(Source: Peter M. Lee, Bayesian Statistics: an introduction, Fourth Edition, 2012)
Whether certain mice are black or brown depends upon a pair of genes, each of which is either \(B\) or \(b\). If both members of the pair are alike, the mouse is said to be homozygous, and if they are different it is said to be heterozygous. The mouse is brown only if it is homozygous \(bb\). The offspring of a pair of mice have two such genes, one from each parent, and if the parent is heterozygous, the inherited gene is equally likely to be \(B\) or \(b\). Suppose that a black mouse results from a mating between two heterozygotes.
Now suppose that this mouse is mated with a brown mouse, resulting in seven offspring, all of which turn out to be black.
Use Bayes’ Theorem to find the probability that the black
mouse was homozygous \(BB\).
Let \(\mbox{G1}\) denote the
event that the black mouse is mated with a brown mice and \(\mbox{G2-7BO}\) the event that the seven
offspring of \(\mbox{G1}\) are black.
Then, assuming that each offspring is independent, \[\begin{eqnarray*}
P(\mbox{G2-7BO} \, | \, \mbox{G1-BB}, \mbox{G1}, \mbox{G0}) & =
& 1, \\
P(\mbox{G2-7BO} \, | \, \mbox{G1-Bb}, \mbox{G1}, \mbox{G0}) & =
& \left(\frac{1}{2}\right)^{7}.
\end{eqnarray*}\] These can be viewed as likelihoods. Using the
theorem of total probability, \[\begin{eqnarray*}
P(\mbox{G2-7BO} \, | \, \mbox{G1}, \mbox{G0}) & = &
P(\mbox{G2-7BO} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G1}, \mbox{G0}) + \\
& & P(\mbox{G2-7BO} \, | \, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0})P(\mbox{G1-Bb} \, | \, \mbox{G1}, \mbox{G0}) \\
& = & P(\mbox{G2-7BO} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G1-black}, \mbox{G0}) + \\
& & P(\mbox{G2-7BO} \, | \, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0})P(\mbox{G1-Bb} \, | \, \mbox{G1-black}, \mbox{G0}) \\
& = & 1\left(\frac{1}{3}\right) +
\left(\frac{1}{2}\right)^{7}\left(\frac{2}{3}\right) \\
& = & \frac{1}{3}\left(1 + \frac{1}{2^{6}}\right).
\end{eqnarray*}\] Thus, using Bayes’ theorem we find \[\begin{eqnarray*}
P(\mbox{G1-BB} \, | \, \mbox{G2-7BO}, \mbox{G1}, \mbox{G0})
& = & \frac{P(\mbox{G2-7B0} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G1}, \mbox{G0})}{P(\mbox{G2-7BO}
\, | \, \mbox{G1}, \mbox{G0})} \\
& = & \frac{1\left(\frac{1}{3}\right)}{\frac{1}{3}\left(1 +
\frac{1}{2^{6}}\right)} \\
& = & \frac{2^{6}}{2^{6} + 1} \ = \ \frac{64}{65}.
\end{eqnarray*}\] Note the form of this equation. We can view
\(P(\mbox{G1-BB} \, | \, \mbox{G1},
\mbox{G0})\) as our prior probability for the event G1-BB, \(P(\mbox{G2-7B0} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})\) our likelihood (the probability of observing the
seven black offspring given that we knew G1-BB) and \(P(\mbox{G1-BB} \, | \, \mbox{G2-7BO}, \mbox{G1},
\mbox{G0})\) our posterior probability for the event G1-BB. We
shall make explicit use of this relationship in the next part of the
question.
Recalculate the same probability by regarding the seven
offspring as seven observations made sequentially, treating the
posterior after each observation as the prior for the next.
Let \(\mbox{G2-BOi}\) denote
the event that the \(i\)th offspring is
black and \(\mbox{G2-iBO}\) denote the
event that there have been \(i\) black
offspring. Thus, \(\mbox{G2-iBO} =
\mbox{G2-BOi} \cap \mbox{G2-(i-1)BO}\). After observing \((i-1)\) offspring, the probability for the
event \(\mbox{G1-BB}\) is \(P(\mbox{G1-BB} \, | \, \mbox{G2-(i-1)BO},
\mbox{G1}, \mbox{G0})\) whilst the probability for the event
\(\mbox{G1-Bb}\) is \(P(\mbox{G1-Bb} \, | \, \mbox{G2-(i-1)BO},
\mbox{G1}, \mbox{G0})\). Given the ancestral history, we assume
that the offspring are independent. In particular, for each \(i = 1, \ldots, 7\), we have the likelihoods
\[\begin{eqnarray*}
P(\mbox{G2-BOi} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1-BB}, \mbox{G1},
\mbox{G0}) \ = \ P(\mbox{G2-BOi} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})\ = \ 1, \\
P(\mbox{G2-BOi} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0}) \ = \ P(\mbox{G2-BOi} \, | \, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0}) \ = \ \frac{1}{2}.
\end{eqnarray*}\] Then, using Bayes’ theorem we have \[\begin{eqnarray*}
P(\mbox{G1-BB} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) & =
& P(\mbox{G1-BB} \, | \, \mbox{G2-B0i}, \mbox{G2-(i-1)BO},
\mbox{G1}, \mbox{G0}) \\
& = & \frac{P(\mbox{G2-B0i} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0})}{P(\mbox{G2-BOi} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0})}, \\
P(\mbox{G1-Bb} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) & =
& P(\mbox{G1-Bb} \, | \, \mbox{G2-B0i}, \mbox{G2-(i-1)BO},
\mbox{G1}, \mbox{G0}) \\
& = & \frac{P(\mbox{G2-B0i} \, | \, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0})P(\mbox{G1-Bb} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0})}{P(\mbox{G2-BOi} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0})}
\end{eqnarray*}\] Thus, noting that the denominator \(P(\mbox{G2-BOi} \, | \, \mbox{G2-(i-1)BO},
\mbox{G1}, \mbox{G0})\) is common to both terms, \[\begin{eqnarray*}
P(\mbox{G1-BB} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) &
\propto & P(\mbox{G2-B0i} \, | \, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0}), \\
P(\mbox{G1-Bb} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) &
\propto & P(\mbox{G2-B0i} \, | \, \mbox{G1-Bb}, \mbox{G1},
\mbox{G0})P(\mbox{G1-Bb} \, | \, \mbox{G2-(i-1)BO}, \mbox{G1},
\mbox{G0})
\end{eqnarray*}\] which are of the form “Posterior \(\propto\) Likelihood \(\times\) Prior”. The constant of
proportionality can be found by noting \[\begin{eqnarray*}
P(\mbox{G1-BB} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) +
P(\mbox{G1-Bb} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) \ = \ 1.
\end{eqnarray*}\] We now add the data about the seven offspring
sequentially. Our initial priors are \(P(\mbox{G1-BB} \, | \, \mbox{G1-black}, \mbox{G0})
= \frac{1}{3}\) for G1-BB and \(P(\mbox{G1-Bb} \, | \, \mbox{G1-black}, \mbox{G0})
= \frac{2}{3}\) for G1-Bb. We now update these beliefs after
observing the event G2-1BO. We have: \[\begin{eqnarray*}
\begin{array}{c|cccc}
\mbox{Event} & \mbox{Prior} & \mbox{Likelihood} &
\mbox{Likelihood $\times$ Prior} & \mbox{Posterior} \\ \hline
\mbox{G1-BB} & \frac{1}{3} & 1 & \frac{1}{3} &
\frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} = \frac{1}{2} \\
\mbox{G1-Bb} & \frac{2}{3} & \frac{1}{2} & \frac{1}{3} &
\frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} = \frac{1}{2}
\end{array}
\end{eqnarray*}\] We now add the second black offspring so that
we want our beliefs after observing the event G2-2BO. We use our
posteriors from the previous stage as our new priors, that is \(P(\mbox{G1-BB} \, | \, \mbox{G2-1BO}, \mbox{G1},
\mbox{G0}) = \frac{1}{2}\) and \(P(\mbox{G1-Bb} \, | \, \mbox{G2-1BO}, \mbox{G1},
\mbox{G0}) = \frac{1}{2}\) act as our priors. We have \[\begin{eqnarray*}
\begin{array}{c|cccc}
\mbox{Event} & \mbox{Prior} & \mbox{Likelihood} &
\mbox{Likelihood $\times$ Prior} & \mbox{Posterior} \\ \hline
\mbox{G1-BB} & \frac{1}{2} & 1 & \frac{1}{2} &
\frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{4}} = \frac{2}{3} \\
\mbox{G1-Bb} & \frac{1}{2} & \frac{1}{2} & \frac{1}{4} &
\frac{\frac{1}{4}}{\frac{1}{2} + \frac{1}{4}} = \frac{1}{3}
\end{array}
\end{eqnarray*}\] We can continue in this vein. A straightforward
inductive argument shows that \[\begin{eqnarray*}
P(\mbox{G1-BB} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) & =
& \frac{2^{i-1}}{1 + 2^{i-1}}, \\
P(\mbox{G1-BB} \, | \, \mbox{G2-iBO}, \mbox{G1}, \mbox{G0}) & =
& \frac{1}{1 + 2^{i-1}}.
\end{eqnarray*}\] Adding the seventh black offspring thus gives
\[\begin{eqnarray*}
\begin{array}{c|cccc}
\mbox{Event} & \mbox{Prior} & \mbox{Likelihood} &
\mbox{Likelihood $\times$ Prior} & \mbox{Posterior} \\ \hline
\mbox{G1-BB} & \frac{32}{33} & 1 & \frac{32}{33} &
\frac{\frac{32}{33}}{\frac{32}{33} + \frac{1}{66}} = \frac{64}{65} \\
\mbox{G1-Bb} & \frac{1}{33} & \frac{1}{2} & \frac{1}{66}
& \frac{\frac{1}{66}}{\frac{32}{33} + \frac{1}{66}} = \frac{1}{65}
\end{array}
\end{eqnarray*}\] We now adopt this explicit “Posterior \(\propto\) Likelihood \(\times\) Prior” approach for the answers to
parts a and b.
Part a. Learning that G1-black.
\[\begin{eqnarray*}
\begin{array}{c|cccc}
\mbox{Event} & \mbox{Prior} & \mbox{Likelihood} &
\mbox{Likelihood $\times$ Prior} & \mbox{Posterior} \\ \hline
\mbox{G1-BB} & \frac{1}{4} & 1 & \frac{1}{4} &
\frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2} + 0} = \frac{1}{3} \\
\mbox{G1-Bb} & \frac{1}{2} & 1 & \frac{1}{2} &
\frac{\frac{1}{2}}{\frac{1}{4} + \frac{1}{2}+0} = \frac{2}{3} \\
\mbox{G1-bb} & \frac{1}{4} & 0 & 0 & 0
\end{array}
\end{eqnarray*}\] Part b. Learning that G1-black
mates with a brown mouse, resulting in seven black offspring. We can
remove the event G1-bb as this has zero prior probability. \[\begin{eqnarray*}
\begin{array}{c|cccc}
\mbox{Event} & \mbox{Prior} & \mbox{Likelihood} &
\mbox{Likelihood $\times$ Prior} & \mbox{Posterior} \\ \hline
\mbox{G1-BB} & \frac{1}{3} & 1 & \frac{1}{3} &
\frac{\frac{1}{3}}{\frac{1}{3} +
\frac{1}{3}\left(\frac{1}{2}\right)^{6}} = \frac{64}{65} \\
\mbox{G1-Bb} & \frac{2}{3} & \left(\frac{1}{2}\right)^{7} &
\frac{1}{3}\left(\frac{1}{2}\right)^{6} &
\frac{\frac{1}{3}\left(\frac{1}{2}\right)^{6}}{\frac{1}{3} +
\frac{1}{3}\left(\frac{1}{2}\right)^{6}} = \frac{1}{65} \\
\end{array}
\end{eqnarray*}\]
What is the probability that an eighth offspring is also
black.
Here we are looking at the predictive
distribution. Let G2-BO8 denote the event that the eighth offspring is
black. The solution is once again to use total probability with the two
events G1-BB and G1-Bb. We have \[\begin{eqnarray*}
P(\mbox{G2-BO8} \, | \, \mbox{G2-7BO}, \mbox{G1}, \mbox{G0}) & =
& P(\mbox{G2-BO8} \, | \, \mbox{G2-7B0}, \mbox{G1-BB}, \mbox{G1},
\mbox{G0})P(\mbox{G1-BB} \, | \, \mbox{G2-7BO}, \mbox{G1}, \mbox{G0}) +
\\
& & P(\mbox{G2-BO8} \, | \, \mbox{G2-7B0}, \mbox{G1-Bb},
\mbox{G1}, \mbox{G0})P(\mbox{G1-Bb} \, | \, \mbox{G2-7B0}, \mbox{G1},
\mbox{G0}) \\
& = & 1\left(\frac{64}{65}\right) +
\frac{1}{2}\left(\frac{1}{65}\right) \ = \ \frac{129}{130}.
\end{eqnarray*}\]