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, \[p(x, N) = \dfrac{1}{\sqrt{2 \pi N}} \exp\left(-\dfrac{x^2}{2N}\right)\]
