Solving the recursion $p_n = p \cdot p_{n-2} + (1-p)p_{n-1}$

Question

Solving the recursion $p_n = p \cdot p_{n-2} + (1-p)p_{n-1}$

$p_n = p \cdot p_{n-2} + (1-p)p_{n-1}$

$p_n = p \cdot p_{n-2} +p_{n-1} - p\cdot p_{n-1}$

$p_n - p_{n-1} = (-p)(p_{n-1} - p_{n-2})$

$= (-p)^{n-1}(p_1 - p_0)$

I am very confused about the last step here...can anyone explain what happened? Also, by "solving a recursion," this means that the equation shouldn't have a $p_{n-1}$ term, correct? Because we don't want the $p_n$ term to be dependent upon the term before it?

Answer 1

You have $$ \begin{align} p_n - p_{n-1} &= (-p)(p_{n-1} - p_{n-2}) \\\\&= (-p)(-p)(p_{n-2} - p_{n-3}) \\\\&= (-p)(-p)(-p)(p_{n-3} - p_{n-4}) \\\\&=\cdots \\\\&= (-p)^{n-1}(p_1 - p_0). \end{align} $$ Then one may sum the preceding identity, terms telescope giving

$$ p_n-p_0=\sum_{k=1}^{n}(p_{k}-p_{k-1})=(p_1 - p_0) \sum_{k=1}^n(-p)^{k-1}. $$

Can you take it from here?