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Math Help - Martingale and absorption probabilities

  1. #1
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    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
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    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...)
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    Moo
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    Re: Martingale and absorption probabilities

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