 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008/m/m4/solution

M_{X} (θ) = E (e^{θX})

where finite. The continuity theorem states that, if moment generating functions are finite on a common open interval containing zero and converge there to the moment generating function of a probability law, the corresponding distributions converge weakly to that law.

Integrating the gamma density and using the gamma integral gives

M_{X} (θ) = \frac{λ ^{α}}{Γ ( α )} \int_{0}^{\infty} x^{α - 1} e^{- (λ - θ) x} d x = (\frac{λ}{λ - θ})^{α}, θ < λ .

For a geometric waiting time, on

1, 2, \dots

, summing the geometric series gives

M_{Y} (θ) = \sum_{j \geq 1} p (1 - p)^{j - 1} e^{j θ} = \frac{p e ^{θ}}{1 - ( 1 - p ) e ^{θ}}, θ < - lo g (1 - p)

when

0 < p < 1

. For

p = 1

Y = 1

deterministically. The waiting times between successive heads are independent and identically distributed, so

N = Y_{1} + \dots + Y_{k}

and

M_{N} = M_{Y}^{k}

. Therefore

M_{2 pN} (θ) = {\frac{p e ^{2 pθ}}{1 - ( 1 - p ) e ^{2 pθ}}}^{k} ⟶ (1 - 2 θ)^{- k} .

Indeed the denominator is

p (1 - 2 θ) + O (p^{2})

and the numerator is

p + O (p^{2})

. On any sufficiently small fixed neighbourhood of zero these moment generating functions exist for all sufficiently small

p

. The limit is the moment generating function of a gamma law with shape $k$ and rate $1/2$ , so

2 pN d Gamma (k, 1/2) .

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