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

Question

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.

Answer 1

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