Negative binomial distribution
The negative binomial distribution
Proof by induction
In the case
it reduces to the Geometric distribution π βΌ N B i n ( 1 , π ) . Now assume the probability mass function above is valid for π βΌ G e o m ( π ) . Let π π βΌ N B i n ( π , π ) be independent so that π βΌ G e o m ( π ) . Then π π + 1 = π π + π βΌ N B i n ( π + 1 , π ) β ( π π + 1 = π₯ ) = β ( π π + π = π₯ ) = π₯ β π = 0 β ( π π + π = π₯ β£ π π = π ) β ( π π = π ) = π₯ β π = 0 β ( π = π₯ β π ) β ( π π = π ) = π₯ β π = 0 ( 1 β π ) π₯ β π π π π ( π + π β 1 π β 1 ) ( 1 β π ) π = π π + 1 ( 1 β π ) π₯ π₯ β π = 0 ( π + π β 1 π β 1 ) = π π + 1 ( π₯ + π π ) ( 1 β π ) π₯ where on the final line we invoked ^P5.
Properties
Let
- Expectation:
π = πΌ β‘ [ π ] = π π π - Variance:
π 2 = V a r β‘ [ π ] = π π π 2 - Moment-generating function:
forπ π ( π‘ ) = ( π 1 β π e π‘ ) π π e π‘ < 1 - Probability-generating function:
π π ( π‘ ) = ( π 1 β π π‘ ) π
Proof of 1β3
Relationship to other distributions
- By the Central limits theorem,
asN B i n ( π , π ) β N β‘ ( π π π , π π π 2 ) π β β