\begin{problem}
The Queen of Neverwas (a.k.a. Nancy Plimpton of Ames, Iowa) was
disappointed that her first two proposals to make there be more girls
than boys in her Queendom both failed.  However, she is considering a
new decree:

\begin{quotation}
Henceforth in Neverwas, every couple must have children until they
have more girls than boys and then they may have no more.
\end{quotation}

Nancy, er, the Queen is very pleased with this plan.  Now the number
of girls will certainly be greater than the number of boys!

Consider a particular couple.  Let the random variable $G$ be the
number of girls they have, and let $B$ be the number of boys.  Assume
that $\expect{G}$ and $\expect{B}$ both exist.

\bparts

\ppart Show that $\expect{G} = \expect{B} + 1$.

\ppart Now let's compute the expected number of boys in a family using
Wald's theorem.

\begin{theorem}[Wald]
Let $C_1, C_2, \dots,$ be a sequence of nonnegative random variables,
and let $Q$ be a positive integer-valued random variable, all with
finite expectations.  Suppose that
\[
\expcond{ C_i}{Q \geq i} = \mu
\]
for some $\mu \in \reals$ and for all~$i \geq 1$.  Then
\[
\expect{C_1 + C_2 + \cdots + C_Q} = \mu \expect{Q}.
\]
\end{theorem}

Let the random variable $Q = B + G$ be the total number of children in
a family.  Let $C_i$ be an indicator for the event that the $i$-th
child is a boy.  Suppose that $\expect{Q}$ has finite expectation.
Use Wald's Theorem to prove that:

\begin{eqnarray*}
\expect{B} = \frac{1}{2} \expect{B + G}
\end{eqnarray*}

\ppart Putting together the conclusions from the two preceding problem
parts, what can you say about $\expect{Q}$, the expected number of
children in a family?  What is wrong with Queen Nancy's decree?

\eparts

\end{problem}

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