Introduction

In the theory of probability and statistics, a Bernoulli trial or Bernoulli Experiment is a random experiment with exactly two mutually exclusive outcomes, “Success” and “Failure” with the probability of success remains same every time the experiment is conducted. The name Bernoulli trial or Bernoulli distribution named after a Swiss scientist Jacob Bernoulli.

Bernoulli Trials

Trials of random experiment are called Bernoulli trials, if they satisfy the following conditions

  1. Each trial of a random experiment has two possible outcomes (like “Success” and “Failures”).
  2. The trials are independent. That is the outcome of one trial has no influence on the outcome of another trial.
  3. The probability of success remains constant from one trial to another.

Examples of Bernoulli trial are

  • Tossing of a coin (“Head” or “Tail”),
  • Sex of newborn baby (“Male” or “Female”),
  • Answer to true/false question (“True” or “False”).

Bernoulli Distribution

Consider a random experiment having two possible outcomes, namely, “Success” (S) and “Failure” (F) with respective probabilities $p$ and $q$. The outcomes success and failures are mutually exclusive. Such a trial is called Bernoulli trial.

The probability distribution of the random variable $X$ representing the number of success obtained in a Bernoulli trial is called Bernoulli distribution. Thus the random variable $X$ takes the value 0 and 1 with respective probabilities $q$ and $p$, i.e., $$ \begin{equation*} P(F) = P(X=0) = q, \text{ and } P(S) = P(X=1) = p. \end{equation*} $$

Definition

The discrete random variable $X$ is said to have Bernoulli distribution if its probability mass function is given by $$ \begin{equation*} P(X=x) = p^x q^{1-x}, \; x=0,1; 0<p,q<1; q=1-p. \end{equation*} $$ Here

  • $P(X=x)\geq 0$ for all $x$.
  • $\sum_{x} P(X=x) = P(X=0) + P(X=1) = q+p =1$.

Hence $P(X=x)$ is a legitimate probability mass function.

Key features of Bernoulli's Distribution

  • There are only two outcomes for a random experiment like success ($S$) and failure ($F$).
  • The outcomes are mutually exclusive.
  • The probability of success is $p$.
  • The random variable $X$ is the total number of success.

Graph of Bernoulli Distribution

Mean of Bernoulli Distribution

The mean (expected value) of Bernoulli random variable $X$ is $E(X) = p$.

Proof

The expected value of Bernoulli random variable $X$ is $$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E(X) &=& \sum_{x=0}^1 x P(X=x) \\ &=& 0\times P(X=0) + 1\times P(X=1)\\ &=& 0\times q + 1\times p = p. \end{eqnarray*} $$

Variance of Bernoulli Distribution

The variance of Bernoulli random variable $X$ is $V(X) = pq$.

Proof

Variance of random variable $X$ is given by $$ V(X) = E(X^2) - [E(X)]^2. $$

Let us find the expected value of $X^2$. $$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E(X^2) &=& \sum_{x=0}^1 x^2 P(X=x) \\ &=& 0\times P(X=0) + 1\times P(X=1)\\ &=& 0\times q + 1\times p = p. \end{eqnarray*} $$ Thus variance of $X$ is $$ \begin{eqnarray*} V(X) &=& E(X^2)-[E(x)]^2\\ &=& p-p^2 = p(1-p)=pq. \end{eqnarray*} $$

Moment Generating Function of Bernoulli Distribution

The moment generating function (M.G.F.) of Bernoulli distribution is given by $M_X(t) = (q + pe^t)$ for $t\in R$.

Proof

The moment generating function of Bernoulli random variable $X$ is $$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) M_X(t) &=& E(e^{tX}) \\ &=& \sum_{x=0}^1 e^{tx}P(X=x)\\ &=& e^0 P(X=0) + e^tP(X=1)\\ &=& q+pe^t. \end{eqnarray*} $$

Probability Generating Function of Bernoulli Distribution

The probability generating function (P.G.F.) of Bernoulli distribution is given by $P_X(t) = q+pt$, $t\in R$.

Proof

The probability generating function of Bernoulli random variable $X$ is given by $$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) P_X(t) &=& E(t^{X}) \\ &=& \sum_{x=0}^1 t^xP(X=x)\\ &=& t^0 P(X=0) + t^1P(X=1)\\ &=& q+pt. \end{eqnarray*} $$

Characteristic Function of Bernoulli Distribution

The Characteristic function of Bernoulli distribution is given by $\phi_X(t) = (q + pe^{it})$.

Proof

The characteristic function of Bernoulli random variable $X$ is $$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \phi_X(t) &=& E(e^{itX}) \\ &=& \sum_{x=0}^1 e^{itx}P(X=x)\\ &=& e^0 P(X=0) + e^{it}P(X=1)\\ &=& q+pe^{it}. \end{eqnarray*} $$

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