 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-206/4/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

For

0 < p < 1

Pr (Y = y) = p^{y} (1 - p)^{1 - y} = exp [y lo g \frac{p}{1 - p} + lo g (1 - p)], y \in {0, 1} .

exp {[y θ - b (θ)] / ϕ + c (y, ϕ)}

, take

θ = lo g \frac{p}{1 - p}, b (θ) = lo g (1 + e^{θ}), ϕ = 1, c (y, 1) = 0.

The support is

{0, 1}

; outside it the probability is zero. The natural parameter of an exponential family is the log odds, and the dispersion parameter is fixed rather than freely estimated. Differentiating the cumulant gives

E Y = b^{'} (θ) = \frac{e ^{θ}}{1 + e ^{θ}} = p, Var (Y) = ϕ b^{''} (θ) = \frac{e ^{θ}}{( 1 + e ^{θ} ) ^{2}} = p (1 - p) .

p = 0

1

the degenerate distributions are limiting boundary cases with natural parameter

- \infty

+ \infty

 New to topics? Read the docs here!