path: root/content/know/concept/binomial-distribution/index.pdc

---
title: "Binomial distribution"
firstLetter: "B"
publishDate: 2021-02-26
categories:
- Statistics
- Mathematics

date: 2021-02-25T21:08:52+01:00
draft: false
markup: pandoc
---

# Binomial distribution

The **binomial distribution** is a discrete probability distribution
describing a **Bernoulli process**: a set of independent $N$ trials where
each has only two possible outcomes, "success" and "failure",
the former with probability $p$ and the latter with $q = 1 - p$.
The binomial distribution then gives the probability
that $n$ out of the $N$ trials succeed:

$$\begin{aligned}
    \boxed{
        P_N(n) = \binom{N}{n} \: p^n q^{N - n}
    }
\end{aligned}$$

The first factor is known as the **binomial coefficient**, which describes the
number of microstates (i.e. permutations) that have $n$ successes out of $N$ trials.
These happen to be the coefficients in the polynomial $(a + b)^N$,
and can be read off of Pascal's triangle.
It is defined as follows:

$$\begin{aligned}
    \boxed{
        \binom{N}{n} = \frac{N!}{n! (N - n)!}
    }
\end{aligned}$$

The remaining factor $p^n (1 - p)^{N - n}$ is then just the
probability of attaining each microstate.

The expected or mean number of successes $\mu$ after $N$ trials is as follows:

$$\begin{aligned}
    \boxed{
        \mu = N p
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-mean"/>
<label for="proof-mean">Proof</label>
<div class="hidden">
<label for="proof-mean">Proof.</label>
The trick is to treat $p$ and $q$ as independent until the last moment:

$$\begin{aligned}
    \mu
    &= \sum_{n = 0}^N n \binom{N}{n} p^n q^{N - n}
    = \sum_{n = 0}^N \binom{N}{n} \Big( p \pdv{(p^n)}{p} \Big) q^{N - n}
    \\
    &= p \pdv{p} \sum_{n = 0}^N \binom{N}{n} p^n q^{N - n}
    = p \pdv{p} (p + q)^N
    = N p (p + q)^{N - 1}
\end{aligned}$$

Inserting $q = 1 - p$ then gives the desired result.
</div>
</div>

Meanwhile, we find the following variance $\sigma^2$,
with $\sigma$ being the standard deviation:

$$\begin{aligned}
    \boxed{
        \sigma^2 = N p q
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-var"/>
<label for="proof-var">Proof</label>
<div class="hidden">
<label for="proof-var">Proof.</label>
We use the same trick to calculate $\overline{n^2}$
(the mean squared number of successes):

$$\begin{aligned}
    \overline{n^2}
    &= \sum_{n = 0}^N n^2 \binom{N}{n} p^n q^{N - n}
    = \sum_{n = 0}^N n \binom{N}{n} \Big( p \pdv{p} \Big)^2 p^n q^{N - n}
    \\
    &= \Big( p \pdv{p} \Big)^2 \sum_{n = 0}^N \binom{N}{n} p^n q^{N - n}
    = \Big( p \pdv{p} \Big)^2 (p + q)^N
    \\
    &= N p \pdv{p} p (p + q)^{N - 1}
    = N p \big( (p + q)^{N - 1} + (N - 1) p (p + q)^{N - 2} \big)
    \\
    &= N p + N^2 p^2 - N p^2
\end{aligned}$$

Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$:

$$\begin{aligned}
    \sigma^2
    &= \overline{n^2} - \mu^2
    = N p + N^2 p^2 - N p^2 - N^2 p^2
    = N p (1 - p)
\end{aligned}$$

By inserting $q = 1 - p$, we arrive at the desired expression.
</div>
</div>

As $N \to \infty$, the binomial distribution
turns into the continuous normal distribution,
a fact that is sometimes called the **de Moivre-Laplace theorem**:

$$\begin{aligned}
    \boxed{
        \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big)
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-normal"/>
<label for="proof-normal">Proof</label>
<div class="hidden">
<label for="proof-normal">Proof.</label>
We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$
around the mean $\mu = Np$:

$$\begin{aligned}
    \ln\!\big(P_N(n)\big)
    &= \sum_{m = 0}^\infty \frac{(n - \mu)^m}{m!} D_m(\mu)
    \quad \mathrm{where} \quad
    D_m(n) = \dv[m]{\ln\!\big(P_N(n)\big)}{n}
\end{aligned}$$

We use Stirling's approximation to calculate the factorials in $D_m$:

$$\begin{aligned}
    \ln\!\big(P_N(n)\big)
    &= \ln\!(N!) - \ln\!(n!) - \ln\!\big((N - n)!\big) + n \ln\!(p) + (N - n) \ln\!(q)
    \\
    &\approx \ln\!(N!) - n \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) - (N\!-\!n) \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big)
\end{aligned}$$

For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation
to get a non-zero result. We take advantage of $N - N p = N q$:

$$\begin{aligned}
    D_0(\mu)
    &= \ln\!(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln\!(p) + N q \ln\!(q)
    \\
    &= \Big( N \ln\!(N) - N + \frac{1}{2} \ln\!(2\pi N) \Big)
    - \Big( N p \ln\!(N p) - N p + \frac{1}{2} \ln\!(2\pi N p) \Big) \\
    &\qquad - \Big( N q \ln\!(N q) - N q + \frac{1}{2} \ln\!(2\pi N q) \Big)
    + N p \ln\!(p) + N q \ln\!(q)
    \\
    &= N \ln\!(N) - N (p + q) \ln\!(N) + N (p + q) - N - \frac{1}{2} \ln\!(2\pi N p q)
    \\
    &= - \frac{1}{2} \ln\!(2\pi N p q)
    = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
\end{aligned}$$

Next, we expect that $D_1(\mu) = 0$, because $\mu$ is the maximum.
This is indeed the case:

$$\begin{aligned}
    D_1(n)
    &= - \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) + \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big) - 1 + 1
    \\
    &= - \ln\!(n) + \ln\!(N - n) + \ln\!(p) - \ln\!(q)
    \\
    D_1(\mu)
    &= \ln\!(N q) - \ln\!(N p) + \ln\!(p) - \ln\!(q)
    = \ln\!(N p q) - \ln\!(N p q)
    = 0
\end{aligned}$$

For the same reason, we expect that $D_2(\mu)$ is negative.
We find the following expression:

$$\begin{aligned}
    D_2(n)
    &= - \frac{1}{n} - \frac{1}{N - n}
    \qquad
    D_2(\mu)
    = - \frac{1}{Np} - \frac{1}{Nq}
    = - \frac{p + q}{N p q}
    = - \frac{1}{\sigma^2}
\end{aligned}$$

The higher-order derivatives tend to zero for $N \to \infty$, so we discard them:

$$\begin{aligned}
    D_3(n)
    = \frac{1}{n^2} - \frac{1}{(N - n)^2}
    \qquad
    D_4(n)
    = - \frac{2}{n^3} - \frac{2}{(N - n)^3}
    \qquad
    \cdots
\end{aligned}$$

Putting everything together, for large $N$,
the Taylor series approximately becomes:

$$\begin{aligned}
    \ln\!\big(P_N(n)\big)
    \approx D_0(\mu) + \frac{(n - \mu)^2}{2} D_2(\mu)
    = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2}
\end{aligned}$$

Taking $\exp$ of this expression then yields a normalized Gaussian distribution.
</div>
</div>


## References
1.  H. Gould, J. Tobochnik,
    *Statistical and thermal physics*, 2nd edition,
    Princeton.