1.3 Bernoulli trials and the binomial distribution

Bernoulli trials and the binomial distribution

A Bernoulli experiment is a random experiment, the outcome of which can be classified in but one of two mutually exclusive and exhaustive ways, mainly, success or failure (e.g., female or male, life or death, nondefective or defective).

A sequence of Bernoulli trials occurs when a Bernoulli experiment is performed several independent times so that the probability of success, say, p , remains the same from trial to trial. That is, in such a sequence we let p denote the probability of success on each trial. In addition, frequently $q=1-p$ denote the probability of failure; that is, we shall use q and $1-p$ interchangeably.

Bernoulli distribution

Let X be a random variable associated with Bernoulli trial by defining it as follows:

X (success)=1 and X (failure)=0.

That is, the two outcomes, success and failure , are denoted by one and zero, respectively. The p.d.f. of X can be written as

and we say that X has a Bernoulli distribution . The expected value of is

and the variance of X is

It follows that the standard deviation of X is $\sigma =\sqrt{p\left(1-p\right)}=\sqrt{pq}.$

In a sequence of n Bernoulli trials, we shall let $X_i$ denote the Bernoulli random variable associated with the i th trial. An observed sequence of n Bernoulli trials will then be an n -tuple of zeros and ones.

Binomial Distribution

In a sequence of Bernoulli trials we are often interested in the total number of successes and not in the order of their occurrence. If we let the random variable X equal the number of observed successes in n Bernoulli trials, the possible values of X are 0,1,2,…, n . If x success occur, where $x=\mathrm{0,1,2,...,}n$ , then n - x failures occur. The number of ways of selecting x positions for the x successes in the x trials is $$\left(\begin{array}{l}n\\ x\end{array}\right)=\frac{n!}{x!\left(n-x\right)!}.$$ Since the trials are independent and since the probabilities of success and failure on each trial are, respectively, p and $q=1-p$ , the probability of each of these ways is $p^x{\left(1-p\right)}^{n-x}.$ . Thus the p.d.f. of X , say $f\left(x\right)$ , is the sum of the probabilities of these $\left(\begin{array}{l}n\\ x\end{array}\right)$ mutually exclusive events; that is,

$$f\left(x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n.$$

These probabilities are called binomial probabilities, and the random variable X is said to have a binomial distribution .

Summarizing, a binomial experiment satisfies the following properties:

• A Bernoulli (success-failure) experiment is performed n times.
• The trials are independent.
• The probability of success on each trial is a constant p ; the probability of failure is $q=1-p$ .
• The random variable X counts the number of successes in the n trials.

A binomial distribution will be denoted by the symbol $b\left(n,p\right)$ and we say that the distribution of X is $b\left(n,p\right)$ . The constants n and p are called the parameters of the binomial distribution , they correspond to the number n of independent trials and the probability p of success on each trial. Thus, if we say that the distribution of X is $b\left(12,14\right),$ we mean that X is the number of successes in n =12 Bernoulli trials with probability $p=\frac{1}{4}$ of success on each trial.

A result that had to follow from the fact that $f\left(x\right)$ is a p.d.f. We use the binomial expansion to find the mean and the variance of the binomial random variable X that is $b\left(n,p\right)$ . The mean is given by

Since the first term of this sum is equal to zero, this can be written as

because $x/x!=1/\left(x-1\right)!$ when $x>0.$

To find the variance, we first determine the second factorial moment $E\left[X\left(X-1\right)\right]$ :

The first two terms in this summation equal zero; thus we find that

$$E\left[X\left(X-1\right)\right]={\displaystyle \sum _{x=2}^n\frac{n!}{\left(x-2\right)!\left(n-x\right)!}{p}^x{\left(1-p\right)}^{n-x}}.$$

After observing that $x\left(x-1\right)/x!=1/\left(x-2\right)!$ when $x>1$ . Letting $k=x-2$ , we obtain

$$E\left[X\left(X-1\right)\right]={\displaystyle \sum _{x=0}^{n-2}\frac{n!}{k!\left(n-k-2\right)!}{p}^{k+2}{\left(1-p\right)}^{n-k-2}}.=n(n-1)p^2{\displaystyle \sum _{x=0}^{n-2}\frac{\left(n-2\right)!}{k!\left(n-2-k\right)!}{p}^k{\left(1-p\right)}^{n-2-k}}.$$

Since the last summand is that of the binomial p.d.f. $b\left(n-\mathrm{2,}p\right)$ , we obtain $$E\left[X\left(X-1\right)\right]=n\left(n-1\right)p^2.$$

Thus, $$\begin{array}{l}{\sigma }^2=Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=E\left[X\left(X-1\right)\right]+E\left(X\right)-{\left[E\left(X\right)\right]}^2\\ =n\left(n-1\right)p^2+np-{\left(np\right)}^2=-n{p}^2+np=np\left(1-p\right).\end{array}$$

Summarizing ,

if X is $b\left(n,p\right)$ , we obtain

$$\mu =np,{\sigma }^2=np\left(1-p\right)=npq,\sigma =\sqrt{np\left(1-p\right)}.$$