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Could someone help me find a way out:

If $X$ is a random variable and $X$ is independent of itself. Show that there is a constant $a$ such that $P(X=a)=1$ if and only if $E[X_1]$ exists.

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    $\begingroup$ @Did: Which implies that X is integrable, so in what sense is it not true?‌ :-) (I agree that "if and only if" could be replaced with "and that"... but still, the statement technically doesn't seem wrong.) $\endgroup$ – ShreevatsaR Aug 10 '13 at 16:58
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    $\begingroup$ @ShreevatsaR Right, thanks for spotting that. Revised version: the mention "if and only if E[X1] exists" is superfetatory (and slightly misleading) in this context, as well as the title. $\endgroup$ – Did Aug 10 '13 at 17:03
  • $\begingroup$ i will try to get the question again and get back to you. All the same thanx so much for bringing my attention to the error in the formulation of the question. $\endgroup$ – Zico Aug 27 '13 at 19:37
  • $\begingroup$ There is a quite elegant way to prove this statement. Let $X_1,X_2,\dots$ be copies of $X$, then by Kolmogorov's 0-1 law it follows that $P(X=a)\in\{0,1\}$. The rest of the claim should now be obvious. $\endgroup$ – julian Jul 9 '17 at 12:40
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$\def\R{\mathbb R} \def\Z{\mathbb Z} \def\P{\mathbb P}$ As was already noticed in the comments the formulation is a bit misleading. Under the assumptions both sides of the equivalence must be true.

$X$ is independent of itself - in particular $\mathbb P(X\in B) = \mathbb P(X\in B, X \in B) = \mathbb P(X\in B) \cdot \mathbb P(X\in B) = \mathbb P(X\in B)^2$ (where $B$ is a borel subset of $\mathbb R$), which means that $\mathbb P(X\in B)$ is just equal to $0$ or $1$ for any $B$.

EDIT (more hints requested):

Now divide $\R$ into a contable number of disjoint intervals e.g. $I_n=[n,n+1)$ for $n\in \Z$. There exists a unique $n$ such that $\P(X\in I_n)=1$ (why?). Then you can divide $I_n$ into two halves and see which half has probability $1$. You can do it inductively and in the limit get a set of probability $1$ and diameter $0$. It must be a point and that's $a$ we were looking for.

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  • $\begingroup$ I will very glad to get the hint $\endgroup$ – Zico Aug 27 '13 at 19:31
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Using the definition of independence one can see that for each $t$, we have $\mathbb P\{X\leqslant t\}\in\{0,1\}$. Define $$t_0:=\inf\{t\in\mathbb R\mid\mathbb P\{X\leqslant t\}=1\}.$$ The infimum exists since $\lim_{t\to +\infty}\mathbb P\{X\leqslant t\}=1$ (hence $\mathbb P\{X\leqslant t\}=1$ for $t$ large enough).

If $t_n\downarrow t_0$, then we can see that $\mathbb P\{X\leqslant t_0\}=1$ and since $\mathbb P\{X\leqslant s\}=0$ for $s\lt t_0$, we have $\mathbb P\{X\lt t_0\}=0$.

We conclude that $\mathbb P\left\{X=t_0\right\}=1$.

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