Binomial Distribution

Binomial Distribution

Consider a series of $n$ (finite) independent Bernoulli trials in which the probability $p$ of success in any trial is constant for each trial. Then the probability of failure is $q=1-p$ in any trial.

Binomial distribution can be used under following conditions:

  • The random experiment is performed repeatedly a finite and fixed number of times, say $n$ (number of trials).
  • Only two outcomes, success and failure, are possible at each trial. (the outcomes are mutually exclusive).
  • All the trials are independent, i.e. the results of any trial is not affected by the preceding trials and does not affect the outcomes of succeeding trials.
  • There is a constant probability $p$ of success in $n$ trials. $q = 1-p$ is called the probability of failure.
  • The random variable is the total number of successes in $n$ trials.

Suppose that a trial is repeated so that we have a series of $n$ trials. Let $p$ be the probability of success and $q=1-p$ be the probability of failure in a single trial. Assume that trials are independent and the probability $p$ of success remains constant in every trial.

Let $X$ denote the number of successes in $n$ trials. Then we have $X = 0,1,2,\cdots,n$.

Then the probability that $x$ trials are successes and remaining $(n-x)$ trials are failure is

$$ \begin{eqnarray*} P(X=x) &=& \binom{n}{x}p\cdot p \cdots (x \text{ times})\times q\cdot q \cdots ((n-x)\text{ times}) \\ &=& \binom{n}{x} p^x q^{n-x}. \end{eqnarray*} $$

Hence, the probability of $x$ successes in a series of $n$ Bernoulli is given by

$$ \begin{eqnarray*} P(X=x) & =& \binom{n}{x} p^x q^{n-x},\\ & & \qquad \; x = 0,1,2, \cdots, n; \; 0 \leq p \leq 1, q = 1-p \end{eqnarray*} $$

which is the p.m.f. of Binomial distribution.

Clearly, (i) $P(X=x)\geq 0$ for all $x$ and (ii) $\sum_{x=0}^n P(X=x)=1$. Hence the $P(X=x)$ is a probability mass function.

In notation, it can be written as $X \sim B(n,p)$ distribution. Here $n$ (number of trials) and $p$ (probability of success) are the parameters of Binomial distribution.

Definition of Binomial Distribution

A discrete random variable $X$ is said to have Binomial distribution with parameter $n$ and $p$ if its probability mass function is

$$ \begin{equation*} P(X=x)= \begin{cases} \binom{n}{x} p^x q^{n-x} & x=0,1,2,\cdots, n\\ & 0 < p < 1, q=1-p\\ 0 & \text{ otherwise }. \end{cases} \end{equation*} $$

Key features of Binomial distribution

  • The random experiment consists of n repeated Bernoulli trial.
  • Each trial has only two possible outcomes like success ($S$) and failure ($F$).
  • All the trials are independent, i.e. the results of any trial is not affected by the preceding trials and does not affect the outcomes of succeeding trials.
  • The probability of success $p$ is constant for each trial.
  • The random variable $X$ is the total number of successes in $n$ trials.

Graph of Binomial Distribution

Graph of Binomial distribution with parameter $n=6$ and $p=0.4$ is

Binomial Distribution
Binomial Distribution

Mean and variance of Binomial Distribution

The mean of Binomial distribution is

$$ \begin{eqnarray*} \text{mean = }\mu_1^\prime &=& E(X) \\ &=& \sum_{x=0}^n x\cdot P(X=x)\\ &=& \sum_{x=0}^n x\cdot \binom{n}{x} p^x q^{n-x}\\ &=& \sum_{x=0}^n x\frac{n!}{x!(n-x)!}p^x q^{n-x}\\ &=& 0+\sum_{x=1}^n \frac{n(n-1)!}{(x-1)!(n-x)!}p^x q^{n-x}\\ &=& np\sum_{x=1}^n \frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1} q^{n-x}\\ &=& np\sum_{x=1}^n \binom{n-1}{x-1}p^{x-1}q^{n-x}\\ &=& np\cdot (q+p)^{n-1}\\ &=& np. \end{eqnarray*} $$

Thus the mean of a binomial distribution is $E(X)=np$.

To find variance of $X$, let us find $E(X^2)$ as

$$ \begin{eqnarray*} \mu_2^\prime=E(X^2) & = & E[X(X-1)+X]\\ &=& \sum_{x=0}^n x(x-1)\cdot P(X=x)+ np\\ &=& \sum_{x=0}^n x(x-1)\cdot \binom{n}{x} p^x q^{n-x}+np\\ &=& \sum_{x=0}^n x(x-1)\frac{n!}{x!(n-x)!}p^x q^{n-x}+np\\ &=& 0+0+\sum_{x=2}^n \frac{n(n-1)(n-2)!}{(x-2)!(n-x)!}p^x q^{n-x}+np\\ &=& n(n-1)p^2\sum_{x=2}^n \frac{(n-2)!}{(x-2)!(n-x)!}p^{x-2} q^{n-x}+np\\ &=& n(n-1)p^2\sum_{x=2}^n \binom{n-2}{x-2}p^{x-2}q^{n-x}+np\\ &=& n(n-1)p^2(q+p)^{n-2}+ np, \\ &=& n(n-1)p^2+ np. \end{eqnarray*} $$

Hence, the variance of Binomial distribution is

$$ \begin{eqnarray*} \text{Variance = } \mu_2 &= &\mu_2^\prime-(\mu_1^\prime)^2\\ &=& n(n-1)p^2+ np-(np)^2\\ &=& np-np^2\\ &=&np(1-p)\\ &=&npq\\ \end{eqnarray*} $$

If $X$ is Binomial random variable with parameter $n$ and $p$, mean = $E(X)=np$ and variance $V(X)= npq$. For Binomial distribution Mean $>$ Variance.

M.G.F. of Binomial Distribution

Let $X\sim B(n,p)$ distribution. Then the m.g.f. of $X$ is $M_X(t) = (q+pe^t)^n$.

Proof

Let $X\sim B(n,p)$ distribution. Then the m.g.f. of $X$ is

$$ \begin{eqnarray*} M_X(t) &=& E(e^{tx}) \\ &=& \sum_{x=0}^n e^{tx}\binom{n}{x} p^x q^{n-x} \\ &=& \sum_{x=0}^n \binom{n}{x} (pe^t)^x q^{n-x} \\ &=& (q+pe^t)^n. \end{eqnarray*} $$

Thus the probability generating function of Binomial distribution is $M_X(t) = (q+pe^t)^n$.

C.G.F. of Binomial Distribution

Let $X\sim B(n,p)$ distribution. Then the c.g.f. of $X$ is $K_X(t) =n\log_e (q+pe^t)$.

Proof

Given that $X\sim B(n,p)$ distribution. Then the c.g.f. of $X$ is

$$ \begin{equation*} K_X(t) = \log_e M_X(t) = n\log_e(q+pe^t). \end{equation*} $$

Thus the cumulant generating function of Binomial distribution is $K_X(t) = n\log_e (q+pe^t)$.

Recurrence relation for cumulants

The recurrence relation for cumulants of Binomial distribution is

$$ \begin{equation*} \kappa_{r+1} = pq \frac{d\kappa_r}{dp}. \end{equation*} $$

Proof

The $r^{th}$ cumulant is given by

$$ \begin{equation*} \kappa_r = \bigg[\frac{d^r}{dt^r} n\log_e(q+pe^t)\bigg]_{t=0} \end{equation*} $$

Therefore,

$$ \begin{eqnarray*} \frac{d \kappa_r}{dp} &=& n\bigg[\frac{d^r}{dt^r} \frac{d}{dp} \log_e(q+pe^t)\bigg]_{t=0} \\ &=& n\bigg[\frac{d^r}{dt^r} \bigg(\frac{e^t-1}{q+pe^t}\bigg) \bigg]_{t=0}. \end{eqnarray*} $$

Also,

$$ \begin{eqnarray*} \kappa_{r+1} & = & \bigg[\frac{d^{r+1}}{dt^{r+1}} n\log_e(q+pe^t)\bigg]_{t=0}\\ &=& n\bigg[\frac{d^{r}}{dt^{r}}\bigg(\frac{pe^t}{q+pe^t}\bigg)\bigg]_{t=0}. \end{eqnarray*} $$

Hence,

$$ \begin{eqnarray*} \kappa_{r+1}-pq \frac{d \kappa_r}{dp} &=& n\bigg[\frac{d^{r}}{dt^{r}}\bigg(\frac{pe^t-pqe^t+pq}{q+pe^t}\bigg)\bigg]_{t=0}\\ &=& n\bigg[\frac{d^{r}}{dt^{r}}\bigg(\frac{p(pe^t+q}{q+pe^t}\bigg)\bigg]_{t=0}.\\ &=& n\bigg[\frac{d^{r}}{dt^{r}}(p)\bigg]_{t=0}=0. \end{eqnarray*} $$

Hence, the recurrence relation for cumulants of Binomial distribution is

$$ \begin{equation*} \kappa_{r+1}=pq \frac{d \kappa_r}{dp}. \end{equation*} $$

P.G.F. of Binomial Distribution

Let $X\sim B(n,p)$ distribution. Then the p.g.f. of $X$ is

$$ \begin{eqnarray*} P_X(t) &=& E(t^x) \\ &=& \sum_{x=0}^n t^x\binom{n}{x} p^x q^{n-x} \\ &=& \sum_{x=0}^n \binom{n}{x} (pt)^x q^{n-x} \\ &=& (q+pt)^n. \end{eqnarray*} $$

Thus the probability generatong function of Binomial distribution is $P_X(t) = (q+pt)^n$.

Additive Property of Binomial Distribution

Let $X_1$ and $X_2$ be two independent Binomial variate with parameters $(n_1, p)$ and $(n_2, p)$ respectively. Then $Y=X_1+X_2\sim B(n_1+n_2, p)$.

Let $X_1$ and $X_2$ be two independent Binomial variate with parameters $(n_1, p)$ and $(n_2, p)$ respectively. Let $Y=X_1+X_2$. Then the m.g.f. of $Y$ is

$$ \begin{eqnarray*} M_Y(t) &=& E(e^{tY}) \\ &=& E(e^{t(X_1+X_2)}) \\ &=& E(e^{tX_1} e^{tX_2}) \\ &=& E(e^{tX_1})\cdot E(e^{tX_2})\qquad (\because X_1, X_2 \text{ are independent })\\ &=& M_{X_1}(t)\cdot M_{X_2}(t)\\ &=& (q+pe^t)^{n_1}(q+pe^t)^{n_2}\\ &=& (q+pe^t)^{n_1+n_2}. \end{eqnarray*} $$

which is the m.g.f. of Binomial variate with parameter $n_1+n_2$ and $p$. Hence, $Y=X_1+X_2\sim B(n_1+n_2, p)$.

Recurrence relation for raw moments

The recurrence relation for raw moments of Binomial distribution is

$$ \begin{equation*} \mu_{r+1}^\prime = p \bigg[ q\frac{d\mu_r^\prime}{dp} + n\mu_r^\prime\bigg]. \end{equation*} $$

Recurrence relation for central moments

The recurrence relation for central moments of Binomial distribution is

$$ \begin{equation*} \mu_{r+1} = pq \bigg[ \frac{d\mu_r}{dp} + nr\mu_{r-1}\bigg]. \end{equation*} $$

Recurrence relation for probabilities

The recurrence relation for probabilities of Binomial distribution is

$$ \begin{equation*} P(X=x+1) = \frac{n-x}{x+1}\cdot \frac{p}{q}\cdot P(X=x), \; x=0,1,2\cdots, n-1. \end{equation*} $$

Endnote

You can read step by step tutorial on Binomial Distribution examples,tutorial will help you to understand how to solve numerical examples based on binomial distribution.

Binomial distribution calculator is used to find the probability and cumulative probabilities for binomial random variable given the number of trials ($n$) and probability of success ($p$).

Binomial Distribution Examples

Binomial Distribution Calculator

Let me know in the comments if you have any questions on Binomial Distribution and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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