% FA 2001, PS 12, problem 1
\begin{problem}
{\bf Waiting for Godot}
A couple plans to have children until they have a boy, whom they'd like to name Godot.  What is the expected number of children that they have, and what is the
variance?

\solution{The probability of having a single child (a boy) is
$\frac{1}{2}$, the probability of having two children (a girl and then
a boy) is $\frac{1}{4}$, and so forth; the probability of having $k$
children is $\frac{1}{2^k}$.  Therefore, the expected number of
children is:

\begin{eqnarray*}
\expect{\text{children}}& = &  \sum_{k=1}^{\infty} k \cdot \frac{1}{2^k} \\
                       & = &   2
\end{eqnarray*}

Now, let $C = 2$ be the expected number of children up to and including
the first boy. We can compute the variance as follows:
\[
\variance{C}
         = \expect{C^2} - \expect{C}^2 
         = \sum_{k=1}^{\infty} k^2 \cdot \left(\frac{1}{2}\right)^k - 2^2 
         = \frac{\frac{1}{2} + (\frac{1}{2})^2}{(1 - \frac{1}{2})^3} - 4 
         = 6 - 4
         = 2
\]
The sum is computed by differentiating the
formula for the sum of an infinite geometric sequence.
}
\end{problem}
