Suppose there are two fair coins and has two random variables associated with it $X_1$ and $X_2$. Each of them is a Bernoulli Random variable as:

$X_1(w) = \begin{cases} 0,\; if\; w = H \\ 1, \; if\; w = T\end{cases}$

$X_2(w) = \begin{cases} 0,\; if\; w = H \\ 1, \; if\; w = T\end{cases}$

Lets take a sequence $X_n = X_1 \;\forall n$ and ranodom variable $X = X_2$.

By definition of almost sure convergence in probability implies:

$P\{w \in \Omega : lim_{n \to\infty} X_n(w) \rightarrow X(w) \} = 1$.

In this example, for $w \in \{H,T\}$ implies $X_n(w) = X(w)$ as $X_1(w) = X_2(w)$ and probability of $P\{H,T\} = 1$.

(because when $w={H}$, $X_n(\{H\}) = X(\{H\}) = 1$ as $X_1(\{H\}) = X_2(\{H\})$ similar arguments hold for $\{T\}$)

Thus $X_n \overset{a.s.}{\rightarrow} X$ ?

Now convergence almost surely implies convergence in probablity. Thus $\underset{n \rightarrow \infty}{lim} P (|X_{n} - X| \ge \epsilon) = 0 \;\;\forall \epsilon >0$ should be true.

But $\underset{n \rightarrow \infty}{lim} P (|X_{n} - X| \ge \epsilon) = P(|X_1 - X_2| \ge \epsilon) \ne 0 \;\;\forall \epsilon >0$

as above probability will be on joint outcomes of $X_1, X_2$.

Now there is some problem in the argument. I am not able to observe.

Let me know if any other clarification is required. Thanks in advance.

  • $\begingroup$ Your question is not very clear to me. According to the definition, a rand. proc. $(X_n)_n$ converges to a rand. var. $X$ in probability if for every $\epsilon>0$, $\lim_n \mathrm{P}[|X_n-X|>\epsilon]=0$. In your case $|X_n-X|=0(\omega)$ and $\mathrm{P}[|X_n-X|>\epsilon]=0$ for all $n\in\mathbb{N}$. Therefore, it converges in probability. Does this answer your question? $\endgroup$ – Pantelis Sopasakis May 8 '17 at 10:49
  • $\begingroup$ @PantelisSopasakis I think $ P\{|X_n -X|>\epsilon\} \ne 0 $ as this probability will be on their joint event i.e. {H,H}, {H,T} etc. $\endgroup$ – hunch May 8 '17 at 11:21
  • $\begingroup$ But you are saying that $X_n=X$ for all $n$. That means that $X_n$ and $X$ are just two labels for the same random variable. Your definition of $X$ gives the impression that you are working on a probability space with underlying set $\Omega=\{H,T\}$. Is that so? On that space only $4$ functions $\Omega\to\{0,1\}$ can be defined. Finally your notation $P (\cup_{n=1}^{\infty} |X_{N+n} - X| \ge \epsilon)$ is unclear. What do you mean by that? $\endgroup$ – drhab May 8 '17 at 12:26
  • $\begingroup$ I suspect you mean to say: $X$ and $X_n$ have the same distribution. That is essentially different from $X=X_n$. $\endgroup$ – drhab May 8 '17 at 12:40
  • $\begingroup$ I have updated the problem. $\endgroup$ – hunch May 8 '17 at 14:26

$\newcommand{\P}{\mathrm{P}}$In order to dispel any doubts let's recall that the probability is a measure and as such it applies to events (sets).

Here we need to evaluate $\P[|X_n-X|>\epsilon]$, where both $X_n$ and $X$ are random variables on $\Omega$, that is, they are functions $X_n:\Omega\ni \omega \mapsto X_n(\omega) \in \{0,1\}$ (the same for $X$).

Their absolute difference, $R(\omega)=|X_n(\omega) - X(\omega)|$ is itself a random variable.

Now the probability $\P[R(\omega)|>\epsilon]=$ is, by definition, the probability of the event $E_{\epsilon} = \{\omega \in \Omega : R(\omega) > \epsilon\}$.

But $R(\omega)=0$ for all $\omega$, so $E_{\epsilon}=\varnothing$ and $\P[E_\epsilon]=\P[\varnothing]=0$ (it is an axiom).

  • $\begingroup$ I am upvoting your answer as this correct. But as drhab comment cleared my doubt, I can not accept this as an answer. Thanks $\endgroup$ – hunch May 9 '17 at 9:23

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