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\begin{document}

\newcommand{\covar}[1]{{\rm Cov}\left[ #1 \right]}

\psol{12}{December 4, 1999 (corrected 12/14)}

\reading F97 Lecture Notes 22, 23 (skipping Wald's Theorem), 24, and H72
notes on expectation.

\iffalse

\textbf{Remember to include a \textbf{collaboration statement} on the
front page of your solutions.}
\fi

\begin{problems}

\problem Eight men and seven women, all single, happen randomly to have
purchased single seats in the same 15-seat row of a theater.

\bparts

\ppart
What is the probability that the first two seats contain a marriageable
couple, i.e., a single man next to a single woman?

\solution{

Here's one way to do it.  There are $\binom{15}{2} = 105$
possible pairs that could occupy the first two seats, but only $8\cdot7 =
56$ of them are marriageable couples.  So the probability is $56/105 =
8/15$.

}

\ppart What is the expected number of pairs of adjacent seats which
contain marriageable couples?  (For example, if the sequence of men and
women in the seats is
\begin{center}
MMMWMMMMWWWWWWM,
\end{center}
then there are four possible couples sitting in the pairs of adjacent
seats that start at the 3rd, 4th, 7th and 14th seats, respectively.)

\solution{

Let the random variable S be the number of pairs of seats containing a
marriageble couple; we want to find $\expect{S}$.  We can solve this problem
by introducing an indicator random variable $S_i$, where $S_i = 1$ if
there is a marriageable couple in seats $i$ and $i+1$, and $S_i = 0$
otherwise.  So we have $S = \sum_{i=1}^{14} S_i$.

Since each seat is like any other, the probability that seats $i$ and
$i+1$ contain a marriageable couple is the same as in the first
part, $8/15$.  Thus
\[
\expect{S_i} = 1\cdot\Pr(S_i=1) + 0\cdot\Pr(S_i=0) = \frac{8}{15}.
\]
Finally, because expectation is linear,
\[
\expect{S} =  \sum_{i=1}^{14} \expect{S_i}  =  14\cdot \frac{8}{15}.
\]

Note that this approach works despite the fact that the indicator random
variables $S_i$ are not independent.  That's the beauty of linearity of
expectation!

}

\eparts

\problem

% modifed from spring 97 Pset 13, 3

There are $n$ MIT students who are taking 6.042 and 6.003 this term.  To
make it easier on themselves, the professors in charge of these classes
have decided to randomly permute their class lists and then assign
students grades based on their rank in the permutation\footnote{\dots just
as many students have suspected \texttt{:-)}}.  Assume all permutations
are equally likely and that the ranking in each class is independent of
the other.

\bparts

\ppart
What is the expected number of students that have a higher rank
in 6.042 than 6.003?

\solution{

Let $X_i$ be random variable whose value is 1 if student i has higher rank
in 6.042 than 6.003 and 0 otherwise.  If student $i$ has rank of $r$ for
6.042 than the probability that student $i$ has higher rank on 6.042,
given that the student has rank $r$ in 6.042, is $(r-1)/n$.  The
probability student $i$ has rank $r$ in 6.042 is $1/n$.  Thus, the
probability that student $i$ has higher rank in 6.042 is
\[
\sum_{1\leq r \leq n}  \frac{1}{n} \frac{r-1}{n} = \frac{n-1}{2n}.
\]
By linearity of expectation, the expected number of students with a higher
rank on 6.042 is
\[
\expect{x} = \expect{x_1}+\expect{x_2}+...+\expect{x_n} = \frac{n-1}{2}.
\]

Another way to solve this problem is to observe that students with higher
rank in one class have lower rank in the other class, so by symmetry, the
expected number of higher- and lower-ranked students in either class is
the same.  The expected number of students with the same rank equals 1 by
linearity of expectation, since probability student ranked the same in
both schools is $1/n$.  These observations now yield same answer as above.

}

\ppart
What is the expected number of students that have a ranking at
least $k$ higher in 6.042 than in 6.003?

\solution{

This part can be done in the same way as the previous one.  Let $X_i$ be
the random variable whose value is 1 if student $i$ is ranked at least $k$
higher in 6.042 than 6.003 and 0 otherwise.  If student $i$ has rank of $r$
$(r>k)$ for 6.042 than the probability that student $i$ has at least $k$
higher rank in 6.042 given student ranked $r$ in 6.042 is $(r-k)/n$.  The
probability student $i$ has rank $r$ in 6.042 is $1/n$.  Thus, the probability
that student $i$ has at least $k$ higher rank in 6.042 is
\[
\sum_{k+1\leq r\leq n} \frac{1}{n} \frac{r-k}{n} = \frac{(n-k)(n-k+1)}{2n^2}.
\]
By linearity of expectation, the expected number of students with a higher
rank on 6.042 is
\[
\expect{x}=\expect{x_1}+\expect{x_2}+...+\expect{x_n} = \frac{(n-k)(n-k+1)}{2n}.
\]

}

\eparts


\problem

Ben Bitdiddle is asked to analyze a game in which a fair coin is tossed
until the first time a head turns up.  If this head occurs on the $n$th
toss, and $n$ is odd, then he wins $\$2^n/n$, but if $n$ is even, he loses
$\$2^n/n$.  Ben observes that the expected dollar win from this game is
the value of the series:
\[
(1/2) \cdot 2 - (1/4) \cdot 2 + (1/8) \cdot 8/3 + \cdots \pm (1/2^n) \cdot 2^n/n
 = 1 - 1/2 + 1/3 - 1/4 + \cdots \pm 1/n.
\]
This is the \emph{alternating harmonic series}---a series whihc is known
to converge to a definite real number $r>0$.  Ben concludes that since $r
>0$, it's to his advantage to play this game.  As usual, Ben's
shoot-from-the-hip analysis is off the mark.  Explain.

\solution{

This series is convergent, but not \emph{absolutely} convergent.  The
definition of expectation requires absolute convergence, so the
expectation in this case is not defined.  The fact that the expectation is
undefined is no mere formality.  The number $r$ is not relevant to the
behavior of the game.  Indeed, we showed in a tutorial problem that by
reordering its terms, the series can be made to converge to any real value
or diverge to $\pm\infty$.

}


\problem 

In the Monte Carlo version of roulette there is only one green
slot---which is numbered 0---as opposed to the Las Vegas version, which
has 2 green slots numbered $0$ and $00$.  If you bet on red (or,
symmetrically, on black) and 0 comes up, one or more of these three
options may be offered:
\begin{enumerate}
\item[(a)] You get $1/2$ of your bet back, and the casino gets the other
half.

\item[(b)] Your bet is put ``in prison.''  If red comes up on the next
turn, you get your bet back (but don't win any money).  If black or 0
comes up, you lose your bet.

\item[(c)] Your bet is put in prison as before.  If red comes up on the
next turn, you get your bet back, and if black comes up on the next turn,
then you lose your bet.  If a 0 comes up on the next turn, then your bet
is put into ``double prison.''  If your bet is in double prison, and if
red comes up on the next turn, then your bet is moved back to ordinary
prison and the game proceeds as before.  If your bet is in double prison,
and if black or 0 come up on the next turn, then you lose your bet.

%(FIGURE).

\end{enumerate}
Note that all three of these options are better than Las Vegas roulette,
in which you simply lose your entire bet if 0 or 00 comes up.  But it
is not clear which is the best choice.  Find the expectation of a
one dollar bet on red under each option.  Which is best?

\solution{

Under option~(a), if red turns up you win 1 dollar, if black turns up
you lose one dollar, and if 0 turns up, you lose 1/2 dollar.  Thus, the
expected winnings are 
\[
1(\frac{18}{37}) + (-1)\frac{18}{37} + (\frac{-1}{2})(\frac{1}{37}) \approx
-0.0135.
\]

Under option~(b), if red turns up you win 1 dollar, if black turns up, you
lose 1 dollar, and if 0 followed by black or 0 turns up, you lose 1
dollar. Thus the expected winnings are
\[
1(\frac{18}{37}) + (-1)(\frac{18}{37}) + (-1)(\frac{1}{37})(\frac{19}{37})
\approx -0.0139.
\]

Under option (c), it is a bit more complicated.  Let $x$ be the
probability that you lose your money given that it is in prsion, and let
$y$ be the probability that you lose given that your money is in double
prison.  Then
\[
x = \frac{18}{37} + \frac{1}{37}y.
\]
Also, 
\[
y = \frac{19}{37} + \frac{18}{37}x.
\]
So, we have 
\[
x = \frac{18}{37} + \frac{1}{37}(\frac{19}{37} + \frac{18}{37}x).
\]
Solving, we have that $x = 685/1351$.  Thus, starting from the beginning,
the probability that you lose your bet is
\[
\frac{18}{37} + \frac{1}{37}x = \frac{25003}{49987}.
\]
Now, the only way you can win is if red comes up on the first turn, which
happens with probability $18/37$.  Thus the expected winnings are
\[
1 \cdot \frac{18}{37} - 1 \cdot \frac{25003}{49987} = -\frac{685}{49987}
\approx -0.0137.
\]

Thus, the simple option~(a) is actually the best.

}

\problem  %revised from S99 Final

Three hundred students take the 6.042 final exam.  The mean on the exam is
$90$ and the lowest score was 30.  You have no other information about the
students and the exam, e.g., you should not assume that the final is worth
100 points.

\bparts

\ppart State a best possible upper bound on the number of students who
scored at least $180$.  \hint Markov.

\solution{Let $R$ be the score.  Apply Markov's Bound to $R-30$:
\[
\prob{R \geq 180} = \prob{R-30 \geq 150} \leq \expect{R-30}/150 = 60/150 = 2/5,
\]
so at most $(2/5)300 = \textbf{120}$ students score $\geq 180$.}

\ppart Give an example set of scores which achieve your bound.

\solution{Of the 300 students, 120 score 180 and 180 score 30.  The mean
is
\[
(120 \cdot 180 + 180 \cdot 30)/300 = 90,
\]
as required.}


\eparts


\problem
Suppose you are playing the game ``Hearts'' with three of your friends.
In Hearts, all the cards are dealt to the players, in this case the four
of you will each have 13 cards.

\bparts

\ppart

What is the expectation and variance of the number of hearts that you
will have in your hand?

\solution{

Let $H$ be the number of hearts in your hand, and let $X_i$ be the
indicator variable that is 1 if card $i$ is a heart and is 0 otherwise, so
$H = \sum_{1\leq i\leq 13} X_i$.  Because $X_i$ is an indicator variable,
its expectation is the probability that card $i$ is a heart, namely, 1/4.
So
\[
\expect{H} = \sum_1^{13} 1/4 = 13/4.
\]

To compute variance, we use the definition:
\begin{eqnarray*}
\variance{\sum_{i=1}^{13} X_i} &=&
    \expect{(\sum_{i=1}^{13} X_i)^2} - \expect{\sum_{i=1}^{13} X_i}^2 \\
        &=& \expect{(\sum_{i=1}^{13} X_i)^2} - (13/4)^2 \\
        &=& \mathrm{E}[X_1^2 + X_1X_2 + X_1X_3 + \ldots + X_1X_{13} + \\
        &&   X_2X_1 + X_2^2 + X_2X_3 + \ldots + X_2X_{13} + \\
        &&        \vdots \\
        &&   X_{13}X_1 + X_{13}X_2 + \ldots + X_{13}^2] - (13/4)^2 \\
        &=& \expect{X_1^2} + \expect{X_1X_2} + \expect{X_1X_3} + \ldots + \expect{X_1X_{13}} + \\
        &&  \expect{X_2X_1} + \expect{X_2^2} + \expect{X_2X_3} + \ldots + \expect{X_2X_{13}} + \\
        &&        \vdots \\
        &&   \expect{X_{13}X_1} + \expect{X_{13}X_2} + \ldots + \expect{X_{13}^2} - (13/4)^2.
\end{eqnarray*}
Now we note that $\expect{X_i^2} = \expect{X_i}$ since $X_i$ is an
indicator variable, so $\expect{X_i^2} = 1/4$.  Also, when $i \neq j$,
$X_iX_j = 1$ if cards $i$ and $j$ are both hearts, and 0 otherwise.  The
probability that both cards are hearts is $(1/4)(12/51)$, and thus
$\expect{X_iX_j} = (1/4)(12/51) = 3/51$.  In the summation above, there
are 13 $\expect{X_i^2}$ terms, and $2\binom{13}{2} = 156$ terms of the
form $\expect{X_iX_j}$.  Thus,
\begin{eqnarray*}
 \variance{\sum_{i=1}^{13} X_i} &=& 13 \cdot \frac{1}{4} + 156 \cdot \frac{3}{51} - (\frac{13}{4})^2 \\
                &\approx& 1.863970588.
\end{eqnarray*}

}

% \textbf{REVISE TO SOLVE W/O COVARIANCE:}

% To compute variance we use the identity:
% \[
% \variance{\sum X_i} = \sum \variance{X_i} + \sum_{i\neq j} \covar{X_i,X_j}.
% \]
% where the second sum is taken over all possible ordered pairs of distinct
% $i$'s and $j$'s.  Now for each indicator variable we have
% \[
% \variance{X_i} = \expect{X_i^2} - \expectsq{X_i} = 1/4 - 1/16 = 3/8,
% \]
% and for each pair of distinct indicator variables we have
% \[
% \covar{X_i, X_j} = \expect{X_i X_j}-\expect{X_i}\expect{X_j} =
% \frac{1}{4} \frac{12}{51} - \frac{1}{16} = \frac{-1}{272}.
% \]
% Putting it all together,
% \begin{eqnarray*}
% \variance{H} & = & 
% \sum_{i=1}^{13}\variance{X_i} + 2\left(\sum_{1<=i<j<=13}\covar{X_i,X_j}\right)\\
%  &  = & 13 \frac{3}{8} + 2\binom{13}{2}\frac{-1}{272} = \frac{585}{136} = 4.30147.
% \end{eqnarray*}


\ppart What is the expectation and variance of the number of suits in
your hand?

\solution{

Similar to the first part: let $N$ denote the number of suits in a hand.
Let $X_s$ be the indicator variable which is 1 if there is a spade in the
hand and 0 otherwise. Define $X_h, X_d, X_c$ analogously for the three
remaining suits: hearts, diamonds, and clubs.  Then $N = X_s + X_h + X_d +
X_c$.

The expectation of each of the indicator variables is the probability
that a hand contains the corresponding suit.  Thus, for $i \in \set{s,h,d,c}$
\[
\expect{X_i} = 1 -  \frac{\binom{52-13}{13}}{\binom{52}{13}} = 0.987209,
\]
because $\binom{52-13}{13}$ is the number of hands that are missing the
particular suit, and $\binom{52}{13}$ is the total number of hands.
\[
\expect{N}=\expect{X_s + X_h + X_d + X_c} = \expect{X_s} + \expect{X_h} +
\expect{X_d} + \expect{X_c} = 4 \cdot 0.987209 = 3.94884.
\]
To compute variance:
\begin{eqnarray*}
 \variance{N} &=& \prob{N=1}(1 - \expect{N})^2 + \prob{N=2}(2- \expect{N})^2 + \\
              & & \qquad \prob{N=3}(3 - \expect{N})^2 + \prob{N=4}(4 - \expect{N})^2.
\end{eqnarray*}
Now we must calculate each of the probabilities.  First,
\[
\prob{N=1} = \frac{4} {\binom{52}{13}} = 6.299 \times 10^{-12},
\]
because there are four possible hands of all one suit out of
$\binom{52}{13}$ possible hands of 13 cards.

\[
\prob{N=2} = \frac{\binom{4}{2}[\binom{26}{13} - 2]}{\binom{52}{13}}
             = 9.827 \times 10^{-5},
\] 
because there are $\binom{4}{2}$ ways to choose which two suits.  There
are $\binom{26}{13}$ ways to choose 13 cards from the 26 possible cards of
those two suits.  Two of those $\binom{26}{13}$ hands actually contain
only one suit.  Thus $\binom{4}{2}(\binom{26}{13} -2)$ is the number of
hands with two suits.

\[
\prob{N=3} = \frac{\binom{4}{3}[\binom{39}{13} -
\binom{3}{2}\binom{26}{13} + 3]}{\binom{52}{13}} = 0.05097,
\]
because there are $\binom{4}{3}$ ways to choose the three suits.  There
are $\binom{39}{13}$ ways to choose 13 cards from the 39 cards of those
three suits.  But that also includes hands with two suits and one suit.
So, using Inclusion-Exclusion, we subtract the
$\binom{3}{2}\binom{26}{13}$ ways to choose two suits from the three suits
and then choose choose 13 cards from the 26 cards of those suits.  Next,
we add back the three ways to choose hands of one suit among the three.
Finally, we divide by the $\binom{52}{13}$ ways to choose a possible hand
of 13.

Of course, 
\[
\prob{N=4} = 1 - (\prob{N=1} +\prob{N=2} + \prob{N=3}) = 0.9489.
\]
This allows us to complete the calculation of the variance:
\[
\variance{N}=0.04874.
\]


%\textbf{REVISE TO SOLVE W/O COVARIANCE:}
% To compute variance, first compute variance of each of the indicator
% variables:
% \[
% \variance{X_i}=\expect{X_i^2} +\expectsq{X_i} = 0.987209 - 0.987209^2 =
% 0.0126273.
% \]
% Note: $\expect{X_i^2} = \expect{X_i}$ since $X_i$ is an indicator variable.


% To compute the covariance of any pair of the indicator variables, we
% need to compute $\expect{X_{i} X_{j}}$, in other words, the
% probability that a hand contains both suits $i$ and $j$. Let $I$ be
% event that the hand contains suit $i$, and $J$ be event that the hand
% contains suit $j$. Then
% \begin{eqnarray*}
% \prob{I \cap J} & = & 1 - \prob{\bar{I} \cup \bar{J}}\\ & = & 1 -
%  \prob{\bar{I}} - \prob{\bar{J}} + \prob{\bar{I} \cap \bar{J}}\\ & = &
%  1 - 2\frac{\binom{52-13}{13}}{\binom{52}{13}} +
%  \frac{\binom{52-13-13}{13}}{\binom{52}{13}} = 0.974434.
% \end{eqnarray*}

% Then covariance is
% \[
% \covar{X_{i},X_{j}} = \expect{X_{i} X_{j}} -
%            \expect{X_{i}}\expect{X_{j}} = 0.974434 - 0.987209^2 =
%            -0.00014723.
% \]

% Putting this all together we get that
% \begin{eqnarray*}
% \variance{N} & = & \variance{X_s} + \variance{X_h} + \variance{X_d} + \variance{X_c}
%        + \sum_{i \neq j \in \set{ s,h,d,c}} \covar{X_{i},X_{j}}\\
%      &  = & 4 \cdot 0.0126273 + 2\binom{4}{2}(-0.00014723) = 0.487424.
% \end{eqnarray*}

}

\eparts


\problem Suppose that $n$ people have their hats returned at random.  Let
$X_i = 1$ if the $i$th person gets his or her own hat back and 0
otherwise.  Let $S_n = \sum_{i=1}^n X_i$, so $S_n$ is the total number of
people who get their own hat back.  Show that

\bparts
\ppart  $\expect{X_i^2} = 1/n$.

\solution{

$X_i = 1$ with probability $1/n$ and 0 otherwise. Thus $X_i^2 = 1$
with probability $1/n$ and 0 otherwise. So $\expect{X_i^2} = 1/n$.

}

\ppart $\expect{X_i X_j} = 1/n(n-1)$ for $i \neq j$.

\solution{

The probability that $X_i$ and $X_j$ are both 1 is $1/n \cdot 1/(n-1) =
1/n(n-1)$.  Thus $X_i X_j = 1$ with probability $1/n(n-1)$, and is zero
otherwise.  So $\expect{X_i X_j} = 1/n(n-1)$.

}

\ppart $\expect{S_n^2} = 2$. \hint Use (a) and (b).

\solution{

\begin{eqnarray*}
\expect{S_n^2} & = & \sum_i \expect{X_i^2} + \sum_i \sum_{j \neq i} \expect{X_i X_j} \\
         & = & n \cdot \frac{1}{n} + n(n-1) \cdot \frac{1}{n(n-1)} \\
         & = & 2.
\end{eqnarray*}

}

\ppart $\variance{S_n} = 1$.

\solution{

\begin{eqnarray*}
\variance{S_n} & = & \expect{S_n^2} - \expectsq{S_n} \\ 
       & = & 2 - (n (1/n))^2 \\
       & = & 2 - 1 \\
       & = & 1.
\end{eqnarray*}

}

\ppart  Using Chebyshev's
Inequality, show that $\prob{S_n \geq 11} \leq .01$ for any $n \geq
11$.

\solution{

\begin{eqnarray*}
\prob{S_n \geq 11} &=& \prob{S_n - \expect{S_n} \geq 11 - \expect{S_n}} \\
                 &=& \prob{S_n - \expect{S_n} \geq 10} \\
                 &\leq& \frac{\variance{S_n}}{10^2} = .01
\end{eqnarray*}

Note that the $X_i$'s are Bernoulli variables but are \emph{not}
independent, so $S_n$ does not have a Bernoulli distribution and the
Bernoulli estimates from Lecture Notes 21 do not apply.

}

\eparts


\problem Psychological diagnosis of sociopaths is notoriously unreliable.
For example, a random psychologist is as likely to diagnose a normal
patient as sociopathic as he is to recognize that the patient is normal.
A random psychologist is actually slightly \emph{less} likely to correctly
diagnose a real sociopath than not---sociopaths are really good liars.

\bparts

\ppart In these circumstances it might seem there was no reliable way to
diagnose sociopaths, but in theory there is a way---if we have a large
enough population of psychologists who reach their judgements
independently.  Explain how to obtain a reliable diagnosis in this case.
Briefly explain why your method will be reliable, citing appropriate
results from class.  A qualitative answer is all that's required---you
need not come up with any numerical estimates or growth rates.

\solution{

Get a reliable test by repeatedly asking randomly chosen psychologists to
do the diagnosis and computing the fraction, $f$, of ``normal'' diagnoses.
Let $s>1/2$ be the probability of a ``normal'' diagnosis on a sociopathic
patient.  Diagnose normal if $f$ is closer to $1/2$ than to $s$.

By diagnosing independently enough times, there is small probability that
$f$ differs from the expected value of a diagnosis by very much (the
expected value is either 1/2 or $s$ depending on whether or not the
patient is normal---we don't know which).  By the Law of Large Numbers, by
Chebyshev's bound, or by the estimation of the binomial distribution as
used in the Gallup Poll example in Lecture Notes, if we diagnose enough
times, we can make this probability as small as desired, so the
probability of a correct diagnosis can be as close to 1 as desired.

}

\ppart Is it plausible to assume that psychologists will make their
diagnoses independently?  Briefly explain.

\solution{

Not really.  It figures that some sociopaths would be particularly good
liars and be able to fool most psychologists.  Conversely, there figure to
be normal people whose anxious or defensive manner makes a sociopathic
diagnosis by many psychologists more likely.  Also, there may be some
specially expert psychologists whose diagnoses are more reliable than
average; this implies if one expert diagnoses ``normal,'' the diagnosis is
likely to be correct, making it more likely that another expert will also
correctly diagnose ``normal,'' so their diagnoses will not be independent
of each other.

}

\eparts


\problem We have two coins: one is a fair coin and the other is a coin
that produces heads with probability 3/4.  One of the two coins is picked,
and this coin is tossed $n$ times.

\bparts

\ppart

Does the Weak Law of Large Numbers allow us to \emph{predict} what limit, if any,
is approached by the expected proportion of heads that turn up as $n$
approaches infinity?  Briefly explain.


\solution{

There is no way to predict the proportion of heads in the long run, since
this will depend upon which of the two coins is picked, and we have no
information about which coin is picked or how it is picked.  So while the
Law of Large Numbers implies that the proportion of heads will, with
probability appraoching one, be near either 1/2 or 3/4, it does not, and
cannot, say anything about which of these numbers is the expected limit.

}

\ppart

How many tosses suffice to make us 95\% confident which coin was chosen?
Explain.


\solution{

To be 95 percent sure which coin was picked, bet that the fair coin was
chosen if the proportion of heads is less than the average of 1/2 and 3/4,
namely 5/8; otherwise bet that the biased coin was chosen.  You will win
the bet if the proportion of heads does not deviate from the probability
of heads by more than 1/8.  A quick calculation using Chebyshev's bound
and the fact that the variance of a Bernoulli variable is at most 1/4,
implies that the probability of a deviation this large is at most
$1/(4n(1/8)^2)$.  This will be less than or equal to .05 if $n > 320$.

In this problem, the number of heads in $n$ flips will be a Bernoulli
distribution, and the estimates from Lecture Notes 21 could also be
applied to yield a tighter bound: $n > 42$.

}

\eparts

\end{problems}

\iffalse

\newpage

\section*{TOPICS:}
\subsection*{Expectation}
\begin{itemize}

\item Definitions.

\item Linearity.

\item of product of random variables (independent and not independent).

\item Expected time to failure.

\item Markov's Theorem.

\end{itemize}

\subsection*{Variance}
\begin{itemize}

\item Definitions.

\item Linearity for independent random variables.

\item Variance of a sum.

\item Variance of Bernoulli distribution.

\item Deviation from the mean: Chebyshev's Theorem and Bound.

\item Weak Law of Large Numbers.

\end{itemize}
\fi

\end{document}
