7.1:

Gambler with initial capital \(a\) plays until his fortune increases \(b\) units or he is ruined. Recall that \(\rho = \frac{q}{p} > 1\), where \(p\) is the probability of success and \(q\) is the probability of failure. Suppose that the chance of success is multiplied by \(1+\theta\) if his initial capital is infinite instead of \(a\). Show that \(0 < \theta < (\rho ^{a} -1)^{-1} < (a(\rho -1))^{-1}\)

Solution: Recall that the probability of of success with initial capital is given by \(S= \frac{\rho^a -1}{\rho^{a+b}-1} \leq\ \rho^{-b}\) where the inequality follows from the probability of success with infinite capital. Then, adapting what we have from the problem itself:

\(S= \frac{\rho^a -1}{\rho^{a+b}-1} (1+\theta)= \rho^{-b}\), then solve for \(\theta\)

This yields that \(\theta = \frac{1-\rho^{-b}}{\rho^{a}-1} < \frac{1}{\rho^a -1}\)

Note that since \(\rho >1\), we have that \(a(\rho -1) < \rho^a -1\) and thus \(\frac{1}{\rho^a -1} < \frac{1}{a(\rho -1)}\) Hence, \(\theta < \frac{1}{a(\rho -1)}\)

7.2:

Gamblers Ruin/Fortune deduced from SLoLN: \(\begin{equation} P( \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} x_i = p-q) = 1\end{equation}\), where \(p-q \neq 0\).

As Peng showed in class:

\(\begin{equation} P(\lim_{n \rightarrow \infty} \frac{\sum X_i}{n}=0)>P(\frac{-a}{n} < \frac{\sum X_i}{n} < \frac{b}{n})=0 \end{equation}\)

So the strong law of large numbers imples here that right hand side will go to zero. I.E. That there will never be continuous play. If play never terminates, observe that \(S_c(a) <1\), the probability that fortune \(c\) will be achieved from initial \(a\). So then the probability of \(c\) succesive wins is bounded by \(\rho ^{-b}\), the probability of achieving \(c\) with unlimited fortune. So then from the Borel Cantelli Lemma, this will tell us that the probability of getting \(c\) succesive wins goes to 0.

8.2:

Suppose we had the sequence of events \(\{ X_0 =0 , \ldots , X_n=0\} = \{Y_0 = \ldots = Y_{n+1}=0 \} \cup \{Y_0 = \ldots = Y_{n+1} = 1 \} \)

Where the equality follows from the fact that either all of the \(Y\)'s are 0 or they are all 1 to achieve equality. Since \(X_n = 0 \implies Y_n = Y_{n+1}\)

So then \(P(X_0 = 0, X_1 = 0, \ldots, X_n = 0) = P(Y_0 = 0, \ldots Y_{n+1} = 0 \} + P(Y_0 = 1, \ldots, Y_{n+1} = 1 \} = p^{n+2} + q^{n+2} \)

So the probability that the next \(X\) is also 0 would be: \(P(X_{n+1} = 0 | X_1 = 0, \ldots ,X_n = 0) = \frac{p^{n+2} +q^{n+2}}{p^{n+1} +q^{n=1}}\)

Which is different from \(P(X_{n+1} = 0 | X_n =0) = \frac{P(X_{n+1} = 0, X_n = 0)}{P(X_n =0)} = \frac{p^3 +q^3}{p^2+q^2} \)

Hence, the \(X\)'s do not form a Markov Chain.

However, we do have that \(X_{n-1} = Y_{n-1} + Y_n\) and \(X_{n+1} = Y_{n+1} + Y_{n+2}\) So we do have that \(X_{n-1}\) and \(X_{n+1}\) being independent. So the given relationship does hold.

However, the last relation might not hold. Consider the markov chain given in the book in 8.13. If there was an unsuccessful attempt on the \(n-1\)st try. Then the probability of success on the \(n+1\)st try given that is not independent.

8.3:

Take a random walk and send to {-1,0,1} determined if the value at the walk was positive, negative,or 0

Alternatively, take a markov chain in 3 states. If it is set up where two events are switching and the last state stays where it is. We can define a function \(f\) that sends the 3 state system to a 2 state system sending the two switching events to the same point and the last event to the other one.

See the picture I will attach later.

8.15: