Solution

(i): Define the person's game as the infinite sequence ${\displaystyle \omega =\{\omega _{1},\omega _{2},...\}}$ where each ${\displaystyle \omega _{k}}$ equals either 1 (corresponding to a win) or 0 (corresponding to a loss).

Define the random variable ${\displaystyle \tau _{k}:\Omega \to \mathbb {N} }$ by

${\displaystyle \tau _{k}(\omega )=\{\#{\text{ times }}\omega _{j}=\omega _{j-1}=1,\,\forall 2\leq j\leq k\}}$ that is, ${\displaystyle \tau _{k}}$ counts how many times the player received two consecutive wins in his first ${\displaystyle k}$ games. Thus, the player will win ${\displaystyle \tau _{k}}$ dollars in the first ${\displaystyle k}$ games. Clearly, ${\displaystyle \tau _{k}}$ is measurable. Moreover, we can compute the expectation:

${\displaystyle E[\tau _{k}(\omega )]=\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}.}$

Now observe what happens as we send ${\displaystyle k\to \infty }$:

${\displaystyle E[\lim _{k\to \infty }\tau _{k}(\omega )]=\lim _{k\to \infty }\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}=\infty }$

Hence the expected winnings of the infinite game is also infinite. This implies that the player will surpass $${\displaystyle A}$ in winnings almost surely.

(ii): Define everything as before except this time ${\displaystyle \tau _{k}(\omega )=\{\#{\text{ times }}\omega _{j}=\omega _{j-1}=\omega _{j-2}=1,\,\forall 3\leq j\leq k\}.}$

Then ${\displaystyle E[\tau _{k}(\omega )]=\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}{\frac {1}{\sqrt {j-2}}}.}$ which gives ${\displaystyle E[\lim _{k\to \infty }\tau _{k}(\omega )]<\infty .}$ Thus we cannot assert that the probability of surpassing any given winnings will equal 1.

Solution

This is just a direct application of Bayes' theorem. Let ${\displaystyle N}$ denote the event that you pulled a normal coin. Let ${\displaystyle T}$ denote the even that you have a tail.

By Bayes,

${\displaystyle P(N|T)={\frac {P(N\cap T)}{P(T)}}={\frac {5/20}{13/20}}=5/13.}$

The probability of seeing a tail on a normal coin, ${\displaystyle P(N\cap T)}$ 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).

Solution

(i) Let ${\displaystyle P}$ be the Markov transition matrix. I claim that for any initial probability distribution, ${\displaystyle \mu }$, then ${\displaystyle E[\mu ]\leq E[\mu P]}$.

Proof of claim: It is sufficient to consider the case where the initial distribution is singular, i.e. ${\displaystyle \mu =\chi _{n}}$. Clearly we can see that ${\displaystyle n=E[\chi _{n}]}$. Then ${\displaystyle E[\chi _{n}P]=2}$ if ${\displaystyle n=1}$ and for ${\displaystyle n\geq 2}$ we have ${\displaystyle E[\chi _{n}P]=1/2(n-1)+1/2(n^{2})\geq n}$.

Now let ${\displaystyle f(n)=1/n}$. We want to compute ${\displaystyle E[f(X_{t})-f(X_{s})|{\mathcal {F}}_{s}]}$ for ${\displaystyle t>s}$.

${\displaystyle E[f(X_{t})-f(X_{s})|{\mathcal {F}}_{s}]=E[f(X_{s}P^{t-s})]-E[f(X_{s})]=E[1/(X_{s}P^{t-s})]-E[1/(X_{s})]\geq 0}$ where the last inequality comes from our claim above. This shows that ${\displaystyle f(X_{s})}$ is a supermartingale.

Solution

(i) Notice that

${\displaystyle |\sum _{n=1}^{\infty }e^{-n}\xi _{n}|\leq \sum _{n=1}^{\infty }e^{-n}={\frac {1}{1-e}}.}$ So the series is bounded. Moreover, it must be Cauchy. Indeed for any ${\displaystyle \epsilon >0}$ we can select ${\displaystyle N}$ sufficiently large so that for every ${\displaystyle n,m>N}$, ${\displaystyle \sum _{k=n}^{m}e^{-k}<\epsilon .}$ Hence, the series ${\displaystyle \sum _{n=1}^{\infty }e^{-n}\xi _{n}}$ converges almost surely.

(ii) To show that ${\displaystyle \xi }$ is supported on a set of Lebesgue measure zero, first recall some facts about the Cantor set.

The Cantor set ${\displaystyle C}$ is the set of all ${\displaystyle x\in [0,1]}$ with ternary expansion ${\displaystyle x=.a_{1}a_{2}\cdots ,\,a_{j}=0{\text{ or }}2}$ (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 ${\displaystyle E}$ consisting of all ${\displaystyle x\in [0,1]}$ with expansion ${\displaystyle x=.a_{1}a_{2}\cdots ,\,a_{j}=0{\text{ or }}2}$ in base ${\displaystyle e}$. There exists an obvious bijection between the elements of ${\displaystyle E}$ and ${\displaystyle \xi }$. Since the Lebesgue measure of ${\displaystyle E}$ is ${\displaystyle \lim _{n\to \infty }({\frac {2}{e}})^{n}=0}$. Hence ${\displaystyle \xi }$ has support on a set of Lebesgue measure zero.

Solution

This is a direct appliction of Central Limit Theorem, Lindeberg Condition.

We know that each random variable ${\displaystyle \xi _{i}}$ has mean ${\displaystyle i^{2}/2}$ and variance ${\displaystyle i^{4}/12}$.

Then ${\displaystyle a_{n}=\sum _{i=1}^{n}i^{2}/2={\frac {n(n+1)(2n+1)}{12}}}$ and ${\displaystyle b_{n}^{2}=\sum _{i=1}^{n}i^{4}/12}$. Then ${\displaystyle (\sum _{i=1}^{n}\xi _{i}-a_{n})/b_{n}}$ converges in distribution to the standard normal provided the Lindeberg condition holds.

Hence we want to check ${\displaystyle \lim _{n\to \infty }{\frac {1}{b_{n}^{2}}}\sum _{i=1}^{n}\int _{\{x:|x-m_{i}|\geq \epsilon b_{n}\}}(x-m_{i})^{2}\,dF_{i}(x)=0}$

Since ${\displaystyle b_{n}}$ grows faster than ${\displaystyle n^{2}}$ then for sufficiently large ${\displaystyle n}$, the domain of each integral is empty. Hence the above equation goes to 0 as ${\displaystyle n\to \infty }$. Thus the Lindeberg condition is satisfied and CLT holds.

Solution

(i) Let ${\displaystyle A=\{\omega \in \Omega |\lim _{t\to 0}X_{t}(\omega )=X(\omega )\}}$. By assumption ${\displaystyle P(A)=1}$. Now we compute the ${\displaystyle L^{2}}$ norm:

${\displaystyle \lim _{t\to 0}E|X_{t}-X|^{2}=\lim _{t\to 0}\int _{A}|X_{t}-X|^{2}\,d\omega +\lim _{t\to 0}\int _{\Omega \setminus A}|X_{t}-X|^{2}\,d\omega .}$

Let us evaluate the first integral on the right-hand side. We can write ${\displaystyle \lim _{t\to 0}\int _{A}|X_{t}-X|^{2}\,d\omega =\lim _{t\to 0}\int _{A}|X_{t}|^{2}\,d\omega +\int _{A}|X|^{2}\,d\omega -2\lim _{t\to 0}\int _{A}|XX_{t}|^{2}\,d\omega }$

${\displaystyle \quad \leq \lim _{t\to 0}E|X_{t}|^{2}+E|X|^{2}-2\int _{A}\lim _{t\to 0}|XX_{t}|^{2}\,d\omega }$ by Fatou's lemma

${\displaystyle =E|X|^{2}+E|X|^{2}-2E|X|^{2}=0}$ (since ${\displaystyle E|X_{t}|^{2}=E|X|^{2}\,\forall t}$).

Now the second term:

${\displaystyle \lim _{t\to 0}\int _{\Omega \setminus A}|X_{t}-X|^{2}\,d\omega \leq \int _{\Omega \setminus A}|X|^{2}\,d\omega +\lim _{t\to 0}\int _{\Omega \setminus A}|X_{t}|^{2}\,d\omega }$ by the triangle inequality.

${\displaystyle \leq (E|X|^{2}+E|X|^{2})P(\Omega \setminus A)=0}$ since ${\displaystyle X,X_{t}}$ all have finite second moments.

Thus we have just shown that under the above assumptions, almost sure convergence implies convergence in mean square.

(ii)