2.1 Bernoulli processes

We can calculate the probability of any finite sequence of the trials such as SSFSFFS by $ppqpqqp=p^4{q}^3$, multiplying together a $p$ for each S and a $q$ for each F.

We will now look at calculating the expected time until we first observe either two successes in a row, or a failure immediately followed by a success using conditional expectation of the outcomes of the first few trials. If a success and a failure are equally likely, which will have the smaller expected time?

Intuitively it feels like these should have the same expected value; but we will see that this is not the case.

The ends of the branches correspond to distinct possiblities whose probabilities sum to 1. We label these by letting $V$ be a rv with values

$V$ = $\{\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}$ if the sequence starts with $\{\begin{array}{c}\hfill \text{F}\hfill \\ \hfill \text{SF}\hfill \\ \hfill \text{SS}\hfill \end{array}$ with probability $\{\begin{array}{c}\hfill q\hfill \\ \hfill pq\hfill \\ \hfill p^2\hfill \end{array}$.

Now consider:

if $V$ = $\{\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}$ then $T$ = $\{\begin{array}{c}\hfill 1+T^{\prime }\hfill \\ \hfill 2+T^{\prime \prime }\hfill \\ \hfill 2\hfill \end{array}$ with probability $\{\begin{array}{c}\hfill q\hfill \\ \hfill pq\hfill \\ \hfill p^2\hfill \end{array}$

where $T^{\prime }$ and $T^{\prime \prime }$ are the further times (numbers of trials) needed to obtain the sequence SS in the first two cases. Of course in the third case we have just achieved SS which is why $T$ = 2. Now we use Theorem 1.3.1 to write

	$\mathrm{E}(T)$	$=$	$\mathrm{E}\{E(T|V)\}$		(2.1)
		$=$	$\mathrm{E}(T|V=1)q+\mathrm{E}(T|V=2)pq+\mathrm{E}(T|V=3)p^2$
		$=$	$\mathrm{\{}1+\mathrm{E}(T^{\prime })\}q+\{\mathrm{\hspace{0.17em}2}+\mathrm{E}(T^{\prime \prime })\}pq+\mathrm{\hspace{0.17em}2}p{}^2$

In the cases starting with F or SF, the further prospects of obtaining SS are exactly the same as starting over again. Therefore:

$T$, $T^{\prime }$ and $T^{\prime \prime }$ all have the same distribution

and these all have the same expectation which we call

$$e=\mathrm{E}(T)=\mathrm{E}(T^{\prime })=\mathrm{E}(T^{\prime \prime }).$$

Substituting for these into (2.1) gives

$$e=(1+e)q+(2+e)pq+2p^2$$

which simplifies to $(1-q-pq)e=q+2pq+2p^2$ and further to $p^{2}e=1+p$ so that

$$e=\frac{1+p}{p^2}.$$

At this point the choice of tree diagram is seen to be such as to ensure that the branches terminate with an F of SS - the point is to avoid an S or an FS. Note also that the values given to $V$ were only used to enumerate the possibilities; its precise values did not enter into the calculation.

So if $T_F$ is the further time to $FS$ given we have had an $F$, then

$$\text{E}(T_F)=q(1+\text{E}(T_F))+p(1)\Rightarrow \text{E}(T_F)=\frac{1}{p}.$$

Hence

$$\text{E}(T)=q\left(1+\frac{1}{p}\right)+p(1+\text{E}(T))\Rightarrow \text{E}(T)=\frac{1}{p}+\frac{1}{q}=\frac{1}{pq}.$$