## Friday, 24 January 2014

### Lecture 4

I have added to the list of Problem Solving Strategies a note about the inclusion-exclusion formula.

I showed you in Example 4.5 a use of the probabilistic method in combinatorics. If you liked that, you might like to read about other examples, for instance in this link.

In Example 4.4 we considered an infinite sequence of tosses of biased coins and saw that if $\sum_{i=1}^\infty p_i<\infty$  then $P(\text{infinite number of heads})=0.$

Now let's prove that $\sum_{i=1}^\infty p_i=\infty \implies P(\text{finite number of heads})=0.$ This is trickier.

We are assuming the coin tosses are independent (which is terminology from Lecture 5, but you can guess intuitively what this means). Let $A_k$ be the event that the $k$th toss is a head. Let $E_n=\bigcap_{k=n}^\infty A_k^c=[\text{no heads after the$(n-1)$th toss}]$. Then for $m > n$

$P(E_n)\leq P(\bigcap_{k=n}^m A_k^c)=\prod_{k=n}^m P(A_k^c) =\prod_{k=n}^m (1-p_k)\leq \prod_{k=n}^m e^{-p_k} =e^{-\sum_{k=n}^m p_k}$.

The right hand side $\to 0$ as $m\to\infty$.

Hence $P(E_n)=0$, and then using Property (v) of $P$ (continuity), we have that

$P(\text{finite number of heads})=P(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k^c)=\lim_{n\to\infty}P(E_n)=0.$

These two results are examples of applications of the Borel-Cantelli lemmas, which appear in the Part II course Probability and Measure.

I did not have time to talk through Example 4.8, but it is another nice use of the inclusion-exclusion formula.