Consider a particle, initially at the origin, jumping around randomly on a 1D lattice. The particle tosses a fair coin, and decides to jump left or right.
A particular trajectory of the particle may look like the following:
Suppose the particle makes n_{+} hops to the right, and n_{-} hops to the left. Then, the total number of steps N = n_{+} + n_{-}, and the position at the end is x = n_{+} - n_{-}.
The process is probabilistic, and the outcome of any single trajectory is impossible to predict. However, let us enumerate the number of ways in which a random walk of N steps, results in n_{+} hops to the right. This is given by, \begin{align*} W(x, N) & = {}^N C_{n_{+}}\\ & = \dfrac{N!}{N-n_{+}!n_{+}!}\\ & = \dfrac{N!}{n_{-}!n_{+}!} \end{align*} The probability p(x, N) of ending up at x after N steps can be obtained by dividing W(x, N) by the total number of paths. Since we can make two potential choices at each step, the total number of paths is 2^N.
p(x, N) = \dfrac{W(x,N)}{2^N}.
For large N, Stirling's approximation is N! \approx \sqrt{2 \pi N} (N/e)^N. For x \ll N, this implies, p(x, N) = \dfrac{1}{\sqrt{2 \pi N}} \exp\left(-\dfrac{x^2}{2N}\right)
Both the distributions have the same shape. However, because one is a discrete distribution, while the other is continuous, they have different normalizations, and hence different actual values of p(x,N).
A particular trajectory of the particle may look like the following:
Suppose the particle makes n_{+} hops to the right, and n_{-} hops to the left. Then, the total number of steps N = n_{+} + n_{-}, and the position at the end is x = n_{+} - n_{-}.
The process is probabilistic, and the outcome of any single trajectory is impossible to predict. However, let us enumerate the number of ways in which a random walk of N steps, results in n_{+} hops to the right. This is given by, \begin{align*} W(x, N) & = {}^N C_{n_{+}}\\ & = \dfrac{N!}{N-n_{+}!n_{+}!}\\ & = \dfrac{N!}{n_{-}!n_{+}!} \end{align*} The probability p(x, N) of ending up at x after N steps can be obtained by dividing W(x, N) by the total number of paths. Since we can make two potential choices at each step, the total number of paths is 2^N.
p(x, N) = \dfrac{W(x,N)}{2^N}.
For large N, Stirling's approximation is N! \approx \sqrt{2 \pi N} (N/e)^N. For x \ll N, this implies, p(x, N) = \dfrac{1}{\sqrt{2 \pi N}} \exp\left(-\dfrac{x^2}{2N}\right)