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.

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 .

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

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

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 .