# Convergence in probability of sequence of random variables

Let $$X \sim \text{Ber}(1/2)$$. I am asked to show that the sequence of random variables $$\{X_n = (1+\frac{1}{n})X\}$$ converges in probability to $$X$$. My attempt:

Let $$\epsilon > 0$$, take $$N = \frac{1}{\epsilon}$$, then for all $$n > N$$ we have $$\begin{cases} & X_n = 0 \text{ wp } 1/2 \\ & X_n = 1+\delta \text{ wp } 1/2 \end{cases}$$ Where $$\delta < \epsilon$$ for all $$n > N$$. Hence $$P(|X_n-X| > \epsilon) = 0$$ and the result follows. Is this enough to prove it? I am familiar with convergence of sequences of functions from real analysis but am having a harder time with random variables. Also, I know that almost surely convergence implies convergence in probability. Can I use that fact here?

• You can use almost sure convergence argument in the answers provided below. If you want direct proof, notice that $\mathbb{P}[|(1 + 1/n)X - X| > \epsilon]$ is the same as $\mathbb{P}[|X(1/n)|>\epsilon]$. Since $X$ can only be $1$ or $0$, taking the limit establishes the result. Mar 3, 2020 at 23:00

A slight nitpick is you should say 'take $$N > \frac{1}{\epsilon}$$' rather than an equality, but that really is just a nitpick. Other than that your proof looks fine.

As you have guessed, you can also use the fact that almost sure convergence implies convergence in probability. Can you see why it's clear that $$X_n \to X$$ almost surely as $$n \to \infty$$?

• Is it because the sequence $1+\frac{1}{n} \to 0$? I'm not so sure I understand the definition of almost sure convergence
– user704448
Mar 2, 2020 at 23:13
• I think you mean $1 + \frac{1}{n} \to 1$, if so then yes. Almost sure convergence states that $P(X_n \to X\ \text{as}\ n \to \infty) = 1$. Since $X_n \to X$ no matter if $X = 0$ or $X = 1$, this holds. Mar 2, 2020 at 23:16
• Oops! Sorry about the typo. I think this makes a lot more sense now!
– user704448
Mar 2, 2020 at 23:22

$$(1+\frac 1 n )X$$ converges to $$X$$ in probability for any random variable $$X$$. You don't require any specific distribution for this. Proof: $$(1+\frac 1 n )X \to X$$ with probability one and almost sure convergence always implies convergence in probability.

You may also use the following simple argument: $$P(|(1+\frac 1 n) X-X| >\epsilon) =P(|X| >n\epsilon) \to 0$$ because the events $$(|X| >n\epsilon)$$ decrease to the empty set.