# Double Random Walks

I'm trying to find the distribution of the second sequence after some time (say $t=10$).

I have that sequence $A$ starts at $100$ and so does sequence $B$.

If sequence $A$ moves from some value $\alpha$ to $\alpha \pm 1$, then state B will move from state $\beta$ to state $\beta \pm \frac{3\beta}{\alpha}$. How would I go about trying to get the distribution?

Edit for clarification, (and I forgot to mention this before), If $\alpha$ hits $0$, the walk is stopped.

Edit 2: As an example question, what is the probability that at time $t=10$, $B$ is greater than $100$?

-
What makes you think this should have a nice expression? – Did Nov 28 '12 at 19:50
@did, it doesn't need to be nice, say that time $t$ goes to infinity, then obviously, since the walk is random, $A$ is expected to be "near" $100$ and my question simply is... where is $B$ expected to be? and whether there is some easy way to calculate that. – picakhu Nov 28 '12 at 19:54
This is an entirely different question (and no, A is not expected to be near 100). – Did Nov 28 '12 at 20:39
@did, that was the intuition behind the actual question. The actual question is as posted (especially edit 2) – picakhu Nov 28 '12 at 21:15
Hi, I would like to ask for a clarification: say $a^{t+1}=a^t+1$. Then $\beta^{t+1}=\beta^t + \frac{3\beta^t}{a^t}$? Or the direction of the movement of $\beta$ is independent and could be e.g. $\beta^{t+1}=\beta^t- \frac{3\beta^t}{a^t}$ – user50600 Dec 3 '12 at 23:16

## 1 Answer

You have i.i.d. random variables $\xi_1,\xi_{2},\xi_{3},\ldots$ that take on values $\pm 1$, in terms of which $$\alpha_{n}=\alpha_{n-1}+\xi_{n}$$ and $$\beta_{n}=\beta_{n-1}\left(1 + \frac{3\xi_{n}}{\alpha_{n-1}}\right).$$ Solving yields $$\alpha_{n}=\alpha_{0}+\sum_{i=1}^{n}\xi_{i}$$ and $$\beta_{n}=\beta_{0}\prod_{i=1}^{n}\left(1+\frac{3\xi_{i}}{\alpha_{i-1}}\right)=\beta_{0}\prod_{i=1}^{n}\left(1+\frac{3\xi_{i}}{\alpha_{0}+\sum_{j=1}^{i-1}\xi_{j}}\right).$$ We can write the law for $\log\beta_{n}$ as a power series expansion in $\alpha_{0}^{-1}$, which we expect to be a good approximation for $n \ll \alpha_{0}$. The first few terms are: $$\begin{eqnarray} \log\beta_{n}&=&\log\beta_{0}+\sum_{i=1}^{n}\log\left(1+\frac{3\xi_{i}}{\alpha_{0}+\sum_{j=1}^{i-1}\xi_{j}}\right) \\ &=& \log\beta_{0}+\sum_{i=1}^{n}\left(\frac{3\xi_{i}}{\alpha_{0}+\sum_{j=1}^{i-1}\xi_{j}}-\frac{1}{2}\left(\frac{3\xi_{i}}{\alpha_{0}+\sum_{j=1}^{i-1}\xi_{j}}\right)^{2}+...\right) \\ &=&\log\beta_{0}+\alpha_{0}^{-1}\sum_{i=1}^{n}3\xi_{n}-\alpha_{0}^{-2}\left(\sum_{j<i\le n}3\xi_{i}\xi_{j} + \frac{1}{2}\sum_{i=1}^{n} 9\xi_{i}^{2}\right)+O(\alpha_{0}^{-3}) \\ &=&\log\beta_{0}+3\alpha_{0}^{-1}(\alpha_{n}-\alpha_{0})-\alpha_{0}^{-2}\left(\frac{3}{2}(\alpha_{n}-\alpha_{0})^{2}+3n \right)+O(\alpha_{0}^{-3}). \end{eqnarray}$$ The first-order term is a symmetric random walk, and the second-order term is a negative drift, suggesting that $\beta$ is more likely to decrease over time.

In this approximation, and assuming $n \ll \alpha_{0}$, we find that $\beta_{n} < \beta_{0}$ whenever $\alpha_{n} \leq \alpha_{0}$. So for odd $n$, $\beta_{n} < \beta_{0}$ with probability $1/2$, while for even $n$, $\beta_{n} < \beta_{0}$ with probability $1/2 + {n\choose{n/2}}2^{-(n+1)}$. For instance, this gives a decrease with probability $0.623046875$ for $n=10$. Testing this numerically, I found a decrease in $622477$ of $10^6$ trials for $\alpha_{0}=100$ and $n=10$, which is consistent with the approximation.

-
this is very helpful! Is there a way to use this to answer the question in Edit 2? (even if there is not, just the knowledge that $\beta$ "decreases" over time is good information.) – picakhu Dec 6 '12 at 19:29
@picakhu: Yes, see the edited answer. – mjqxxxx Dec 6 '12 at 21:06
I don't quite see how you got that $\beta_{n} < \beta_{0}$ whenever $\alpha_{n} \leq \alpha_{0}$. – picakhu Dec 6 '12 at 22:17
@picakhu: If you examine the sign of the quadratic approximation for $\log\beta_{n} - \log\beta_{0}$, you find a decrease for large $\alpha_{n}$ (which isn't meaningful, since we're assuming small $n$) and for $\alpha_{n}-\alpha_{0} < n/\alpha_{0}$. In the range we're working in, $n/\alpha_{0}$ is between $0$ and $1$, so since $\alpha_{n}$ is an integer, it just has to be $\le 0$. – mjqxxxx Dec 6 '12 at 22:27
That makes a lot of sense, thanks! – picakhu Dec 6 '12 at 22:32