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Suppose we have a probability measure $$\mu: \mathcal{S} \times \mathbb{R}\rightarrow [0,1],$$where $\mathcal{S}$ is some countable set. In some personal, handwritten lecture notes I'm reading it is then stated that we "take $(A,B)$ such that $(A,B) \sim \mu $ and consider $\mathbb{E}[B]$".

I can't parse this sentence. So far, I've come across the "$\sim$" notation only where one side was a random variable and the other a density function. But in this case that does not seem to apply.

So I'm not sure what $B$ is; it seems to be some kind of projection of $\mu$ onto $\mathbb{R}$. How can I define $B$ precisely/rigorously?

temo
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  • The keyword to search for might be "marginal distribution". (I am absolutely not sure about this) – Giuseppe Negro Mar 17 '19 at 12:34
  • Note that in my answer, i disregard that your $\mu$ has domain $\mathcal{S}\times \mathbb{R}$ and not the product sigma algebra on that space. – John Mar 18 '19 at 15:20
  • @Martin I don't quite understand what you mean by this comment. $\mathcal{S}$ and $\mathbb{R}$ carry some $\sigma$-algebras, and then we take the product of it; call it for brevity $\mathcal{F}$. So $\mu$ is w.r.t to the measurable space $(\mathcal{S}\times \mathbb{R},\mathcal{F})$. I implicitly read your answer with this in mind and it seemed to work. But you did not mean this? – temo Apr 10 '19 at 03:26
  • @temo Yes, but I was just making sure that you understood that I worked with $\mu$ taking arguments in the product sigma-algebra (of $\mathcal{S}$ and $\mathbb{R}$ respectively) and not the product space $\mathcal{S}\times \mathbb{R}$ as you originally wrote was the domain of $\mu$. I think, we agree, but I'm just not used to working with measures having domain being a measurable space, as we only need the sigma-algebra and not the space itself to define a measure. – John Apr 10 '19 at 10:18

2 Answers2

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Normally when you have a probability measure $\nu$ on some measure space $(\mathcal{S},\mathcal{F},\nu)$ and you write "let $X$ be a random variable with $X\sim \nu$", then it just means that $X$ is a measurable mapping from some probability space $(\Omega,\mathcal{H},P)$ such that $X(P)=\nu$. In other words $\nu$ is the distribution of $X$.

In your case you are given a probability measure $\mu$ on the product space $\mathcal{S}\times \mathbb{R}$. When we now say that $(A,B) \sim \mu$, we simply mean that $A,B$ are random variables/elements defined on some common probability space $(\Omega,\mathcal{H},P)$ such that the pushforward measure $(A,B)(P)=\mu$, that is $\mu$ is the simultaneous distribution of $A$ and $B$. Here $(A,B)(P)$ is the pushforward measure of $P$ with the bundle map $\omega\mapsto(A(\omega),B(\omega))\in\mathcal{S}\times\mathbb{R}$. Constructing the bundle $(A,B)$ like this, will yield that the marginal distributions are given by $A\sim \pi_1(\mu)$ and $B \sim \pi_2(\mu)$

Similarly we can arrive at the following representation of the expectation of $B$: \begin{align*} E(B) &= \int_{\Omega} B(\omega) \, dP(\omega) \\ &= \int_{\Omega} \pi_2(A(\omega),B(\omega)) \, dP(\omega)\\ &=\int_{\mathcal{S}\times \mathbb{R}} \pi_2(a,b) \, d(A,B)(P)(a,b)\\ &= \int_{\mathcal{S}\times \mathbb{R}} \pi_2(a,b) \, d\mu(a,b) \\ &= \int_{ \mathbb{R}} b \, d \pi_2(\mu)(b) \end{align*} where we used the abstract change of variable theorem, and that $\pi_2$ is the coordinate projection onto the second coordinate, that is $\pi_2 :\mathcal{S}\times \mathbb{R}\to \mathbb{R}$ given by $\pi_2(s,x) =x$ for any $(s,x)\in\mathcal{S}\times \mathbb{R}$.

One way to construct $A$ and $B$ rigorously is to let the common probability space be given by $(\mathcal{S}\times \mathbb{R}, \mathcal{B}(\mathcal(S))\otimes \mathcal{B}(\mathbb{R}),\mu)$ and now define the random variables/elements as $A=\pi_1$ and $B=\pi_2$.

John
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My guess is that $\ \mu\ $ is a form of probability mass-distribution function, and "$\ (A,B) \sim \mu\ $" means that $\ A\ $ and $\ B\ $ are random variables jointly distributed according to $\ \mu\ $: $$ \mathrm{Prob}\left( \left\{A = s\right\}\ \&\ \left\{y<B\le x\right\}\right) = \mu\left(s,x\right) - \mu\left(s,y\right)\ .$$ The distribution function $\ F_B\ $ of $\ B\ $ would then be given by $\ F_B\left(x\right) = \displaystyle{\sum_{s\in\mathcal{S}}}\mu\left(s,x\right) \ $, and $\ \mathbb{E}\left[B\right]\ $ by the Lebesgue-Stieljes integral $$\mathbb{E}\left[B\right] = \int_{-\infty}^\infty x\ dF_B\left(x\right) .$$

lonza leggiera
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