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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?

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  • $\begingroup$ 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' $f$? $\endgroup$
    – Taylor
    Jan 9, 2016 at 17:05
  • $\begingroup$ @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? $\endgroup$ Jan 9, 2016 at 17:12
  • $\begingroup$ @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.) $\endgroup$ Jan 9, 2016 at 17:15
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    $\begingroup$ Yep, assuming it isn't weird. $P(f(X) \le c) = P(f^{-1}(f(X)) \le f^{-1}(c)) = P(X \le f^{-1}(c)) = P(Y \le f^{-1}(c)) = P(f(Y) \le c)$. $\endgroup$
    – Taylor
    Jan 9, 2016 at 17:21
  • $\begingroup$ And they're (mathematically) equivalent. I guess it helps if you think of $T$ and $\frac{Z}{\sqrt{V/v}}$ being hypothetical/unobserved. $\endgroup$
    – Taylor
    Jan 9, 2016 at 17:23

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$5.5$ 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").

This immediately implies that obviously equality in distribution must hold, since equality almost surely of two random variables implies that their distribution functions must be equal.

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.

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