# Almost sure convergence of random variables

$(X_n)$ is a sequence of random variables having the following distribution:

$$P(X_n=1)=1- \frac{1}{n},\; P(X_n=0)=\frac{1}{n}$$ (we don't assume that those variables are independent). $X$ is some random variable. We consider variables $Y_n=XX_n$. Does $(Y_n)$ converge almost surely? Does it converge in probability?

I think that $Y_{n} \to X$ a.s., because $X_n \to 1$ a.s.

Using Chebyshev's inequality we have for $\epsilon > 0$:

$$P(\left|X_{n}-1 \right| \ge \epsilon ) \le \frac{\text{E}(X_n-1)}{\epsilon} \le -\frac{1}{\epsilon n}$$

Seeing that $$-\sum\frac{1}{\epsilon n}=-\infty,$$ we conclude that $$\sum P(\left|X_{n}-1 \right| \ge \epsilon ) <+\infty.$$ Hence $X_n \to 1$ a.s.

Convergence in probability stems from almost sure convergence.

Is everything there okay?

• math.stackexchange.com/q/1325295 Same class?
– Did
Jun 14 '15 at 20:26
• Who knows ;) Maybe it's just the same problem... Jun 14 '15 at 20:42
– Did
Jun 14 '15 at 22:55
• It's not a homework... I've found this problem on the list of exemplary exercies before the exam... Jun 15 '15 at 0:04

Chebyshev's inequality says that for any non-negative random variable $Z$ and any $\epsilon > 0$, we have $$P(Z \ge \epsilon) \le \frac{E[Z]}{\epsilon}.$$
You are applying this with $Z = |X_n - 1|$, so you get $$P(|X_n - 1| \ge \epsilon) \le \frac{E[|X_n - 1|]}{\epsilon}.$$ Note the absolute value bars on the right side, which you apparently dropped. Now $E[|X_n - 1|] = 1/n$, not $-1/n$, and you get $P(|X_n - 1| \ge \epsilon) \le \frac{1}{n \epsilon}$. This is true but a bit silly, since you can get a better bound without Chebyshev: for $\epsilon \le 1$ you have $P(|X_n - 1| \ge \epsilon) = P(X_n = 0) = 1/n$, and for $\epsilon > 1$ you have $P(|X_n - 1| \ge \epsilon) = 0$. But this also doesn't help you apply Borel-Cantelli.
In fact, you cannot prove from the given information that $X_n \to 1$ a.s., because that can be false. Suppose that the random variables $X_n$ were independent (the statement of the problem doesn't assume this, but also doesn't rule it out). Then you can use the second Borel-Cantelli lemma to show that $P(X_n = 0 \text{ i.o.}) = 1$ and also $P(X_n = 1 \text{ i.o.}) = 1$. Hence the sequence diverges almost surely.
It is true that $X_n \to 1$ in probability. You can use Chebyshev for this (if you use it correctly) but the "better bound" I mention above seems easier.