# Using Markov's Inequality to show that $E(X_i) \rightarrow 0$ implies $P(X_i=0) \rightarrow 1$

I am trying to use Markov's inequality to show that for a sequence of positive random variables $$X_1, X_2, ....$$ with values in $$N=\left\{0,1,2,...\right\}$$ and $$\lim_{i\rightarrow\infty}E[X_i]=0$$, it holds that $$\lim_{i\rightarrow\infty}P[X_i=0]=1$$.

Intuitively, I understand this concept due to the law of large numbers, since as $$i$$ approaches infinity the expected value converges to the true mean 0, and the probability that $$X_i$$ equals the true mean converges to 1.

However, I am stumped as to how to use Markov's Inequality, $$P(X\geq c)=E(X)/c$$, to prove this. Any help is appreciated.

• Using markov's inequality to show $\sum_i P(|X_i| > \epsilon) < \infty$ for all $\epsilon > 0$ would prove the result, by the first Borel-Cantelli Theorem. Perhaps the fact $E[X_i] \to 0$ can be used to do that combined with Markov's. – Xiaomi Oct 29 '18 at 0:14
• Are you told $E[X_i]$ is strictly decreasing in $i$? As I think it would make the proof easier. – Xiaomi Oct 29 '18 at 0:37
• No, the above information is all that is given. Not sure I understand how to use that inequality to prove the result... how is it helpful if $\varepsilon$ must be greater than 0? And what is the purpose in summing the probabilities across i? – Candace McKeag Oct 29 '18 at 0:49
• A direct consequence of Borel Cantelli theorem says that if that summation is finite for any positive $\epsilon$ then $X_i$ converges to 0 almost surely. – Xiaomi Oct 29 '18 at 1:09
• @xiaomi - This has nothing to do with sums, it's about limits. There are no sums in the problem statement. (I misread it at first too!) - OP: Note that since $X_i$ is on the integers, setting $c = 1$ and showing that $P(X_i \geq 1) \rightarrow 0$ does the job; look at the r.h.s. of the Markov expression and substitute. – jbowman Oct 29 '18 at 1:45

Since $$X_1,X_2,X_3,...$$ are non-negative random variables, for all $$n \in \mathbb{N}$$ you have:

$$\mathbb{E}(X_n) = \sum_{k=1}^\infty \mathbb{P}(X_n \geqslant k) \geqslant \mathbb{P}(X_n \geqslant 1) = 1-\mathbb{P}(X_n=0).$$

(This is Markov's inequality with $$c=1$$.) Re-arranging this inequality, and imposing the upper bound of one on probabilities, gives you:

$$1 - \mathbb{E}(X_n) \leqslant \mathbb{P}(X_n=0) \leqslant 1 \quad \quad \quad \text{for all } n \in \mathbb{N}.$$

We can now establish the desired result by using the squeeze theorem, by taking the limit of this sequence of inequalities. In particular, since $$\lim_{n \rightarrow \infty} \mathbb{E}(X_n) = 0$$ we get the squeezed limit:

$$1 \leqslant \lim_{n \rightarrow \infty} \mathbb{P}(X_n=0) \leqslant 1,$$

which establishes that $$\lim_{n \rightarrow \infty} \mathbb{P}(X_n=0) = 1$$. $$\blacksquare$$