About random walk 1D

Question

I just don't understand why is betha expressed in this way. I don't understand the "conditioning on the initial transition" . Hope you help me thanks

Answer 1

Let $$E = \text{"the event that it returns to $0$"} $$ $$F = \text{"the event that it goes to the right"}$$

$$F^c = \text{"the event it goes to the left"}$$

We let $p$ denote the probability it has a chance of going to the right and $(1 - p)$ of going to the left. Notice we do not assume it ever stays at $0$. This is a simple random walk.

By the law of total probability, this is precisely $$P(E) = P(E|F)p + P(E|F^c)(1-p).$$

The "initial transition" refers to "the first step" of the random walk. It just breaks it up into manageable pieces instead of "conditioning on nth steps", we want to start somewhere, so why not the first step?

Answer 2

Bayes' Theorem states that the probability of event A given event B:

$P(A|B) = \frac{P(A \bigcap B)}{P(B)}$

Thus, $P(A\bigcap B) = P(A|B)*P(B)$

Since The first step $X_1$ is either 1 or -1, for all events "ever return to 0," they must be either of the following:

a) $X_1 = 1 \ \bigcap "ever \ return \ to \ 0"$

b) $X_1 = -1 \ \bigcap "ever \ return \ to \ 0"$

These two conditions are mutually exclusive, so we can sum their respective probability:

a) P($X_1 = 1 \ \bigcap "ever \ return \ to \ 0"$) = $P("ever \ return \ to \ 0" | X_1 = 1)*P(X_1 = 1)$

b) P($X_1 = -1 \ \bigcap "ever \ return \ to \ 0"$) = $P("ever \ return \ to \ 0" | X_1 = -1)*P(X_1 = -1)$

Thus the expression you provided.