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Let $X,Y: \Omega \rightarrow S$ be two random variables. Now, I was wondering what exactly does it mean that $X,Y$ are identically distributed? Does this mean that $P(X=Y)=1$? Identically distributed random variables are apparently pretty common in probability theory, but I could not find a definition for general $S.$

My intuition tells me that $P(X=Y)=1$ is an even stronger condition, but I am not sure about this.

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  • $\begingroup$ Identically distributed means that the CDFs $F_X(\cdot)$ and $F_Y(\cdot)$ are the same. It does not mean that $P(X=Y)=1$. Nor does it mean that $X$ and $Y$ are independent. $\endgroup$ – Dilip Sarwate Oct 25 '15 at 19:44
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    $\begingroup$ $\Pr(X=Y)$ is very much not right. For example let $X$ and $Y$ be the number of heads in two independent tosses of a fair coin. Use instead $\Pr(X\le t)=\Pr(Y\le t)$ for all $t$. $\endgroup$ – André Nicolas Oct 25 '15 at 19:45
  • $\begingroup$ @AndréNicolas But $P(X=Y)=1$ implies identical distribution, right? $\endgroup$ – user167575 Oct 25 '15 at 19:46
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    $\begingroup$ Yes, but it is a much stronger condition, and not a useful one. $\endgroup$ – André Nicolas Oct 25 '15 at 19:47
  • $\begingroup$ @user167575 $P(X{=}Y)=1$ implies identical distribution, but identical distribution does not necessarily mean $P(X{=}Y)=1$. $\endgroup$ – Graham Kemp Oct 25 '15 at 20:27
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Two $\mathbb{R}$ or $\mathbb{R}^n$-valued variables being identically distributed means that they have the same cumulative distribution function. This is a rather weak condition; in particular it does not even imply that the two variables are defined on the same underlying probability space.

Even if two identically distributed variables are defined on the same probability space, $P(X=Y)$ could be pretty much anything. For instance, suppose $\Omega=[0,1]^2$, $\mathcal{F}$ is the Borel $\sigma$-algebra on $[0,1]^2$, and $P$ is the Lebesgue measure. Then $X((x,y))=x$ and $Y((x,y))=y$ are two identically distributed random variables, and $P(X=Y)=0$ (the Lebesgue measure of the diagonal of the square).

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  • $\begingroup$ So $P(X=Y)=1$ does not imply an identical distribution? $\endgroup$ – user167575 Oct 25 '15 at 19:47
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    $\begingroup$ @user167575 No, it does. I haven't said anything about that case. $\endgroup$ – Ian Oct 25 '15 at 19:47
  • $\begingroup$ Indeed, for a general $S$-valued random variable (which is not necessary $\mathbb R$-valued) the cumulative distribution function does not exist. In this case identically distributed means that the probability of all events are the same, i.e. that $\Pr(X\in A)=\Pr(Y\in A)$ for any measurable set $A\subseteq S$. $\endgroup$ – Marcel Oct 25 '15 at 19:52
  • $\begingroup$ @Marcel Good point, I've revised slightly. $\endgroup$ – Ian Oct 25 '15 at 20:14

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