Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Background. Let $Y_1,Y_2,\ldots$ be i.i.d. random variables such that $$P(Y_i<-1) = 0,$$ $$P(Y_i<0) > 0,\quad P(Y_i>0)>0,$$ $$E[Y_i] = \mu > 0\qquad \text{($\mu$ is finite)}.$$ Now define the discrete-time random process $(X_n)$ with $$X_n = \prod_{i=1}^n (1+\frac{1}{2}Y_i).$$ Real-life motivation. Our initial capital is one dollar. At bet $i$, we make the percentage return $Y_i$ on our investment. At every bet, we bet half of our available capital. Our capital after $n$ steps is $X_n$.

Question. What is the probability that the drop from "peak to bottom" of the process $X_n$ ever becomes greater than, say, 20%? More precisely, please help me calculate $$P\left( \sup_{t\in[0,\infty)}\left[ \sup_{s\in[0,t]} \frac{X_s-X_t}{X_s} \right] > 0.2 \right).$$

Question rephrased. The above quantity can be rewritten to the following:

$$P\left( \sup_{t\in[0,\infty)}\left[ \sup_{s\in[0,t]} \left\{ 1 - \prod_{i=s+1}^t (1+\frac{1}{2}Y_i) \right\} \right] > 0.2 \right).$$

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Since the $Y_i$ are iid it seems obvious the answer is $1$ over infinite time.

Choose some $k \lt 0$ such that $q = \Pr(Y_i \lt k) \gt 0$.

Then $\Pr(X_{i} \lt (1+\frac{k}{2}) X_{i-1}) = q$.

So if $n \gt \log(0.8)/\log(1+\frac{k}{2})$ then $\Pr(X_{s+n} \lt 0.8 X_s) \ge q^n \gt 0$.

So the probability your event does not occur in the first $mn$ steps is less than or equal to $(1-q^n)^m$ and that reduces towards $0$ as $m$ increases without limit.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.