Conditional probability 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 $q = 1 − p$.

What is $P(X_4 = 4 | X_8 = 6)$?

Thank you, any help would be appreciated.

Answer 1

Work backwards. Between $X_4$ and $X_8$, to go from $4$ to $6$, you need three "up's" and one "down," in no particular order.

Since you're working backwards, you go down with probability $p$ and up with probability $(1-p)$.

This will give you a total probability of $\displaystyle \binom{4}{1}p^1(1-p)^3=4p(1-p)^3$.

Answer 2

Note that for any $A,B \in \mathcal F$ such that $\mathbb P(B) > 0$ we have Bayes rule: $\mathbb P(A|B) = \mathbb P(B|A)\mathbb P(A) \frac{1}{\mathbb P(B)}$

In our case $A = \{ X_4 = 4 \}, B = \{ X_8 = 6 \}$, so that:

$\mathbb P(A) = p^4$, $\mathbb P(B) = {8 \choose 1} p^7 (1-p)$ and lastly $\mathbb P(B|A) = \mathbb P(X_4 = 6 | X_0 = 4) = \mathbb P(X_4 = 2 | X_0 = 0) = {4 \choose 1} p^3 (1-p)$

This means $\mathbb P(X_4 = 4 | X_8 = 6) = \frac{1}{2}$