BINOMIAL AND NORMAL RANDOM VARIABLES

5.1 BINOMIAL RANDOM VARIABLES

Let X be the number of successes in n independent trials of some experi­ment whose outcome is “success” or “failure” when the probability of success in each trial is p. Such a random variable often appears in practice (for example, the number of heads in n tosses) and is called a binomial random variable. More formally we state

DEFINITION 5.1.1 Let (FJ, г = 1, 2, be mutually independent

with the probability distribution given by

(5.1.1) F,- = 1 with probability/?

= 0 with probability 1 — p = q.

Then the random variable X defined by

П

(5.1.2) X = X Y{

i=1

is called a binomial random variable. Symbolically we write X ~ Bin, p).

Note that Fj defined in (5.1.1) is distributed as 5(1, p), which is called a binary or Bernoulli random variable.

theorem 5.1.1 For the binomial random variable X we have

(5.1.3) P(X = k) = Cnkpkqn~k,

(5.1.4) EX = np, and

(5.1.5) VX = npq.

Proof. The probability that the first k trials are successes and the remain­ing n — k trials are failures is equal to phqn~k. Since k successes can occur in any of Cl combinations with an equal probability, we must multiply the above probability by Cl to give formula (5.1.3). Using (5.1.2), the mean and variance of X can be obtained by the following steps:

(5.1.6)

EYi = p

for every і

(5.1.7)

EY* = p

for every і

(5.1.8)

1

•ec

II

= pq

for every і

(5.1.9)

n

EX = £ EYi

i= 1

= np

by Theorem 4.1.6

(5.1.10)

n

FX = X VYi

i= 1

= npq by Theorem 4.3.3.

Note that the above derivation of the mean and variance is much simpler than the direct derivation using (5.1.3). For example, in the direct derivation we must compute

П

EX = X kCnkpkqn~k.

k=

EXAMPLE 5.1.1 Let X be the number of heads in five tosses of a fair coin. Obtain the probability distribution of X and calculate EX and VX.

Подпись: (5.1.11) P(X = k) = Cl image172 Подпись: 32

In this example we have n = 5 and p = 0.5. Therefore by (5.1.3) we have

Evaluating (5.1.11) for each k, we have (5.1.12) P(X = 0) = P(X = 5) = 0.03125

P(X = 1) = P(X = 4) = 0.15625 P(X = 2) = P(X = 3) = 0.3125.

Using (5.1.4) and (5.1.5), we have EX = 2.5 and VX = 1.25.

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