Is it possible that following series of random variables exists?

$\displaystyle P({\omega \in \Omega : lim_{n -> \infty} X_{n}=0})=0.5$

Thx for any help

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- Jan 8th 2011, 11:48 AMslavertExistence of series of random variables
Is it possible that following series of random variables exists?

$\displaystyle P({\omega \in \Omega : lim_{n -> \infty} X_{n}=0})=0.5$

Thx for any help - Jan 8th 2011, 12:34 PMMoo
Hello,

No, have a look at Kolmogorov's 0-1 law :) - Jan 9th 2011, 02:03 AMslavert
Please explain me why,

in Kolmogorov's 0-1 law we have $\displaystyle A \in F$ instead of $\displaystyle \omega \in \Omega$

It contradicts ituition, it seems to me that $\displaystyle P(X_{n}=0)=1/2 and P(X_{n}=n)=1/2$ satisfy that. - Jan 9th 2011, 09:02 AMMoo
I think considering $\displaystyle \omega$ or the measurability of the rv's doesn't change much thing... I'll have to give a deeper look into the definition of a tail event but I'm quite tired.

Anyway, I think your example doesn't fit what you want...

Let's assume the rv's are independent.

We want to find $\displaystyle \displaystyle P(\lim_{n\to\infty} X_n=0)$.

We have $\displaystyle \forall \epsilon>0,\displaystyle \sum_n P(X_n>\epsilon)=\sum_n P(X_n=n)=\infty$.

This condition, added to the independence, lets us use Borel-Cantelli's lemma (part II) and we get that $\displaystyle \displaystyle \forall \epsilon>0,P(\limsup_n \{X_n>\epsilon\})=1 \Leftrightarrow \forall \epsilon>0,P(\liminf_n \{X_n<\epsilon\})=0 \quad (\star)$

Now, (for the first equality, we can take $\displaystyle \epsilon \in\mathbb Q^+$ because it's dense in $\displaystyle \mathbb R^+$)

$\displaystyle \displaystyle \begin{aligned} P(\lim_n X_n=0)&=P(\forall \epsilon \in \mathbb Q^+,\exists N\in\mathbb N,\forall n>N,X_n<\epsilon) \\

&=P(\bigcap_{\epsilon\in\mathbb Q^+}\bigcup_{N\in\mathbb N}\bigcap_{n>N}\{X_n<\epsilon\}) \\

&=P(\bigcap_{\epsilon\in\mathbb Q^+} \liminf_n \{X_n<\epsilon\}) \\

&<P(\liminf_n \{X_n<\epsilon_0\}),\text{ for a given } \epsilon_0\in\mathbb Q^+ \end{aligned}$

By $\displaystyle (\star)$, this final probability is 0. Hence $\displaystyle \displaystyle P(\lim_n X_n=0)=0$.

I agree though that this is kind of counterintuitive. So if there's a mistake somewhere just tell me!