# Thread: Martingale and absorption probabilities

1. ## Martingale and absorption probabilities

Let $X_t$ be the fortune of a gambler at time t. It is assumed that $(X_t : t=0,1,2...)$ is a discrete time Markov process with state space {0,1,2,...N} where N is a positive integer. For $1 \leq i \leq N-1$

$\mathbb{P}(X_{t+1} = i-1 | X_t = i)=\frac{1}{2}= \mathbb{P}(X_{t+1}=i+1 | X_t=i)$

States 0 and N are absorbing so that

$\mathbb{P}(X_{t+1} = 0 | X_t = 0)=1=\mathbb{P}(X_{t+1}=N | X_t=N)$

You may assume that $(X_t)$ is a martingale with respect to itself.

(i) Use the Martingale Convergence Theorem to show that with probability 1 the process is eventually absorbed at 0 or N.

I cannot see how the MCT can show that the process is eventually absorbed at 0 or N as it only shows that $X_t \rightarrow X$ for some $X$

2. ## Re: Martingale and absorption probabilities

Hello,

Just a quick thought.

Write the definition of a.s. convergence : $P(\omega : \lim_n X_n(\omega)=X(\omega)=1 \quad (*)$

Let $\Omega_a = \{\omega ~:~ X(\omega)=a\}$. We want to show, through the MCT, that $P(\Omega_a)=0$ for any $a\neq 0,N$.
The aim is to prove that by contradiction (I think that's the way...). So let's suppose $P(\Omega_a)=\delta>0$.

$\lim_n X_n(\omega)=X(\omega)$ means that $\forall \epsilon>0, \exists m\in\mathbb N,~s.t.~\forall n>m,~|X_n(\omega)-X(\omega)|<\epsilon$.
Assuming that $\epsilon<1$ (it's true for any epsilon and we usually choose a small epsilon), it means that $X_n(\Omega)=X(\omega)$, since Xn takes only integer values.

And then try to finish by considering the probability (*) and the fact that $P(\Omega_a)>0$. It should be possible to finish the proof (but I've merely written what I was thinking, so I'm not sure...)

3. ## Re: Martingale and absorption probabilities

Maybe you can also consider liminf of Xn=0 or N and use Kolmogorov's 0-1 law ?