Let us consider $\{Z_n\}_{n\in\mathbb{N}}$ a sequence of independent Bernoulli random variables of parameter $1/n$. So $P(Z_n=1)=1/n$ for all $n$. Let $X_n=Z_1+\dots+ Z_n$. Then $X_1\le X_2\le\dots$ is a nondecreasing sequence of random variables, so it admits an almost sure limit. I claim that this limit is $\infty$. Indeed, we consider the independent events $A_n=\{Z_n=1\}$. Since $$\sum_n P(A_n)=\sum_n 1/n=\infty,$$ by Borel Cantelli lemma, $P(\limsup A_n)=1$, so for almost every $\omega\in\Omega$ (our possibility space), there exist infinite indices $n$ for which $Z_n(\omega)=1$, and so $\lim_n X_n(\omega)=\infty$ almost surely. On the other hand, every $X_n$ is a nonnegative fandom variable with finite mean 1, so we can use Markov inequality to deduce that $\{X_n\}$ is a tight sequence: $$P(|X_n|>M)<1/M$$ for every $M$. By Prokhorov theorem, we can find a subsequence converging in distribution to some real value, and this contradicts the almost sure convergence to $\infty$. What am I missing?
A question about tightness and almost sure convergence
probability theory
Best Answer
$$P(|X_n| >M) \leq 1/M$$
seems to be false.
Indeed,
$$P(|X_n|> M) \leq E|X_n|/M$$
and $E[Z_n]= 1/n$, so $E[X_n] = 1+1/2+ \dots + 1/n \neq 1$
In particular, your mistake is that you claim that $X_n$ has mean $1$.