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One learns in a probability course that a (real) random variable is a measurable mapping of some probability space $(\Omega,\mathcal{A},\mathbf{P})$ into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. But as soon as one gets into topics that are a little advanced, the space $(\Omega,\mathcal{A},\mathbf{P})$ is not mentioned unless it is absolutely necessary. After a long time of frustration, I have become quite comfortable with this language. But some things still trouble me. The following kind of reasoning comes up in the book I'm reading:

The author says that $(X_i)_{i\in I}$ is a family of random variables and specifies the distribution of each random variable. Then he phrases some (random) proposition $A((X_i)_{i_\in I})$ (this is a little imprecise, I hope you get the meaning) and talks about $\mathbf{P}[A((X_i)_{i_\in I}) \text{ holds}]$.

My question: Let $(\Omega',\mathcal{A}',\mathbf{P}')$ be another probability space and $(Y_i)_{i\in I}$ random variables such that, for each $i\in I$, the distribution of $Y_i$ is the same as the distribution of $X_i$. Is it then obvious that $\mathbf{P}[(A(X_i)_{i_\in I}) \text{ holds}]=\mathbf{P}'[(A(Y_i)_{i_\in I}) \text{ holds}]$?

Now my guess is that this is true, but needs a proof, which is not completely trivial in case $I$ is infinite, at least not for a beginner. However, in the book this problem isn't discussed at all. So did I miss something?

Edit:

I'm not sure whether the question was correctly understood, so I'll rephrase it a little.

Let $(\Omega,\mathcal{A},\mathbf{P})$ and $(\Omega',\mathcal{A}',\mathbf{P}')$ be two probability spaces, $I$ a set, and $(X_i)_{i\in I}$ and $(Y_i)_{i\in I}$ families of random variables on $(\Omega,\mathcal{A},\mathbf{P})$ and $(\Omega',\mathcal{A}',\mathbf{P}')$ respectively such that, for each $i\in I$, the distribution of $X_i$ is equal to the distribution of $Y_i$. Let $J$ be a countable subset of $I$ and $B_j$ a Borel set for each $j\in J$. The question is:

Is it obvious that $\mathbf{P}\left[\bigcup_{j\in J}\{X_j\in B_j\}\right]=\mathbf{P}'\left[\bigcup_{j\in J}\{Y_j\in B_j\}\right]$?

The sets $B_j$ and the union over $J$ are just an example. What I mean, but cannot formalize: Let $A\in\mathcal{A}$ and $A'\in\mathcal{A}'$ such that there is an expression for $A$ in terms of the $X_i$, and $A'$ is given by the same expression replacing $X_i$ by $Y_i$ for each $i$. Is it obvious that $\mathbf{P}[A]=\mathbf{P}'[A']$?

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If I'm reading your question correctly, the answer is no: you need to know that the joint distributions are the same. If the variables are independent then this is automatic. –  Qiaochu Yuan Feb 21 '11 at 2:08
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If it is an uncountably infinite collection of random variables (often the case in stochastic process theory), then even independent is not enough. Edit: Now I read Shai Covo's answer, he is also making this point. –  George Lowther Feb 21 '11 at 2:35
    
@Qiaochu @George: I hope I've clarified the question. Do your comments still apply? –  Stefan Walter Feb 21 '11 at 12:00
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@Stefan: yes. You need to know the joint distribution or you don't know anything. You can find a counterexample with two Bernoulli random variables (taking one pair to be independent and the other to be identical). –  Qiaochu Yuan Feb 21 '11 at 12:06
    
@Qiaochu: So to prove it in case the families are independent and $I$ is countably infinite, do I have to know that there is a unique product measure and that this has to be the joint distribution of an independent family of random variables? Infinite product measures are in chapter 14 of my book, while the proof for which I think one needs this statement is in chapter 2! –  Stefan Walter Feb 21 '11 at 12:40
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2 Answers

The distributions of each $X_i$ and each $Y_i$ are far from being sufficient to decide anything about the families $(X_i)_i$ and $(Y_i)_i$.

Assume for instance that $X_1$, $X_2$, $Y_1$ and $Y_2$ are all uniform $\pm1$ Bernoulli random variables and that $X_1=X_2=Y_1=-Y_2$. Then the event $[X_1=1\ \mbox{or}\ X_2=1]$ has probability $\frac12$ while the event $[Y_1=1\ \mbox{or}\ Y_2=1]$ has probability $1$.

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The question is not very clear, but here is some idea. Let $U$ be a uniform$[0,1]$ random variable, and define a random process $X=\{X_t: t \in [0,1]\}$ by $X_t = \mathbf{1}(t=U)$, where $\mathbf{1}$ is the indicator function. Then, $X$ is identical in law to the zero process $Y$ defined by $Y_t = 0$ for all $t \in [0,1]$; that is ${\rm P}[X_{t_1} = 0, X_{t_2}=0,\ldots,X_{t_n}=0]=1$ for any choice of $n \geq 1$ and $0 \leq t_1 < \ldots \leq t_n \leq 1$. However, $X$ and $Y$ are quite different: $X$ is not continuous, its $\sup$ is equal to $1$, etc.

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I'm not familiar with stochastic processes, so can't completely understand your answer. But I think you're talking about the converse of what I meant. I hope my question is clearer now, my apologies. –  Stefan Walter Feb 21 '11 at 12:05
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