Let $(\Omega,\mathcal{F},\mathbf{P})$ denote a probability space, $(S,\mathcal{M})$ denote a measurable space, and $X : (\Omega,\mathcal{F},\mathbf{P}) \rightarrow (S,\mathcal{M})$ denote a measurable function (thought of as a random variable). Then there is a pushforward measure induced on $(S,\mathcal{M})$ (thought of as the probability distribution of $X$), which we could denote $X_*(\mathbf{P}),$ following Wikipedia.

However, I like to imagine that $X$ "knows" that its domain is the whole probability space $(\Omega,\mathcal{F},\mathbf{P}).$ Hence $X$ is "aware of" the probability measure on its domain, and hence we shouldn't have to mention $\mathbf{P}$ in the notation for the pushforward. This coincides better with how I like to write and talk: I will usually just speak of the "probability distribution" of $X$, without mentioning $\mathbf{P}$ at all.

Question. Is there accepted notation for the pushforward measure that doesn't mention $\mathbf{P}$? Something like $\mathrm{distr}(X)$ or $\mathbf{D}(X)$, for example.

  • $\begingroup$ It's been a while since I've thought about these things, but I think there's some confusion with the way you've set things up. The probability space and the measure on that space are different objects, but you seem to conflate them. You don't push forward the space, you push forward the measure on that space. And $X$ may "know" its domain, but it doesn't "know" the measure on that domain. $\endgroup$
    – Potato
    Jul 28 '15 at 4:47
  • $\begingroup$ Also, usually $\mathcal M$ is reserved for the collection of measurable sets on space, not the space itself. $\endgroup$
    – Potato
    Jul 28 '15 at 4:48
  • $\begingroup$ @Potato, I intentionally conflate them here because IMO there is no danger in this particular context. $\endgroup$ Jul 28 '15 at 4:49
  • $\begingroup$ But the question you're asking seems to arise exactly from the kind of confusion that comes from conflating them. It's simply not true that $X$, defined as a function, contains any information about the measure on its domain. $\endgroup$
    – Potato
    Jul 28 '15 at 4:52
  • 2
    $\begingroup$ I'm a working probabilist, and I have to tell you that the ordered triple notation $(\Omega, \mathcal{F}, P)$ is absolutely universal in this field - anything else looks weird. I'm sorry you don't like it! $\endgroup$ Jul 28 '15 at 5:05

Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation.

Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on $(M, \mathcal{M})$, which we might denote as $\mu$, defined by $\mu(A) = P(X^{-1}(A))$.

The measure $\mu$ is sometimes called the distribution or law of $X$, and as such I've often seen it denoted by $$\operatorname{Law}(X)$$ or $\operatorname{law}(X)$. See for instance Definition 1.10 of this paper by myself and collaborators, in which we use this notation. We didn't invent it; I don't know who did, but it is quite common in the field, and I think it would be generally understood by probabilists.

For instance, if we had some other measure $\nu$ on $M$ laying around, we could write something like "$\operatorname{Law}(X) \ll \nu$" if it so happened that the one measure was absolutely continuous to the other. In principle you could write something like "$(\operatorname{Law}(X))(B) = 2/3$" but in practice you would simply write "$P(X \in B) = 2/3$" instead. If you plan to use the measure extensively, you should give it a name (e.g. "Let $\mu = \operatorname{Law}(X)$").

As Potato correctly points out, this notation makes it look like the measure only depends on $X$, when of course it also depends on the underlying probability measure $P$. But it is quite standard in probability theory to tacitly assume that everything in sight depends implicitly on the underlying probability space, and thus to suppress it from notation. I guess one could take a categorical interpretation as goblin does, and say that the underlying probability space is "part of" the object $X$, but I don't think most probabilists think that way.


Given a measureable $f:(X,\mathcal{M},\mu)\to(Y,\mathcal{N})$ (a measure space to a measurable space) one can define a measure $\nu=f_{*}\mu$ on $(Y,\mathcal{N})$ by $$ \nu(B)=\mu(f^{-1}(B)). $$ Both $f$ and $\mu$ are necessary to define it, so $f_{*}\mu$ seems like good notation to me.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.