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Suppose a random sequence is defined by $X_n := n B_n$, where $B_n$ is a Bernoulli sequence such that $\mathbb{P}(B_n = 1) = 1/n$. I am interested in the convergence properties of this random process and am not sure how to interpret the results.

To show pointwise (sure) convergence, we need to show that $\lim_{n\rightarrow\infty} X_n(\omega) = X(\omega), \ \forall \omega \in \Omega$. In this case, $\Omega = \{0,1\}$, i.e., success or fail of the Bernoulli trial. Further, $X_n(\omega = 0) = 0$ and $X_n(\omega = 1) = n$, from which we see that the cae of $\omega = 1$ does not yields convergence as clearly $n$ diverges.

For convergence in probability, we need to show that $\lim_{n\rightarrow\infty} \mathbb{P}(|X_n - X| > \epsilon) = 0$, however I am not sure what $X$ should be here. If I assume $X = 0$, then

$$ \mathbb{P}(|X_n - 0| > \epsilon) = \mathbb{P}(X_n > \epsilon) = \mathbb{P}\bigg(B_n > \frac{\epsilon}{n}\bigg) = \frac{1}{n}, $$ since $n,\epsilon > 0$, so the above expression reduces to $\mathbb{P}(B_n = 1)$. Plugging this back in gives $\lim_{n\rightarrow\infty} 1/n = 0$, so $X_n$ does converge in probability.

The same procedure can be done for mean-square (MS) convergence as well, from which I find that $X_n$ diverges in that sense as well.

Am I doing something wrong here when trying to calculate the convergence properties? And if not, what is the intuition behind why $X_n$ does not converge to any random variable?

Edit 1: Cumulative Distribution Functions of $X_n$

enter image description here

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If $B_n$'s are independent (or just pairwise independent), then $X_n = n$ infinitely often almost surely. In other words, almost all realization $X_n$, $n = 1, 2, \cdots$, does not converge. This is because $\sum_n \mathbb{P}(B_n = 1)$ does not converge. Those probabilities are "too large". Therefore, the converse of Borel-Cantelli tells you that $B_n = 1$ infinitely often.

On the other hand, $X_n$ converges to zero in probability.

$X_n$ does not converge in mean square---it's unbounded in mean-square, $E[X_n^2] = n$.

$X_n$ converges in distribution to the point-mass at zero (as your plot of the CDF's shows). Note that convergence in distribution is a different type of notion than the previous three.

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  • $\begingroup$ Thank you. This is exactly what I thought as well. However, when looking at the CDF's $F_n$, I see that as $n \rightarrow \infty$, the distribution becomes basically a straight line at 1, and in the limit $X_n = 0$ with probability 1. See the edit that I made. Can you verify this? $\endgroup$ Sep 10 '20 at 7:05
  • $\begingroup$ $X_n$ converges in distribution to the point mass at zero. Convergence in distribution is different from a.s./in probability/q.m. convergence. E.g. X_n's need not even be defined on the same probability space for convergence in distribution to hold. $\endgroup$
    – Michael
    Sep 10 '20 at 7:10

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