I have the equation $$\alpha_n x_n = 1-p + p \alpha_n x_{n-1},$$ with $\alpha_n, x_n,p \in (0,1)$ for all $n\geq0$.

I would like to express $x_n$ as a function of $(\alpha_n)$ and $p$ for every $n$. I know a formula when $(\alpha_n)$ is constant but not in general.

  • 1
    $\begingroup$ $x_n = (1 - p) / \alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration. $\endgroup$ – user295959 Jan 24 at 16:03

You can observe that


so you can deduce the following formula

$x_n=p^n x_0+(1-p)\sum_{k=1}^n\frac{p^{n-k}}{\alpha_k}$

The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.

$x_{n+1}= px_{n}+\frac{1-p}{\alpha_{n+1}}=$

$= p^{n+1} x_0+(1-p)\sum_{k=1}^n\frac{p^{(n+1)-k}}{\alpha_k} +\frac{1-p}{\alpha_{n+1}} =$

$= p^{n+1} x_0+(1-p)\sum_{k=1}^{n+1}\frac{p^{(n+1)-k}}{\alpha_k} $


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