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GEOMETRIC DISTRIBUTION

Module by: Ewa Paszek

Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

GEOMETRIC DISTRIBUTION

To obtain a binomial random variable, we observed a sequence of n Bernoulli trials and counted the number of successes. Suppose now that we do not fix the number of Bernoulli trials in advance but instead continue to observe the sequence of Bernoulli trials until a certain number r, of successes occurs. The random variable of interest is the number of trials needed to observe the rth success.

Let first discuss the problem when r =1. That is, consider a sequence of Bernoulli trials with probability p of success. This sequence is observed until the first success occurs. Let X denot the trial number on which the first success occurs.

For example, if F and S represent failure and success, respectively, and the sequence starts with F,F,F,S,…, then X =4. Moreover, because the trials are independent, the probability of such sequence is

P( X=4 )=( q )( q )( q )( p )= q 3 p= ( 1−p ) 3 p. P( X=4 )=( q )( q )( q )( p )= q 3 p= ( 1−p ) 3 p.

In general, the p.d.f. f( x )=P( X=x ) f( x )=P( X=x ) , of X is given by f( x )= ( 1−p ) x−1 p, f( x )= ( 1−p ) x−1 p, x=1,2,... x=1,2,... , because there must be x -1 failures before the first success that occurs on trail x. We say that X has a geometric distribution.

Recall that:

for a geometric series, the sum is given by ∑ k=0 ∞ a r k = ∑ k=1 ∞ a r k−1 = a 1−r , ∑ k=0 ∞ a r k = ∑ k=1 ∞ a r k−1 = a 1−r , when | r |<1 | r |<1 .

Thus, ∑ x=1 ∞ f( x )= ∑ x=1 ∞ ( 1−p ) k−1 p= p 1−( 1−p ) =1 , ∑ x=1 ∞ f( x )= ∑ x=1 ∞ ( 1−p ) k−1 p= p 1−( 1−p ) =1 , so that f( x ) f( x ) does satisfy the properties of a p.d.f..

From the sum of geometric series we also note that, when k is an integer,

P( X>k )= ∑ x=k+1 ∞ ( 1−p ) x−1 p= ( 1−p ) k p 1−( 1−p ) = ( 1−p ) k = q k , P( X>k )= ∑ x=k+1 ∞ ( 1−p ) x−1 p= ( 1−p ) k p 1−( 1−p ) = ( 1−p ) k = q k ,

and thus the value of the distribution function at a positive integer k is

P( X≤k )= ∑ x=k+1 ∞ ( 1−p ) x−1 p=1−P( X>k )=1− ( 1−p ) k =1− q k . P( X≤k )= ∑ x=k+1 ∞ ( 1−p ) x−1 p=1−P( X>k )=1− ( 1−p ) k =1− q k .

Example 1

Some biology students were checking the eye color for a large number of fruit flies. For the individual fly, suppose that the probability of white eyes is 1 4 1 4 and the probability of red eyes is 3 4 3 4 , and that we may treat these flies as independent Bernoulli trials. The probability that at least four flies have to be checked for eye color to observe a white-eyed fly is given by

P( X≥4 )=P( X>3 )= q 3 = ( 3 4 ) 3 =0.422. P( X≥4 )=P( X>3 )= q 3 = ( 3 4 ) 3 =0.422.

The probability that at most four flies have to be checked for eye color to observe a white-eyed fly is given by

P( X≤4 )=1− q 4 =1− ( 3 4 ) 4 =0.684. P( X≤4 )=1− q 4 =1− ( 3 4 ) 4 =0.684.

The probability that the first fly with white eyes is the fourth fly that is checked is

P( X=4 )= q 4−1 p= ( 3 4 ) 3 ( 1 4 )=0.105. P( X=4 )= q 4−1 p= ( 3 4 ) 3 ( 1 4 )=0.105.

It is also true that

P( X=4 )=P( X≤4 )−P( X≤3 )=[ 1− ( 3 4 ) 4 ]−[ 1− ( 3 4 ) 3 ]= ( 3 4 ) 3 ( 1 4 ). P( X=4 )=P( X≤4 )−P( X≤3 )=[ 1− ( 3 4 ) 4 ]−[ 1− ( 3 4 ) 3 ]= ( 3 4 ) 3 ( 1 4 ).

In general,

f( x )=P( X=x )= ( 3 4 ) x−1 ( 1 4 ),x=1,2,3,... f( x )=P( X=x )= ( 3 4 ) x−1 ( 1 4 ),x=1,2,3,...

To find a mean and variance for the geometric distribution, let use the following results about the sum and the first and second derivatives of a geometric series. For −1<r<1 −1<r<1 , let g( r )= ∑ k=0 ∞ a r k = a 1−r . g( r )= ∑ k=0 ∞ a r k = a 1−r .

Then g'( r )= ∑ k=1 ∞ ak r k−1 = a ( 1−r ) 2 , g'( r )= ∑ k=1 ∞ ak r k−1 = a ( 1−r ) 2 , and g''( r )= ∑ k=2 ∞ ak( k−1 ) r k−2 = 2a ( 1−r ) 3 . g''( r )= ∑ k=2 ∞ ak( k−1 ) r k−2 = 2a ( 1−r ) 3 .

If X has a geometric distribution and 0<p<1 0<p<1 , then the mean of X is given by

E( X )= ∑ x=1 ∞ x q x−1 p= p ( 1−q ) 2 = 1 p , E( X )= ∑ x=1 ∞ x q x−1 p= p ( 1−q ) 2 = 1 p , (1)

using the formula for g'( x ) g'( x ) with a=p a=p and r=q r=q .

Note:

for example, that if p =1/4 is the probability of success, then E( X )=1/( 1/4 )=4 E( X )=1/( 1/4 )=4 trials are needed on the average to observe a success.

To find the variance of X, let first find the second factorial moment E[ X( X−1 ) ] E[ X( X−1 ) ] . We have

E[ X( X−1 ) ]= ∑ x=1 ∞ x( x−1 ) q x−1 p= ∑ x=1 ∞ pqx( x−1 ) q x−2 = 2pq ( 1−q ) 3 = 2q p 2 . E[ X( X−1 ) ]= ∑ x=1 ∞ x( x−1 ) q x−1 p= ∑ x=1 ∞ pqx( x−1 ) q x−2 = 2pq ( 1−q ) 3 = 2q p 2 .

Using formula for g''( x ) g''( x ) with a=pq a=pq and r=q r=q . Thus the variance of X is

Var( X )=E( X 2 )− [ E( X ) ] 2 ={ E[ X( X−1 ) ]+E( X ) }− [ E( X ) ] 2 = = 2q p 2 + 1 p − 1 p 2 = 2q+p−1 p 2 = 1−p p 2 . Var( X )=E( X 2 )− [ E( X ) ] 2 ={ E[ X( X−1 ) ]+E( X ) }− [ E( X ) ] 2 = = 2q p 2 + 1 p − 1 p 2 = 2q+p−1 p 2 = 1−p p 2 .

The standard deviation of X is σ= ( 1−p )/ p 2 . σ= ( 1−p )/ p 2 .

Example 2

Continuing with example 1, with p =1/4, we obtain

μ= 1 1/4 =4, σ 2 = 3/4 ( 1/4 ) 2 =12, μ= 1 1/4 =4, σ 2 = 3/4 ( 1/4 ) 2 =12,

and σ= 12 =3.464. σ= 12 =3.464.

see also:

Binomial Distribution

see also:

Poisson Distribution

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