Approximation of probability that the walker is at the origin after $2n$ steps I'm reading Lawler's "Lecture on comtemporary probability". There are $2$ parts in the book that I don't understand:


*

*"In order for the walker to be at the origin after $2n$ steps, the walker will have had to have taken an even number of steps in each component. The probability of doing this is about $2^{1-d}$". I understand the first statement, but why the second is true?

*Why is the approximation the product of these $2$ probabilities?
 A: 1) To be looked at is the event that for each direction the numbers of steps taken in that direction is even. This however under the extra condition that the total number of steps is even (we are looking at times of the form $2n$).
If $X_i$ denotes the number of steps in direction $i\in\{1,\dots,d\}$ then: $$P(X_1\text{ is even}\wedge\cdots\wedge X_d\text{ is even}\mid X_1+\cdots+X_1\text{ is even})=$$$$\frac{P(X_1\text{ is even}\wedge\cdots\wedge X_d\text{ is even})}{P(X_1+\cdots+X_d\text{ is even})}\sim\frac{2^{-d}}{2^{-1}}=2^{1-d}$$
A: Based on @drhab answer, this is my explanation for the second question.
Let $A$ be the event "$S_n = 0$", $B$ is the event "$n$ is even", $C$ is the event "each $X_i$ is even". Then $C \subset B$.
Then we have:


*

*$\mathbb P(S_{2n} = 0) = \mathbb P(A|B)$

*The probability in the first question is $\mathbb P(C | B)$

*The probability that all component is $0$ is $\mathbb P(A|C)$
Your second question is thus equivalent to why $\mathbb P(A|B) = \mathbb P(A|C) \times \mathbb P(C|B)$? Base on the relation of $A,B,C$ and definition of conditional probability, you can prove this is true.
