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What is the general approach to solving this recurrent equation given that $p(x)$ and $q(x)$ are not constant and do not depend on $n$ and $p(x)+q(x) \neq 1$. Please just give me some hints, don't solve it for me. I know this is similar to Binomial probability of $x$ successes in $n$ trials with a changing probability of success and solved using generating functions or z-transforms.

$p(x)$ can be seen as a probability of success after $n-1$ trials and $x-1$ successes and $q(x)$ as a probability of failure after $n-1$ trials and $x$ successes.

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I think this may be a hard problem for general $p$ and $q$. If $p$ and $q$ were to depend on $n$ as well as $x$ then the problem is open, even in the case of a linear dependency on $n$ and $x$; that case is research problem 6.94 in Concrete Mathematics. If you give some more information about the specific $p$ and $q$ you're interested in you might have a better chance at a good answer. –  Mike Spivey Nov 3 '11 at 5:36
    
OK< I looked it up in Woodbury(1949), Rutherford(1954) and Gani(2006), but still at loss. I think for constnt $p,q$ this is called Kolmogorov backward-forward equation –  sigma.z.1980 Nov 3 '11 at 5:49
    
I mean yes, I can bound both $p$ and $q$, but this would be less interesting –  sigma.z.1980 Nov 3 '11 at 5:51
    
$Mike: shouldn't it be some determinant of eigenmatrix or something? –  sigma.z.1980 Nov 3 '11 at 5:51
    
Care to accept an answer? –  Did Jan 10 '12 at 22:33
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2 Answers

up vote 1 down vote accepted

You didn't specify the initial conditions which is a big problem. (And $x$ for an integer index is not a variable choice that makes me happy.) I will assume that $A(n,k)$ is zero if $k$ is negative or bigger than $n$. I will assume that the recurrence holds for all $n\ge1$ and all $x$.

Under this assumptions, let $A_k(z)=\sum\limits_{n=0}^{\infty}A(n,k)t^n$, convert the recurrence to a functional equation and this gives you a product formula for the generating function.

If you sum the recurrence, you get

$$A_k(t)=p(k)A_{k-1}(t)t+q(k)A_k(t)t,$$

so

$$A_k(t)=\frac{p(k)t}{1-q(k)t}A_{k-1}(t).$$

In particular, this works for binomial coefficients and Stirling numbers.

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this sounds reasonable, but could you give a little bit more details on how to do this conversion. In Migdal(2010) the solution is in the form of the product of diagonal entries of the eigenmatrix –  sigma.z.1980 Nov 22 '11 at 22:18
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Here is a probabilistic interpretation.

Let $r(x)=p(x)+q(x)$. Consider the random walk $(X_n)_{n\geqslant0}$ on the integer line which, when at $x$, stays at $x$ with probability $p(x)/r(x)$ and moves to $x-1$ with probability $q(x)/r(x)$. Then, writing $\mathbb E_x$ for the expectation when $X_0=x$, one has, for every $n\geqslant0$ and every $x$, $$ \color{blue}{A(n,x)=\mathbb E_x\left(A(0,X_n)\cdot\prod\limits_{k=0}^{n-1}r(X_k)\right)}. $$ Proof: If $X_0=x$ almost surely, the Markov chain $(X_{k+1})_{k\geqslant0}$ is distributed like $(X_{k})_{k\geqslant0}$ for a random $X_0$ with distribution $r(x)^{-1}(p(x)\delta_x+q(x)\delta_{x-1})$. $\ \Box$

To go further, one could want to specify the initial condition $(A(0,x))_x$.

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