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 $100$?
 A: 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.
