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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 $100$?

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    What makes you think this should have a nice expression?2012-11-28
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    @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.2012-11-28
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    This is an entirely different question (and no, A is not expected to be near 100).2012-11-28
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    @did, that was the intuition behind the actual question. The actual question is as posted (especially edit 2)2012-11-28
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    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}$2012-12-03
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    You have $\alpha_{n+1}=\alpha_{n}+\sigma_{n}$ and $\beta_{n+1}=\beta_{n}\left(1 + 3\xi_{n}/\alpha_{n}\right)$, where $\sigma_{n},\xi_{n}\in\{+1,-1\}$. Are $\sigma_{n}$ and $\xi_{n}$ equal, or independent, or dependent on each other in some other way?2012-12-04
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    @nvlass, they are the same so if $\alpha $ increases, $\beta$ increases.2012-12-05
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    @mjqxxxx refer to reply to nvlass.2012-12-05

1 Answers 1

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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$\beta$ 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.

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    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.)2012-12-06
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    @picakhu: Yes, see the edited answer.2012-12-06
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    I don't quite see how you got that $\beta_{n} < \beta_{0}$ whenever $\alpha_{n} \leq \alpha_{0}$.2012-12-06
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    @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 $\le 0$.2012-12-06
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    That makes a lot of sense, thanks!2012-12-06