 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3/26j/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Let

P = N (t) \sim Poisson (x)

, where

x = λ t

. Deleting odd arrivals leaves

M = ⌊ P /2 ⌋

. Put

Q = 1_{{P odd}}

, so

M = (P - Q) /2

E z^{P} = e^{x (z - 1)}

q := E Q = \frac{1 - e ^{- 2 x}}{2}, E [P (- 1)^{P}] = - x e^{- 2 x}, E (PQ) = \frac{x + x e ^{- 2 x}}{2} .

Thus

Var Q = (1 - e^{- 4 x}) /4

and

Cov (P, Q) = x e^{- 2 x}

. Therefore

E M = \frac{2 x + e ^{- 2 x} - 1}{4}, Var M = \frac{4 x - 8 x e ^{- 2 x} - e ^{- 4 x} + 1}{16} .

To establish strict underdispersion for every

x > 0

, subtract the mean:

16 (Var M - E M) = f (x) := 5 - 4 x - (8 x + 4) e^{- 2 x} - e^{- 4 x} .

Here

f (0) = 0

and

f^{'} (x) = 4 [- 1 + 4 x e^{- 2 x} + e^{- 4 x}] < 0

for

x > 0

, because

4 x e^{- 2 x} < 1 - e^{- 4 x}

is equivalent to

2 x < sinh (2 x)

. Hence

Var M (t) < E M (t) (t > 0, λ > 0),

which excludes a doubly-stochastic Poisson representation by part (iii).

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