Geometric Distribution

1 Geometric Distribution

If we conduct a series of independent Bernoulli trials , with probability p of success on each trial, where we stop on the first success, what is the probability of exactly x failures before the first success?

This is exactly the probability of the sequence ...0, 0, 0, ..., 0, 1 ..., where there are x zeroes, i.e

P( X = x ) ...= ...p ( 1 – p )^x , ......x ≥ 0

This is called the geometric distribution. This is a valid probability distribution, i.e.

∞
∑		p ( 1 – p )x
0

...= ...1

From the result for the geometric series .
[ Put S ...= ...∑₀^∞ (1 – p)^x . .........(1 – p) S ...= ...∑₁^∞ (1 – p)^x ...= ...S – 1 ......⇒ ......S ...= ...1/p ]

2 Moments

The of the Geometric Distribution is given by

Φ( t ) ...= ...

∞
∑		...p ( 1 – p ) x t x
x = 0

...= ......p ⁄ ( 1 – ( 1 – p )t

Φ'( t ) ...= ...( 1 – p ) p ( 1 – ( 1 – p )t )^–2

Φ''( t ) ...= ...2 ( 1 – p )² p ( 1 – ( 1 – p )t )^–3

Φ'( 1 ) ...= ...( 1 – p ) ⁄ p , ............Φ''( 1 ) ...= ...2 ( 1 – p )² ⁄ p²

E( X ) ...= ...( 1 – p ) ⁄ p

var( X ) ...= ...2 ( 1 – p )² ⁄ p² ...+ ...( 1 – p ) ⁄ p ...– ...( 1 – p )² ⁄ p²

...........................= ...

	( 1 – p )2 + p( 1 – p )
	...p2 ...

...= ...

	1 – p
	...p2 ...

E( X ) ...= ...( 1 – p ) ⁄ p var( X ) ...= ...( 1 – p ) ⁄ p2

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