Let $T = X(Y/n)^{-1/2}$ with $X \sim N \left(0,1 \right)$ and $Y \sim Gamma \left(\frac{n}{2},\frac{1}{2}\right)$ with $ n \geq 3$. The Gamma and Normal distributions are independent.

Gamma using the following parametrization: $f(x;k,\theta) = \frac{\theta^{k}x^{k-1}e^{-\theta x}}{\Gamma (k)}$

I'm trying to find the expected value, variance and probability distribution of T, but I'm confused.

Should I be using the product distribution formula (https://en.wikipedia.org/wiki/Product_distribution) in this case?

Also, is the expected value just 0, as, due to independence $\mathbb{E}[X] = 0$ and $\mathbb{E}[T] = \mathbb{E}[X]\cdot\mathbb{E}[Y]$?

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    $\begingroup$ Yes, your reasoning about the mean is correct. You can find the variance similarly -- remember the formula for variance in terms of expectations and use the same independence result. And yes, the convolution formula in the linked article is one way to derive the pdf. $\endgroup$ – djs Oct 21 '16 at 14:11
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    $\begingroup$ What approach you use should depend on what you know. One clever approach is suggested by the question at stats.stackexchange.com/questions/52906. Note that the expectation will indeed be zero, according to your correct argument, provided the expectation exists. It does not exist when $n \le 1$. This is one of the reasons your problem restricts $n$ to be relatively large--but it does not free you from the need to demonstrate the expectation exists when $n\ge 3$. (In fact, $n\gt 2$ suffices.) $\endgroup$ – whuber Oct 21 '16 at 14:11

To avoid possible confusion, I note that the OP is using the "shape-rate" parametrization for the Gamma distribution (even though the letters $k,\theta$ are more often used to denote the shape-scale parametrization).

For the given parametrization, we have that

$$Y \sim Gamma \left(\frac{n}{2},\frac{1}{2}\right) = \chi^2(n)$$

i.e. it is a chi-square distribution with $n$ degrees of freedom.

Then we have that

$$T = \frac {X}{\sqrt{Y/n}}$$

is a random variable that is the ratio of a standard normal over a chi-distribution divided by its degrees of freedom. This ratio is the definition of a very well-known distribution (if you don't recognize it, you can see a detailed derivation here, that uses the ratio distribution formula), and that has $n$ degrees of freedom. And as @whuber noted, we need $n>1$ for the mean to exist, and $n>2$ for the variance to exist.

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