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Let $X_i$, $i \geq 1$ be i.i.d random variables. Let $P_j=P(X_i = j)$ and suppose that $$\sum_{j=-1}^{\infty} P_j=1$$. That is the possible values of the $X_i$ are $-1,0,1,\dots$. If we take $$S_0=0, S_n=\sum_{i=1}^{n}X_i$$ then the sequence of random variables $S_n, n\geq 0$ is called a left skip free random walk. Also, $E[X_i] < 0$ (downward drift). We need to prove that $$P(S_n <0, \forall n \geq 1)= P(S_n \neq 0, \forall n \geq 1)$$

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2 Answers 2

up vote 3 down vote accepted

Hint: This is a deterministic result. To wit, let $(s_n)_{n\geqslant0}$ denote an integer valued sequence such that $s_0=0$, $s_{n+1}-s_n\geqslant-1$ for every $n$ and $s_n\to-\infty$ when $n\to\infty$, then: $$ (\forall n\geqslant1,\ s_n\lt0)\iff(\forall n\geqslant1,\ s_n\ne0). $$

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yes, it has nothing to do with probability, how to prove it ? –  podu Feb 18 '13 at 1:42
Well, you do not seem interested, are you? –  Did Feb 18 '13 at 6:46

We have $$\mathbb{P}(\forall n: S_n \not=0) = \mathbb{P}(\forall n: S_n<0) + \mathbb{P}(\exists k \, \forall n \geq k: S_n>0, \forall n < k: S_n<0)$$ (This follows straight from the assumption that steps to the left are only possible with stepsize 1. If we have $S_k(\omega)>0$ for any $k$, we can only avoid reaching zero by staying on the (strict) positive real axis, i.e. $S_n(\omega)>0$ for all $n \geq k$.)

So the claim follows if $$\mathbb{P}(\exists k \, \forall n \geq k: S_n>0, \forall n < k: S_n<0)=0 \tag{1}$$ holds.

Clearly $$\mathbb{P}(\exists k \, \forall n \geq k: S_n>0, \forall n < k: S_n<0) \leq \mathbb{P}( \exists k \, \forall n \geq k: S_n > 0) \tag{2}$$

By the strong law of large numbers we have $$\mathbb{P} \left( \frac{S_n}{n} \to \mathbb{E}X_1 \right) = 1$$ hence in particular (since $\mathbb{E}X_1 <0$) $$\mathbb{P} \left( \exists k \, \forall n \geq k: S_n < 0 \right) = 1$$ and therefore $$\mathbb{P}( \exists k \, \forall n \geq k: S_n > 0) \leq 1- \mathbb{P} \left( \exists k \, \forall n \geq k: S_n < 0 \right)=0$$

From (2) we conclude (1).

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