    Commented Jan 13, 2015 at 19:40
  • $\begingroup$ Yes, I think you have a scaled non-central t. $\endgroup$
    – Glen_b
    Commented Jan 13, 2015 at 19:42

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