\documentclass{2800hw}
\usepackage{amsmath,amssymb,latexsym}
\usepackage{verbatim}

\renewcommand{\S}{\mathcal{S}}
\renewcommand{\Pr}{\mathcal{P}}

\begin{document}

\topic{More probability}
\homework{2}

\maketitle

\begin{exercises}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item
Alice, Bob and Carol share an apartment. One day, their fire alarm goes off and
the fire brigade arrives to find the apartment empty and a pan left smoking on
the lit stove. A neighbor tells the fire marshal that Bob is likely to be the
culprit, since he is the most absent-minded and forgets to turn off the stove
half the time he uses it. Alice forgets to turn it off only a third of the time
she uses it, and Carol only a quarter of the time.

\begin{enumerate}
\item Not knowing how often each of the three uses the kitchen to cook, the
fire marshal assigns a uniform prior of 1/3 to the possibility that each was
cooking. In the fire marshal's opinion, what is the probability Bob left the
stove on?

\item Dave, who is a close friend of the trio, knows that Alice does most of
the cooking in the apartment, Carol hardly ever cooks (1\% of the time), and Bob
cooks occasionally (15\% of the time). According to Dave, what is the
probability Bob left the stove on?
\end{enumerate}

\noindent Show the reasoning and calculations for your answers.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item
If $B \subseteq A$ and $C \subseteq A$, and none of $P(A)$, $P(B)$ and $P(C)$
is zero, then show that
\[
  \frac{P(C)}{P(B)} \ = \ \frac{P(C \mid A)}{P(B \mid A)}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item
We often want to perform two experiments and wish to reason about the combined
outcomes of the two.  For example, the first experiment could be a coin toss,
and the second a dice roll, and we might ask questions like `` what is the
probability that the coin was heads and the die showed
3''.

When we consider these events, we are implicitly constructing a new probability
space whose outcomes are pairs of outcomes from the original experiments.  We
typically assign probabilities to these outcomes by multiplying probabilities
from the original experiments.  In this problem we will formalize that
procedure.

Suppose that $(\S_1, \Pr_1)$ is a probability space (that is, $\S_1$ is a set, and
$\Pr_1$ assigns a probability to each event of $\S_1$, satisfying the Kolmogorov
axioms).  Suppose also that $(\S_2, \Pr_2)$ is a probability space.  Suppose also
that $\S_1$ and $\S_2$ are finite.

We build a new probability space $(\S, \Pr)$ that models the outcome of both
experiments as follows: The outcomes in $\S$ will just be pairs of outcomes of
the two given probability spaces:
\begin{displaymath}
\S = \S_1 \times \S_2 = \{(x,y) \mid x \in \S_1 \textrm{ and } y \in \S_2\}
\end{displaymath}

We choose the probability of a combined event by multiplying the
probabilities of the component events:
\begin{displaymath}
\Pr(A) = \sum_{(x,y) \in A} \Pr_1(\{x\}) \cdot \Pr_2(\{y\})
\end{displaymath}

\begin{enumerate}
\item % a
Prove that $(\S, \Pr)$ is a probability space.

\item % b
Write down the two events ``the outcome of the first experiment is $x$'' and
``the outcome of the second experiment is $y$'' using set notation.  These should
be events of the space $(\S, \Pr)$ (in other words, subsets of $\S$).

\item % c
Generalizing slightly, if $A_1$ is an event of the probability space $\S_1$, use
set notation to express the event $\hat{A}_1$ representing the fact that $A_1$
occured in the first experiment.  Similarly, if $A_2$ is an event of $\S_2$,
write down the event $\hat{A}_2$ expressing the the fact that $A_2$ occured in
the second experiment.

\item % d
Prove that $\hat{A}_1$ and $\hat{A}_2$ from part (c) are independent.
\end{enumerate}

\end{exercises}

\end{document}
