# A normal divided by $\sqrt{\chi^2(s)/s}$

If $Z \sim N(0,1)$ and $X \sim \chi^2(s)$. If $Z$ and $X$ are independently distributed then the variable $Y = \frac{Z}{\sqrt{X/s}}$ follows a $t$ distribution with degrees of freedom $s$.

What happens if $Z \sim N(a,b)$?

Does it simply become a location scale t distribution with mean = a, s.d. = $\sqrt{b}$ still with s degrees of freedom?

If $Z \sim \mathcal N(a, b)$, then note that $Z' = \frac{Z - a}{\sqrt b} \sim \mathcal N(0, 1)$. Then $Y = \frac{Z'}{\sqrt{X / s}} \sim \mathcal T(s)$.
Now, your proposed location-scale t distribution is $\sqrt b Y + a$. But that is $$\sqrt b Y + a = \frac{\sqrt b Z'}{\sqrt{X / s}} + a = \frac{\sqrt b Z' + a \sqrt{X / s}}{\sqrt{X / s}} \ne \frac{\sqrt b Z' + a}{\sqrt{X / s}} = \frac{Z}{\sqrt{X / s}}.$$
The variable that you care about is instead a scaled noncentral t-distribution, with $s$ degrees of freedom, noncentrality parameter $a / \sqrt b$, and then scaled by $\sqrt b$: $$\frac{Z}{\sqrt{X / s}} = \frac{\sqrt b Z' + a}{\sqrt{X / s}} = \sqrt{b} \left( \frac{Z' + a/\sqrt{b}}{\sqrt{X / s}} \right) .$$
• Apologies if I'm misunderstanding, but your stated location-scale t distribution is $\sqrt b Y + \frac{a}{\sqrt b}$. But that is $$\sqrt b Y + \frac{a}{\sqrt b} = \frac{\sqrt b Z'}{\sqrt{X / s}} + \frac{a}{\sqrt b} = \frac{\sqrt b Z' + \frac{a}{\sqrt b} \sqrt{X / s}}{\sqrt{X / s}} \ne \frac{\sqrt b Z' + a}{\sqrt{X / s}} = \frac{Z}{\sqrt{X / s}}.$$ Im sure I am making an error, any help would be massively appreciated Commented Oct 31, 2016 at 15:24
• @CharlesPrice That's my point: $\frac{Z}{\sqrt{X / s}}$ is not a location-scale t distribution at all, but a noncentral t distribution, which is not the same thing. I just edited a bit to try to make that clearer. Commented Oct 31, 2016 at 15:25