Recently, I was studying random walks in 1 dimension, and wanted to find a general solution for the expected time for a randomly walking point at the origin to reach the boundaries of $a$ and $b$ with $a<0<b$, and a probability of the point increasing of $p$, with $p\in(0,1)$. The function $E(a,b,p)$ should give me the expected value.
Therefore, I proved the recursion that $E(a,b,p)=1+pE(a-1,b-1,p)+(1-p)E(a+1,b+1,p)$, with $E(a,0,p)=E(0,b,p)=0$.
Someone told me the solution for $E$ I was looking for was: $$E(a,b,p)=\frac{b(1-p)^{b-a}-ap^{b-a}-(b-a)p^{-a}(1-p)^{b}}{(2p-1)(p^{b-a}-(1-p)^{b-a})}$$
The solution is correct, as it matches all the data I have collected. How would you show this was the case? I have tried multiple methods and have not been able to succeed.
The typical approach to solving a problem like this is:
Define a sequence $A_{i, j} = E(-i, j, p)$ (I'm swapping the sign of the first term so that we're working solely in non-negative numbers).
Define a generating series $G(x, y) = \sum_{i = 0}^\infty \sum_{j = 0}^\infty A_{i, j}$.
Use the given boundary values and recursion to manipulate the series expression for $G$ until you get something with a known power series expansion, and use that to determine the values of $A_{i,j}$.
However, having been given a potential answer, you can also just confirm that it satisfies your requirements. For example, if we substitute $b = 0$, we get:
$$\begin{eqnarray}E(a, 0, p) & = & \frac{0(1 - p)^{0-a} - a p^{0 - a} - (0 - a) p^{-a}(1 - p)^0}{(2p-1)(p^{0-a} - (1-p)^{0 - a}} \\ & = & \frac{-a p^{-a} + a p^{-a}}{(2p-1)(p^{-a} - (1-p)^{-a}} \\ & = & 0 \end{eqnarray}$$
as required.