Let $(W_i)_{0 \leq i \leq n}$ be a simple asymmetric random walk which starts from 0 (that is the probability to go up at each step is equal to $p \neq \frac{1}{2}$). Define $W^*_n = \max\limits_{0 \leq i \leq n} W_i$. Assume that $n$ and $a$ are even positive integers, $b$ is an even integer, $b \leq a \leq n$, and $2a - b \leq n$.
$P(W^*_n \geq a, W_n = b) = \frac{n!}{\left(\frac{n-b}{2} +a\right)! \left(\frac{n+b}{2} - a\right)!} \left(\frac{1}{2}\right)^{\frac{n+b}{2}} \left(1-p\right)^{\frac{n-b}{2}}$
(b) Compute $P(W^*_n < a, W_n = b) $.
For part a I find an argument which quoting some stackexchange author which was very helpful for the symmetric case however I wanted to ask for the given asymmetric case does it still applicable? Argument is as follows:
"If $W^*_n \geq a$, you need to reflect the random walk the first time you hit $a$. Now $S_n$ will hit $b$ if you go down $a - b$ steps after the first time you hit $a$, this is equivalent to asking that the reflected walk goes up $a - b$ times, i.e., $S_n = a + (a - b) = 2a - b$."
this is for proving part a. Can I just say $P(W_n^* < a, S_n = b) = P(S_n = b) - P(W_n^* \geq a - 1, S_n = b) $ using the solution from part a and general path to hit b without restrictions whatever that be. Thank you