From binomial distribution to Gaussian and Poisson distributions. Plus 2D random walk

Mainly for referencing.

1.Derivation of Gaussian distribution from binomial distribution

$P(k,n)=\frac{n!}{k!(n-k)!}p^kq^{n-k}$as n $\rightarrow\infty$ and $p$ remains finite.

• Taking log:$ln(p)=ln(n!)-ln(k!)-ln((n-k)!)+kln(p)+(n-k)ln(q)$
• Using Stirling approximation :

$ln(n!) \approx nln(n)+n+\frac{1}{2}ln(2\pi n)$

$ln(p) \approx [nln(n)+n+\frac{1}{2}ln(2\pi n)]+[kln(k)+k+\frac{1}{2}ln(2\pi k)]+[(n-k)ln((n-k))$$+n+\frac{1}{2}ln(2\pi (n-k)]+kln(p)+(n-k)ln(q)$

(1) For terms of order $ln(n)$, we evaluate them at mean value for $k=\langle k\rangle=np$:$\frac{1}{2}ln(2\pi n)-\frac{1}{2}ln(2\pi k)-\frac{1}{2}ln(2\pi (n-k))$$\implies \frac{1}{2}ln(2\pi n)-\frac{1}{2}ln(2\pi np)-\frac{1}{2}ln(2\pi np)$$\implies \frac{1}{2}ln(2\pi npq)$which is $ln(\frac{1}{\sqrt{2\pi \sigma^2} })$ , where $\sigma^2=npq$.

(2) For terms $kln(p)$ and $(n-k)ln(n-k)$ again we plug in $k=\langle k\rangle=np$:

$kln(p) +(n-k)ln(n-k) \implies np ln(p)+nqln(q)$

(3) For terms $nln(n)$, $kln(p)$ and $(n-k)ln(n-k)$ we can Taylor expand to 2nd order and plug in results from (2) to obtain:$nln(n)+ kln(p)+(n-k)ln(n-k) \implies -kln(p)-nqln(q)+kln(q)-npln(q)-\frac{1}{2npq}(k-nq)^2$which is $-\frac{1}{2\sigma^2} (k-np)^2$.

Thus from (1) and (3):$ln(p)=ln(\frac{1}{\sqrt{2\pi \sigma^2} })+-\frac{1}{2\sigma^2} (k-np)^2$$p(k)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}(\frac{k-\langle k\rangle}{\sigma})^2}$

2.Derivation of Poisson distribution from binomial distribution

 $P(k,n)=\frac{n!}{k!(n-k)!}p^kq^{n-k}$, with the assumption that as $n\rightarrow \infty$, $p>>1$ and $np \rightarrow 0$.

• Let’s define $\mu= np$. Thus, $p=\frac{\mu}{n}$, $q=(1-\frac{\mu}{n})$.

We can rewrite the binomial distribution:$p(k,n)=\frac{n!}{k!(n-k)!}(\frac{\mu}{n})^k(1-\frac{\mu}{n})^{n-k}$$\implies \frac{n!}{(n-k)!k!} \frac{\mu^k}{n^k} (1-\frac{\mu}{n})^n (1-\frac{\mu}{n})^{-k}$

• Now we take the limit of the distribution as $n\rightarrow \infty$ by looking at each term involving n:
  (0) Term $\frac{\mu^k}{k!}$ is irrelevant as we take limit of n.
  (1)$\lim_{n\to \infty} \frac{n!}{(n-k)!n^k}$Since $\frac{n!}{(n-k)!n^k}=\frac{n(n-1) \dots (n-k)(n-k-1)\dots1 }{(n-k)(n-k-1)\dots 1} \frac{1}{n^k}=\frac{n(n-1) \dots (n-k+1)}{n^k}$ $\implies \frac{n}{n}\frac{n-1}{n} \dots\frac{n-k+1}{n}$
  If we take limit as $n \to \infty$, each of the n terms above goes to 1. Thus $\lim_{n\to \infty} \frac{n!}{(n-k)!n^k}=1$.
  (2)$\lim_{n\to \infty} (1-\frac{\mu}{n})^n$Using definition of e:$e=\lim_{n\to \infty} (1+\frac{1}{x})^x$We let$x=-\frac{n}{\mu}$The limit of interest becomes$\implies \lim_{x\to \infty} (1+\frac{1}{x})^{-\mu x}=e^{-\mu}$
  (3)$\lim_{n\to \infty} (1-\frac{\mu}{n})^{-k} =1^{-k}=1$

Thus$p(k,n)=\frac{\mu!e^{-\mu}}{k!}$which is the Poisson distribution.

3. 2D random walk stimulation in Python:

import numpy as np
import random 
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
import math 

def two_d_randomwalk(x,y):
    theta = np.random.random()*2*math.pi #scaling the angle 2pi
    x+=math.cos(theta)
    y+=math.sin(theta)
    return (x,y)

#One random walk
x,y=0,0
steps = 10000
a=np.zeros((steps,2))
r=[0]

for i in range(steps):
    #position coordinates
    x,y=two_d_randomwalk(x,y)
    a[i:]=x,y 
    #final distance squared
    distance_sq=x**2+y**2
    r.append(distance_sq)

    
lc=LineCollection(zip(a[:-1],a[1:]),array=z, cmap=plt.cm.hsv)
fig, ax=plt.subplots(1,1)
ax.add_collection(lc)
ax.margins(0.1)
plt.show()


fig = plt.figure()
ax = fig.add_subplot(111)
plt.xlabel('Number of steps')
plt.ylabel('Distance squared')
ax.plot(r)
plt.show()