Conditional expectation of a simple random walk

Question

Let $(X_n)$ be a simple random walk that starts from $X_0 = 0$ and on each step goes up one with probability $p$ and down one with probability $q = 1 − p$.

I need to calculate:

$E[X_8 | X_4 = -2] $

I have already calculated that $E[X_4] = 4(p-q)$. Would the answer just be $-2 + E[X_4]$ seeing as now essentially $-2$ is the starting point? Any help would be appreciated.

Answer 1

Yes, conditioned on $X_4=-2$, the remainder of the walk is a simple random walk starting at $-2$ and going up with probability $p$ and down with probability $1-p$ at every step.

Answer 2

$$E(X_n|X_{n-1}=t)=p(t+1)+(1-p)(t-1)=t+(2p-1)$$

$$E(X_n|X_{n-1})=X_{n-1}+(2p-1)$$

by Tower Property $$E(X_{n+1}|X_{n-1})\overset{Tower}{=}E\left(E(X_{n+1}|(X_{n-1},X_n))|X_{n-1})\right)$$ $$=E\left(E(X_{n+1}|X_n)|X_{n-1})\right) =E\left(X_{n}+(2p-1)|X_{n-1})\right) $$ $$(2p-1)+E\left(X_{n}|X_{n-1})\right)=2(2p-1)+X_{n-1} $$

so

$$E(X_{n+k}|X_{n})=k(2p-1)+X_{n}=k(p-q)+X_{n}$$

$$E(X_{8}|X_{4}=-2)=4(2p-1)+(-2)=8p-4-2=8p-6$$

$$E(X_4)=8p-4$$