Limits and Events in Probability

Question

Let $ \{A_n\} $ be an infinite series of events in a probability space s.t: $$\sum_{n=1}^\infty P(A_n) < \infty $$

Prove that there exists a series of non-negative numbers $ \{L(M)\}_{M \in \mathbb{N}}$, $L(M) \ge 0$ such that

$$\lim_{M\to \infty}L(M) = 0$$

and for each $n$, probability $L(M)$ is an upper bound to the probability that at least $M$ events out of the first $n$ events will happen, i.e. show that:

$$P\left\{ w \mid \exists n_1 < n_2 < ... < n_M \leq n \text{ s.t } w \in \bigcap_{i=1}^M {A_n}_i\right\} \leq L(M).$$

Hint: for each $M, n$, use markov inequality for the random variable $\sum_{k=1}^n 1_{A_k}$.

To be honest I've no idea what am I suppose to do, could use some help to get me going.

Thanks!

Answer 1

The Markov inequality states that for a non-negative random variable $X$ and $M>0$ we have that $$ \mathbb P (X\geq M) \leq \frac{\mathbb E[X]}M. $$ Now fix $n \in \mathbb N$. Then for $X= \sum_{k=1}^n 1_{A_k}$ we have for all $M \in (0, n]$ that \begin{align} &\mathbb P ( \text{at least $M$ events occur out of the first $n$})= \mathbb P \biggl(\sum_{k=1}^n 1_{A_k} \geq M\biggr)\\ & \leq \frac 1M \mathbb E \biggl[\sum_{k=1}^n 1_{A_k}\biggr] = \frac 1M \sum_{k=1}^n \mathbb P (A_k)=:L(M) \end{align} So, $L(M)$ defined as above is an upper bound and we still have to show that $L(M)$ tends to zero as $M \to \infty$. Note that if $M \to \infty$ we also have to have $n \to \infty$ and we get $$ \lim_{M \to \infty} L(M) = \lim_{M \to \infty}\frac 1M \sum_{k=1}^\infty \mathbb P (A_k) = 0. $$