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Does $X_n$ iid $\sim Bern(1/2)$ converge to anything?

I feel like iid $Bern(1/2)$ random variables does not converge to anything since it takes value $1$ and $0$ each with probability $1/2$. Therefore, the resulting sequence will be oscillating between $1$ and $0$.

However, if we define $X_n(\omega) $ as follows,

\begin{equation} X_n(\omega)= \begin{cases} 0, & \text{if}\ \omega=T\\ 1, & \text{if}\ \omega = H \end{cases} \end{equation}

assuming a fair coin, and check $Pr(\omega\in \Omega: \lim\limits_{n\to\infty}X_n(\omega) = X(\omega))$, we see that $ \lim\limits_{n\to\infty}X_n(H) = X(H)$ and $ \lim\limits_{n\to\infty}X_n(T) = X(T)$. Therefore, $Pr(\omega\in \Omega: \lim\limits_{n\to\infty}X_n(\omega) = X(\omega)) = 1$. So we do have almost sure convergence.

Which one is true? Any help is appreciated! Thanks in advance!!

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    $\begingroup$ If $X_n(\omega) = X(\omega)$ for all $n$, how do you have independence? $\endgroup$ – Trevor Gunn Nov 25 '17 at 5:33
  • $\begingroup$ @TrevorGunn but even remove this statement... it won't affect the argument. $\endgroup$ – metricspace Nov 25 '17 at 5:37
  • $\begingroup$ It depends on what kind of convergence you mean. A sequence of iid random variables with distribution $X$ converges to $X$ in probability (en.wikipedia.org/wiki/…). $\endgroup$ – Qiaochu Yuan Nov 25 '17 at 5:37
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    $\begingroup$ A sequence of iid bernoulli random variables will almost surely not converge (nor does it converge in probability). A constant sequence is not independent but of course will converge. $\endgroup$ – Trevor Gunn Nov 25 '17 at 5:39
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    $\begingroup$ The classic example is to write a number $\omega \in [0,1]$ as $0.X_1(\omega)X_2(\omega)\cdots$ in base $2$. The set of numbers that have more than one representation in base $2$ is countable (measure $0$). $\endgroup$ – Trevor Gunn Nov 25 '17 at 5:50

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