I am stuck on the exercise below from an exam I took. I am going through it again and trying to figure it out because I couldn't during the exam.
In a), from the definition of a martingale I already proved that the $\mathbb{E}[M_n] < \infty$ but couldn't show that $\mathbb{E}[M_{n+1} \mid X_1, \ldots, X_n] = M_n$
For b) and c) however I couldn't get much done.
Any help would be greatly appreciated!
Thanks in advance.
PS: I am interested in understanding the exercise rather than simply getting the solution.
To understand the loss of the casino $M_n,$ first observe that the moment $X_n=-1,$ everyone leaves the Casino empty-handed--that is, the casino doesn't lose anything. In other words, it is a fresh start for the casino every time we see $X_n=-1$ (except for the fact that the Casino has gained $n$ unit of money by this time). So to calculate the loss of the Casino at step $n,$ the only thing that matters is the last consecutive sequence of $+1,$ that is, $K_n.$
With this information, now let us see how much the casino loses given $K_n$ many consecutive $+1.$ In the first step (since the beginning of the misfortune of the casino), there is one gambler and the gambler pays $1$ unit of money and gets $1/p$ amount of money back. The casino has lost $1/p-1$ unit. In the next step, there is another gambler, who wins. So now the loss of the casino is $$\frac{1}{p}\left(\frac{1}{p}-1\right)+(1/p-1)=\frac{1}{p^2}-1.$$ If this misfortune continues for $K_n$ time, it is clear that the Casino has lost $$M_n=\sum_{i=1}^{K_n}\left(\frac{1}{p}\right)^i-n.$$
For part $b)$, I usually prefer the 'conditioning trick'. The idea here is that every time you see $X_n=-1,$ the past does not matter and you have to start all over again. So it makes sense to condition on the 'first time you see $-1$'. In detail, let us think about what can we say about $ET$ when $X_1=-1.$ This means, we already have wasted one step and we have to wait for $T$ more time, that is, $$E(T|X_1=-1)=1+E(T).$$
Continuing this idea, we can write $$E(T|X_1=1=X_2=\ldots=X_{k-1}, X_k=-1)=k+E(T), \quad k\le L.$$ And finally on the event $C=(X_1=\ldots=X_L=1),$ we have $E(T|C)=L.$ Combining all this we get $$E(T)=(1-p)\sum_{j=0}^{L-1}p^{j}(j+1+E(T))+Lp^L=\frac{p^{-L}-1}{1-p}.$$
Also, once I have got the above answer it immediately suggests that $T$ has come connected with $M_n.$ And, notice that $T$ is a stopping time and $M_n$ is a martingale, therefore (given that $E(T)<\infty$) we obtain $$E(M_{T})=E(M_1)=0.$$ On the other hand, we also have that $$E(M_T)=\frac{p^{-L}-1}{1-p}-E(T).$$ This gives the expression for $E(T).$
For $(c)$, repeat the part $a)$ and $(b)$ when each new gambler has initial amount $s^{n-1}.$ it would be a good idea to see what is the loss of the casino. In fact, writing the correct loss would immediately tell you what to do next!