My notes say the following:
Example (Random Walk)
A particle moves $n$ steps on a number line. The particle starts at $0$, and at each step it moves $1$ unit to the right or to the left, with equal probabilities. Assume all steps are independent. Let $Y$ be the particles position after $n$ steps. Find the PMF of $Y$.
Consider each step to be a Bernoulli trial, where right is considered a success and left is considered a failure.
The number of steps the particle takes, say $X$, to the right is a $Bin(n, 1/2)$ random variable, that is $x \sim Bin(n, 1/2)$.
If $X = j$, then the particle has taken $j$ steps to the right and $n – j$ steps to the left, giving a final position of $j – (n – j) = 2j – n$.
So we can express $Y$ as a one-to-one function of $X$, namely, $Y = 2X – n$.
The PMF of $Y$ can then be found from the PMF of $X$:
$P(Y = k) = P(2X – n = k) = P(X = (n + k)/2) = {n \choose \frac{n + k}{2}} \frac{1}{2^n}$
How does one conclude that $P(X = (n + k)/2) = {n \choose \frac{n + k}{2}} \frac{1}{2^n}$?
I would greatly appreciate it if people could please take the time to clarify this.
Best Answer
Note that the PMF of a binomial distribution is$$\Pr\{X=k\}=\binom{n}{k}p^k(1-p)^{n-k}$$where $p$ is the probability of success in a single experiment.
P.S.
It is worth noting that $n$ and $k$ in your question must be both even or both odd simultaneously.