Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 29L / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

For

y = 0, 1, \dots

,

P_{θ} (Y_{i} = y) = e^{- θ y} (1 - e^{- θ}),

so

Y_{i}

is geometric with success probability

1 - e^{- θ}

. The log likelihood is

n lo g (1 - e^{- θ}) - n θ \overline{Y} .

Its score equation gives

e^{- θ} = \frac{Y}{1 + Y},

and hence

θ_{n} = lo g (\frac{1 + Y}{Y}) .

When

\overline{Y} = 0

, the extended MLE is

+ \infty

; this event has probability tending to zero.

ii

Solution

Put

r = e^{θ}

. The geometric moments are

m := E Y_{i} = \frac{1}{r - 1}, Var (Y_{i}) = \frac{r}{( r - 1 ) ^{2}} .

For

g (y) = lo g ((1 + y) / y)

, one has

g (m) = θ

and

g^{'} (m) = - \frac{1}{m ( 1 + m )} = - \frac{( r - 1 ) ^{2}}{r} .

The central limit theorem for

\overline{Y}

followed by the delta method therefore gives

n (θ_{n} - θ) R i g h t a rro wN (0, \frac{( e ^{θ} - 1 ) ^{2}}{e ^{θ}}) .

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