In these lecture slides (pages 13-21) on optional stopping theorem (OST) and martingale I have found a great example of the gambler's ruin problem and what happens when the criterions of the OST are not met, which I am trying to understand.
Let $X_i$ be i.i.d Bernoulli r.v. with $P(X=1)=P(X=-1)$ and $S_n= \sum_{i=1}^n X_i + K$ where K is the our initial balance. We leave when we hit it big and $S_n= N$ or we have gone bust, $S_n=0$. Therefore, the stopping time is
$$T_{0,N}= \inf\{n\in\mathbb{N}: S_n = 0 \lor S_n = N\},$$
which is a reasonable stopping time under which the OST implies $\mathbb{E}(S_{T_{0,N}})=\mathbb{E}(S_0)=K$. Furthermore the probablity of ruin $P_0 = (N-K)/N$ and of hitting it big is $P_N = K/N$. They then go on to show that for the martingale $Z_n = S_n^2-n$ and the same stopping time the expectation is $\mathbb{E}(T_{0,N}) = K(N-K)$
The author then wants to show a case where the OST does not hold. For this they assume that $S_n$ is not bounded and defines the stopping time $\tau$ as
$$\tau = \inf\{n\in\mathbb{N}: S_n = K+1\}$$
and want to show that $\mathbb{E}(\tau)=\infty$. For any $a>0$ let
$$T_{-a,K+1}= \inf\{n\in\mathbb{N}: S_n = -a \lor S_n = K+1\}.$$
For this stopping time it holds that $T_{-a,K+1}\leq \tau$, since
$$\{n\in\mathbb{N}: S_n = -a \lor S_n = K+1\}\subset\{n\in\mathbb{N}: S_n = K+1\}.$$
What I do not understand is the following calculation done by the author
$$\mathbb{E}(T_{-a,K+1}|S_0 = K) = \mathbb{E}(T_{0,K+a+1}|S_0 = K+a) = K + a.$$
Assuming the above equation, it follows that since $T_{-a,K+1}\leq \tau$ and $a$ was arbitraily chosen this implies $\mathbb{E}(\tau) =\infty$.
I tried using the law of total expectation to justify the conditional probablity
$$ \mathbb{E}(T_{-a,K+1}) = \mathbb{E}(\mathbb{E}(T_{-a,K+1} |S_0 = K)),$$ which then would produce the desired result for the martingale $Z_n$, since the starting point would be $\hat{K}= K + a$ and $N= K+a+1$ leading to $\mathbb{E}(T_{0,K+a+1}) = \hat{K}(N-\hat{K}) = (K+a)(K+a+1-K-a) = K+a$. However I do not understand how they justify using this result, which was shown for a different martingales.
Thank you for your time.
Best Answer
To see this, it's best to think of the net profit process $Y_t = S_t - S_0$. Then, one sees that the stopping times $T_{-a, K+1}$ and $T_{0, K + a +1}$ both count the amount of time before we have a net profit of $1$ or $-K-a$.
Formally, you can show that the conditional distributions of these two stopping times conditioned on $S_0 = K$ and $S_0=K+a$, respectively, are the same. That is:
\begin{align} P(T_{-a,K+1} = t \, | S_0 = K) &= P(S_t = K+1 \text{ or }S_t = -a | S_0 = K ) \\ &= P(S_t - S_0 = 1 \text{ or } S_t - S_0 = -a-K) \\ &= P(S_t - S_0 = 1 \text{ or } S_t - S_0 = -a-K \, | \, S_0=K+a) \\ &= P(S_t = 1+ K +a \text{ or } S_t = 0 \, | S_0 = K+a) \\ \end{align}
where we used the fact that $S_t - S_0$ is independent of $S_0$ in the second and third lines.
To obtain the second equality, use your previous result that $E(T_{0,N}) = K(N-K)$ and replace $K \mapsto K+a$ and $N \mapsto a+K+1$ to obtain: \begin{align} E(T_{0,a+K+1}) = (K+a)(a+K+1 - (K+a)) = K+a \end{align}