Suppose I started with $W_1=1$ and decided on a ratio $0<\alpha<1$ such that I invest $\alpha W_n$ of my earning in the next round. I either lose my investment, or win it with probability $p=9/10$.

So that $W_{n+1}=W_n(1+\alpha)$ with probability $p$ or $W_{n+1}=W_n(1-\alpha)$ with probability $1-p$.

Given that $\alpha > 1-2^{-10}$ (close to one I suppose) why does it become likely that $W_n$ tends to zero? In particular, how do we use the law of large numbers to show that there exists a sequence $b_n$ which tends to zero such that $\lim P(W_n \leq b_n) =1$ as n tends to infinity.

I don’t know how to approach this problem because the distribution isn’t something like the sum of binomial terms like in the ordinary gamblers ruin problem. Moreover I’m not sure whether it’s necessary to find the distribution. Do we find a recursive formula and use that to derive the distribution? I’m quite lost.

I don't think the solution is to work out a distribution. The mean winnings at step $n$ are $\alpha W_n p - \alpha W_n (1-p)$ = $2 \alpha W_n p -1$. This is the mean of $W_{n+1}-W_{n}$. Maybe this helps?


Let us write $\xi_1$, $\xi_2$, ... for the $\pm1$-valued random variables that denote "success" or failure": they are iid, with $$ P(\xi = 1) = p = 9/10 \quad\text{and}\quad P(\xi = -1) = 1-p = 1/10.$$ Now, we can write $$ W_n = W_{n-1} (1 - \alpha \xi_{n-1}) = \cdots = \prod_{k=0}^{n-1} (1 - \alpha \xi_k). $$ (Here I have chosen $W_0 = 1$ as the base-case, rather than $W_1 = 1$. It is standard in probability/Markov chains to start from $0$, indicating that no steps have been made; $W_1$ is the distribution after $1$ step.)

Now, were $\alpha$ very small, one might argue that $$ \prod_{k=0}^{n-1} (1-\alpha\xi_k) \approx \prod_{k=0}^{n-1} e^{-\alpha\xi_k} = \exp(-\alpha\sum_{k=0}^{n-1}\xi_k). $$ Now we're looking at a sum of iid rvs, which is always good news! But, it's only an approximation. Instead, let's take a logarithm: $$ \log W_n = \sum_{k=0}^{n-1} \log(1 - \alpha\xi_k) \equiv \sum_{k=0}^{n-1} \lambda_k, $$ where we write $\lambda_k = \log(1-\alpha\xi_k)$ for the log. This is an exact equality. Again, the $\lambda_k$ are iid rvs, with $$ P(\lambda = \log(1+\alpha)) = p = 9/10 \quad\text{and}\quad P(\lambda = \log(1-\alpha)) = 1-p = 1/10. $$ We can now apply a LLN to the $\lambda_k$.

Can you finish from here? I add further details below, hidden, but I encourage you to think for yourself first! :)

Now we must convert "$W_n \to 0$" into something about $\log W_n$. But this is simply that the log tends to $-\infty$. By the LLN, you know that $\sum_{k=0}^{n-1} \lambda_k \approx n E(\lambda)$. It remains to calculate this expectation, finding the condition on $\alpha$ that will ensure it is negative. I leave these details to you.

To be more precise, the statement "$W_n \le b_n$" is equivalent to "$\log W_n \le \log b_n$". It is perhaps easier to consider $-\log W_n$, which will tend to $+\infty$. Now, but the LLN, we know that $$ P( -\log W_n > \tfrac12 n E(\lambda) ) \to 1 \quad \text{as $n \to \infty$}. $$ So we desire $b_n$ so that $-\log b_n = \tfrac12 n E(\lambda)$, ie $b_n = \exp(-\tfrac12 n E(\lambda))$.

(In fact, checking the details yourself, you can see that $b_n = \exp(- (1-\epsilon) n E(\lambda) )$ works for any $\epsilon > 0$. By the CLT, we in fact can let $\epsilon = \epsilon_n$ depend on $n$ with $\epsilon_n \to 0$ providing $\epsilon_n \sqrt n \to \infty$ as $n \to \infty$. This is not needed for this question, though.)

For more details on the law of large numbers, and how it can be applied in a general see these lecture notes by James Norris, in particular Section 21.

  • $\begingroup$ thanks. Two questions: The law of large numbers as I learnt it basically stated that for a binomial distribution, the integral of the distribution is mostly around a small region around the mean. How does this translate to your use of it? second, I can't see how this relates to the sequence $b_n$? $\endgroup$ – Dis-integrating Feb 11 at 14:18
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    $\begingroup$ The LLN means that the sum concentrates about its mean. In particular, it won't deviate far. See the referenced lecture notes (Section 21) for further information :) $\endgroup$ – Sam T Feb 11 at 14:26
  • $\begingroup$ Ah I see. The version that was proved to me (disgusting proof) was a special case of the LLN. I couldn’t really use it here $\endgroup$ – Dis-integrating Feb 17 at 4:17
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    $\begingroup$ I see. Hopefully the link is helpful! :) $\endgroup$ – Sam T Feb 17 at 15:07

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