$\alpha_n x_n = 1-p + p \alpha_n x_{n-1}$

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.

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

You can observe that

$$x_n=px_{n-1}+\frac{1-p}{\alpha_n}$$

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 $$n+1$$.

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

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

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