1 Introduction to stochastic processes 1.2 Some basic definitions 2 Repeated trials and simple random walks

1.3 Conditional expectation

Whilst not specific to stochastic processes, the following theorem will be central to how analyse a number of stochastic processes during the course.

Theorem 1.3.1 (Tower property).

For random variables $U$ and $V$ ,

\mathrm{E}\left(\mathrm{E}(U|V)\right)=\mathrm{E}(U).

Illustration. Suppose $U$ is income and $V$ is age for a certain population. The idea of average (or expected) income is straight forward. So is the idea of the expected income for a given age - a conditional expectation, formed from the conditional distribution of income given age. If this conditional expectation of income is then averaged over age, we will just get the expected income in the whole population. This may seem indirect, but conditional distributions are useful for characterising populations. We shall find this simple result very useful.

Assume we have 5 people, two in their twenties who earn $\pounds 100$ and $\pounds 150$ per week; and 3 in their thirties who earn $\pounds 150$ , $\pounds 180$ and $\pounds 220$ per week.
Let $U$ denote income in pounds per week, and $V$ denote age ( $V=0$ denotes in twenties; $V=1$ denotes in thirties) of a randomly chosen person.

\mbox{E}(U|V=0)=\frac{100+150}{2}=125~{}~{}~{}\mathrm{P}(V=0)=\frac{2}{5}

\mbox{E}(U|V=1)=\frac{150+180+220}{3}=\frac{550}{3}

	$\displaystyle\mbox{E}\left\{\mbox{E}(U\|V)\right\}$	$\displaystyle=$	$\displaystyle\left(\frac{100+150}{2}\right)\frac{2}{5}+\left(\frac{150+180+220% }{3}\right)\frac{3}{5}$
		$\displaystyle=$	$\displaystyle\frac{100+150+150+180+200}{5}=160$

Check this is $\mbox{E}(U)$ :

\mbox{E}(U)=\frac{100+150+150+180+220}{5}=160.

Proof.

For discrete variables, assuming that $U$ and $V$ have a joint distribution:

\mathrm{P}(U=i,V=j)\,=\,p_{\,i\,j}

so that the marginal distribution of $V$ is

\mathrm{P}(V=j)\,=\,\sum_{i}\,p_{\,i\,j}\,=\,p_{\,.\,j}

and the conditional distribution of $U$ given $V$ is

\mathrm{P}(U=i|V=j)\,=\frac{p_{\,i\,j}}{p_{\,.\,j}}.

Then the conditional expectation of $U$ given $V$ is formed from this as

\mathrm{E}(U|V=j)=\left\{\sum_{i}i\,\frac{p_{\,i\,j}}{p_{\,.\,j}}\right\},

the value depending on the value $j$ of $V$ . Now form the expectation of this quantity with respect to the distribution of $V$ , to get:

\mathrm{E}\left(\mathrm{E}(U|V)\right)=\sum_{j}\,\left\{\sum_{i}i\,\frac{p_{\,% i\,j}}{p_{\,.\,j}}\right\}\,p_{\,.\,j}\,=\,\sum_{j}\,\sum_{i}i\,p_{\,i\,j}\,=% \,\sum_{i}i\,\sum_{j}p_{\,i\,j}\,=\,\sum_{i}\,i\,p_{\,i\,.}=\mathrm{E}(U).

∎

Exercise 1.3.2.

Calculate the expectation of the number of times you need to toss a fair coin until you first throw a head. Remember to condition on the outcome of the first toss.

Let $T$ be the time until the first head and $V$ be the outcome of the first throw.

Figure 1.2: Link, Caption: none provided

So, by Theorem 1.3.1,

\mbox{E}(T)=\mathrm{E}(\mathrm{E}(T|V))=\left(\frac{1}{2}\right)1+\left(\frac{% 1}{2}\right)(1+\mbox{E}(T))=1+\frac{1}{2}\mathrm{E}(T).

Solving this linear equation in $\mbox{E}(T)$ gives $\mbox{E}(T)=2$ .

Corollary 1.3.3.

If we consider an event $A$ and random variable $V$

\mathrm{P}(A)=\mathrm{E}\left(\mathrm{P}(A|V)\right).

Remark.

We interpret the right-hand side as

\mbox{E}\left(\mathrm{P}(A|V)\right)=\sum_{j}\mathrm{P}(A|V=j)\mathrm{P}(V=j).

Proof.

Use Theorem 1.3.1 with

U=\left\{\begin{array}[]{cl}1&\mbox{if $A$},\\ 0&\mbox{otherwise},\end{array}\right.

and note that $\mbox{E}(U)=\mathrm{P}(A)$ , and $\mathrm{E}(U|V=j)=\mathrm{P}(A|V=j)$ . ∎