A person plays an infinite sequence of games. He wins the th game with probability , independently of the other games.
(i) Prove that for any , the probability is one that the player will accumulate dollars if he gets a dollar each time he wins two games in a row.
(ii) Does the claim in part (i) hold true if the player gets a dollar only if he wins three games in a row? Prove or disprove it.
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(i): Define the person's game as the infinite sequence
where each
equals either 1 (corresponding to a win) or 0 (corresponding to a loss).
Define the random variable
by
that is,
counts how many times the player received two consecutive wins in his first
games. Thus, the player will win
dollars in the first
games. Clearly,
is measurable. Moreover, we can compute the expectation:
Now observe what happens as we send
:
Hence the expected winnings of the infinite game is also infinite. This implies that the player will surpass $
in winnings almost surely.
(ii): Define everything as before except this time
Then
which gives
Thus we cannot assert that the probability of surpassing any given winnings will equal 1.
There are 10 coins in a bag. Five of them are normal coins, one coin has two heads and four coins have two tails. You pull one coin out, look at one of its sides and see that it is a tail. What is the probability that it is a normal coin?
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This is just a direct application of Bayes' theorem. Let
denote the event that you pulled a normal coin. Let
denote the even that you have a tail.
By Bayes,
The probability of seeing a tail on a normal coin,
is 5/20 since there are five tails on normal coins out of all 20 faces. The probability of seeing a tail is 13 out of 20 (5 normal + 2*4 double).
(i) Let
be the Markov transition matrix. I claim that for any initial probability distribution,
, then
.
Proof of claim: It is sufficient to consider the case where the initial distribution is singular, i.e.
. Clearly we can see that
. Then
if
and for
we have
.
Now let
. We want to compute
for
.
where the last inequality comes from our claim above. This shows that
is a supermartingale.
Let be i.i.d. random variables with .
(i) Prove that the series converges with probability one.
(ii) Prove that the distribution of is singular, i.e., concentrated on a set of Lebesgue measure zero.
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(i) Notice that
So the series is bounded.
Moreover, it must be Cauchy. Indeed for any
we can select
sufficiently large so that for every
,
Hence, the series
converges almost surely.
(ii) To show that
is supported on a set of Lebesgue measure zero, first recall some facts about the Cantor set.
The Cantor set
is the set of all
with ternary expansion
(in base 3). This corresponds to the usual Cantor set which can be thought of the perfect symmetric set with contraction 1/3.
Instead, consider the set
consisting of all
with expansion
in base
. There exists an obvious bijection between the elements of
and
. Since the Lebesgue measure of
is
. Hence
has support on a set of Lebesgue measure zero.
This is a direct application of Central Limit Theorem, Lindeberg Condition.
We know that each random variable
has mean
and variance
.
Then
and
. Then
converges in distribution to the standard normal provided the Lindeberg condition holds.
Hence we want to check
Since
grows faster than
then for sufficiently large
, the domain of each integral is empty. Hence the above equation goes to 0 as
. Thus the Lindeberg condition is satisfied and CLT holds.
(i) Let be random variables defined on a probability space . Assuming that for all , prove that implies , i.e. under the above assumptions, almost sure convergence implies convergence in mean square.
(ii) Let be a random process with the property that and are finite and do not depend on (such a process is called wide-sense stationary). Prove that the correlation function is continuous if the trajectories of are continuous.
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(i) Let
. By assumption
. Now we compute the
norm:
Let us evaluate the first integral on the right-hand side. We can write
by Fatou's lemma
(since
).
Now the second term:
by the triangle inequality.
since
all have finite second moments.
Thus we have just shown that under the above assumptions, almost sure convergence implies convergence in mean square.
(ii)