Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 5J (Statistical Modelling)

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- Solution 5J

Solution

For

X = x

, the last flip must be the

r

th head. Among the preceding

x + r - 1

flips, exactly

x

are tails and

r - 1

are heads. There are

(x x + r - 1)

choices for their positions, and every resulting sequence has probability

(1 - p)^{x} p^{r}

. The negative binomial stopping argument therefore gives

P (X = x) = (x x + r - 1) (1 - p)^{x} p^{r}, x = 0, 1, \dots,

so

X

has the failures-before-the-

r

th-success negative binomial distribution.

To put the mass function into exponential family form, set

θ = lo g (1 - p), - \infty < θ < 0.

Since

p = 1 - e^{θ}

,

P_{θ} (X = x) = (x x + r - 1) exp {θ x + r lo g (1 - e^{θ})} = h (x) exp {θT (x) - A (θ)},

where

T (x) = x, A (θ) = - r lo g (1 - e^{θ}), h (x) = (x x + r - 1) .

Thus the natural parameter of an exponential family is

θ

, and the Fisher-Neyman factorization theorem shows that

T (X) = X

is a sufficient statistic. This is the negative binomial exponential family.

The exponential-family derivative identities now give

E [X] = A^{'} (θ) = \frac{r e ^{θ}}{1 - e ^{θ}} = \frac{r ( 1 - p )}{p}

and

var (X) = A^{''} (θ) = \frac{r e ^{θ}}{( 1 - e ^{θ} ) ^{2}} = \frac{r ( 1 - p )}{p ^{2}} .

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