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Given random variables $X_1, X_2, \ldots \stackrel{iid}{\sim} P(X_i = 1) = p = 1 - q = 1 - P(X_i = -1)$ where $p > q$ in a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^X$,

define $S = (S_n)_{n \ge 0}$ where $S_n = a + \sum_{i=1}^{n} X_i$ where $a$ is a positive integer.

Let $b$ be a positive integer s.t. $0 < a < b$, and let $T:= \inf\{n: S_n = b\}$.

It can be shown that $T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time.

Show $E[T] < \infty$ by finding an upper bound for $P(T=k)$.

If $T=k$ (I guess $k \ge 1$), then there is at least one -1 (step down) $(*)$ in each of the vectors of size b, provided all the indices are less than $k$ (assuming I am interpreting 6.2 here right):

$$(X_{b+1}, X_{b+2}, ..., X_{2b-1}, X_{2b})$$ $$(X_{2b+1}, X_{2b+2}, ..., X_{3b-1}, X_{3b})$$ $$(X_{3b+1}, X_{3b+2}, ..., X_{4b-1}, X_{4b})$$ $$\vdots$$ $$(X_{nb+1}, X_{nb+2}, ..., X_{(n+1)b-1}, X_{(n+1)b})$$ $$\vdots$$

It seems then that based on 6.1, $P(T=k) \le (1-p)^{k/b - 1}$, which seems to make (I just chose p = 0.6 and b = 10

$$E[T] = \sum_{k=1}^{\infty} k P(T=k) < \infty \ QED$$

Is that right?

Why do we have $k/b - 1$? It seems like finding the supremum over n of $(n+1)b < k$ much like in 6.1 where apparently the exponent there was based on the supremum over m of $3m + 3 < k$

Why is $(*)$ true? Edit: Okay, I get it now.

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BCLC
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2 Answers2

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Since both $a$ and $b$ are integers, the smallest possible value for $T$ is the positive integer $b-a\ ;$ this corresponds to a consecutive sequence of all $+1$:s after having started from level $a\ .$ The probability for this happening is $$\mathbb{P}(T=b-a) = p^{b-a} \ .$$

There is no largest possible value for $T$, since is is possible for the random walk to go up a little, then down a little, then up a little, ..., all the while staying between levels $a$ and $b\ ;$ this is due to increments, $X\ ,$ being independent and identically distributed.

If $T = (b-a)+2$ then there is a $-1$ located somewhere among the $(b-a)+2$ increments; the other $(b-a)+1$ increments all being equal to $+1\ .$ This is a binomial scenario, whose probability can be written $$\mathbb{P}(T = (b-a)+2) = {(b-a)+2\choose 1}qp^{(b-a)+1}\ .$$

If $T = (b-a)+2k$ (where $k>1$) there are a $k$ number of $-1$:s located somewhere among the $(b-a)+2k$ increments. $$\mathbb{P}(T = (b-a)+2k) = {(b-a)+2k\choose k}q^{k}p^{(b-a)+k}\ .$$

Anders Muszta
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  • Thanks Anders Muszta. Why must there be a or k # of -1's? – BCLC Dec 19 '15 at 21:22
  • Remember that $T$ is the first time when the random walk ($S$) reaches level $b\ .$ – Anders Muszta Dec 20 '15 at 22:47
  • Yeah I didn't really get that in the link above. If b = 2 and a = 1, we can have $X_1 = 1$...? – BCLC Dec 20 '15 at 23:24
  • As I understand it, yes you can have $X_{1} = 1$ when starting from level $1$ and hoping to reach level $2$; this would be the "best" scenario in this case. – Anders Muszta Dec 21 '15 at 00:13
  • Ah wait never mind got it. Tried out some examples of a and b. Thanks. Thinking further... – BCLC Dec 21 '15 at 01:13
  • Why does k need to be > 1? If we have = 1 then we have the previous proposition right? Did you mean the last part to be a second case rather than a generalisation of the proposition? – BCLC Dec 21 '15 at 01:14
  • Ah okay, I think I get it now. So $P(T=j) = P(X=k)$ where $j = b-a + 2k$ and $X$ is binomially distributed with parameters $n,q$? What's n though? Editing question now... – BCLC Dec 21 '15 at 01:24
  • Edited. Thanks Anders Muszta. How is it? – BCLC Dec 21 '15 at 12:35
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Let $j := (b-a) + 2k$ for $k \ge 1$. Then $k = 1 \iff j=\frac{1-(b-a)}{2}$

$$E[T] = \sum_{j=0}^{\infty} j P(T=j) = \sum_{j=\frac{1-(b-a)}{2}}^{\infty} j P(T=j) = \sum_{k=1}^{\infty} k P(T=b-a+2k) := \sum_{k=1}^{\infty} k P(X=k)$$

where $X \sim Bi(b-a+2k,q)$

Thus,

$$E[T] = \sum_{k=1}^{\infty} k P(X=k) = E[X] = (b-a+2k)(q) < \infty$$

BCLC
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