Negative Binomial Distribution

1 Negative Binomial Distribution

Conduct a sequence of independent Bernoulli trials , with probability p of success on each trial until we obtain a pre-defined number r of successes. What is the probability of exactly x failures before getting the r successes? Such a random variable is said to have a negative binomial distribution:

P ( X = x ) ...= ...p^r

	–r
	x

(–1)^x ( 1 – p )^x , ......x ≥ 0

where

	–r
	x

......:= ......

	–r ( –r – 1 ) ⋯ ( –r – x + 1 )
	x!

...........................= ...(–1)^x

	r ( r + 1 ) ⋯ ( r + x – 1 )
	x!

...........................= ...(–1)^x

	r + x – 1
	x

So we can write the alternative form

P ( X = x ) ...= ...( ^{r +}_x^{x – 1} ) p^r ( 1 – p ) ^x , ......x ≥ 0

Note Taylor series expansion: ...( 1 – t )^–r ...= ...∑₀^∞ ( ^-_x^r ) ( –t ) ^x .
So ...∑ p^r ( ^-_x^r ) ( –1 ) ^x ( 1 – p ) ^x ...= ...p^r ∑ ( ^-_x^r ) ( – ( 1 – p ) ) ^x ...= ...p^r p^–r ...= ...1

2 Alternative Definition

Consider the total number of trials to get r successes, let this be the random variable Y, then Y = X + r, and

P ( Y = y ) ...= ...P ( X = y – r )

...........................= ...( ^y_y^-_-¹_r ) p^r ( 1 – p ) ^{y – r}

...........................= ...( ^y_r^-_-¹₁ ) p^r ( 1 – p ) ^{y – r} , ......y ≥ r

Which is also sometimes called the negative binomial

3 Moments

The p.g.f. of the Negative Binomial Distribution is given by

Φ( t ) ...= ...

∞
∑		pr –r x (–1) x ( 1 – p ) x t x
x = 0

........................= ...p^r

∞
∑		–r x ( –t ( 1 – p ) ) x
x = 0

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

	p ( 1 – t ( 1 – p ))	r

Φ'( t ) ...= ...

	r pr ( 1 – p )
	( 1 – t ( 1 – p ) ) r + 1

Φ''( t ) ...= ...

	( r + 1 ) r pr ( 1 – p )2
	( 1 – t ( 1 – p ) ) r + 2

Φ'( 1 ) ...= ...

	r ( 1 – p )
	...p ...

Φ''( 1 ) ...= ...

	( r + 1 ) r ( 1 – p )2
	...p 2 ...

E( X ) ...= ...

	r ( 1 – p )
	...p ...

var( X ) ...= ...

	( r + 1 ) r ( 1 – p )2 ...+ ...r p ( 1 – p ) ...– ...r 2 ( 1 – p ) 2
	...p 2 ...

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

	r ( 1 – p )
	...p 2 ...

E( X ) ...= ...r ( 1 – p ) ⁄ p var( X ) ...= ...r ( 1 – p ) ⁄ p 2

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