# Converge of Scaled Bernoulli Random Process

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 $$X_n$$ 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.
• 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? Sep 10 '20 at 7:05
• $X_n$ 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. Sep 10 '20 at 7:10