# Equality vs. Equality in Distribution ($t$-distribution for example)

A technical question that came up to mind as I was reading up on linear models today.

Consider the $t$-distribution with $\nu$ degrees of freedom ($t_\nu$) for example. Let's say $T \sim t_{\nu}$; that is, the random variable $T$ follows this distribution.

Does it mean that $T$ must equal $\dfrac{Z}{\sqrt{V/\nu}}$ for some $Z \sim \mathcal{N}(0, 1)$ and $V \sim \chi^2_{\nu}$?

Or does it only mean that $T \overset{d}{=} \dfrac{Z}{\sqrt{V/\nu}}$ (that is, they are equal with respect to distribution, meaning that their characteristic functions are equal)?

Or are these two actually one and the same in this case?

Obviously, if $T = \dfrac{Z}{\sqrt{V/\nu}}$, then they have the same distribution, but obviously equality in distribution does not imply that the variables themselves are equal.

To clarify what I'm asking: is the definition saying that any $t$-distributed random variable can be written as $\dfrac{Z}{\sqrt{V/\nu}}$ or merely that any $t$-distributed random variable is merely identically distributed to $\dfrac{Z}{\sqrt{V/\nu}}$? (Equality does imply identically distributed, but the converse is obviously not true.)

On top of this, if $X \overset{d}{=} Y$, does it mean that for any "reasonable" function $f: \mathbb{R} \to \mathbb{R}$, $f(X) \overset{d}{=} f(Y)$? If so, why is this and are there any sources where I can find a proof of this?

• They're the same. Equality of random variables means they have the same distribution function, characteristic function, and all that. And for the second part, do you mean $Ef(X) = Ef(Y)$ for any 'reasonable' $f$? Jan 9, 2016 at 17:05
• @Taylor No. For example, let's say for example, I have two random variables which are identically distributed. If I perform a transformation on both of them, should I expect that they have the same distribution after the transformation? Jan 9, 2016 at 17:12
• @Taylor To clarify what I'm asking about the $t$-distribution, yes, I agree with what you said. But is the definition saying that any $t$-distributed random variable can be written as $\dfrac{Z}{\sqrt{V/\nu}}$ or merely that any $t$-distributed random variable is merely identically distributed to $\dfrac{Z}{\sqrt{V/\nu}}$? (Equality does imply identically distributed, but the converse is obviously not true.) Jan 9, 2016 at 17:15
• Yep, assuming it isn't weird. $P(f(X) \le c) = P(f^{-1}(f(X)) \le f^{-1}(c)) = P(X \le f^{-1}(c)) = P(Y \le f^{-1}(c)) = P(f(Y) \le c)$. Jan 9, 2016 at 17:21
• And they're (mathematically) equivalent. I guess it helps if you think of $T$ and $\frac{Z}{\sqrt{V/v}}$ being hypothetical/unobserved. Jan 9, 2016 at 17:23

$$5.5$$ years after posting this question, I've since taken measure-theoretic probability and can answer this question.
The very definition of a random variable $$T \sim t_{\nu}$$ is $$T = \dfrac{Z}{\sqrt{V/\nu}}$$ for some $$Z \sim \mathcal{N}(0, 1)$$ and $$V \sim \chi^2_\nu$$ independent, with probability one ("almost surely").
Regarding the more general statement, suppose $$X \overset{d}{=}Y$$. For $$f(X) \overset{d}{=} f(Y)$$ to hold, $$f$$ must be a measurable map so that $$f(X)$$ and $$f(Y)$$ are valid random variables. So we assume $$f$$ is measurable.
Based on Taylor's comment, we may write for some set $$C$$ that $$\mathbb{P}(f(X) \in C) = \mathbb{P}(X \in f^{-1}(C)) = \mathbb{P}(Y \in f^{-1}(C)) = \mathbb{P}(f(Y) \in C)$$ hence the statement holds.