 Solution

ID: past-exam-of-the-options-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/n2/e/ii/solution

For

∣ s ∣ \leq 1

, all the generating-function expectations are absolutely bounded. Conditional on

X = k

, independence gives

E [s^{T} ∣ X = k] = s \prod_{i = 1}^{k} E [s^{T_{i}}] = s G_{T} (s)^{k} .

k = 0

the product is one, consistent with the empty-family case. The law of total expectation now yields

G_{T} (s) = s k = 0 \sum \infty p_{k} G_{T} (s)^{k} = s G_{X} (G_{T} (s)) = s exp {λ (G_{T} (s) - 1)} .

T

ensures

G_{T} (1) = 1

. The conditional-expectation proof avoids assuming a finite expected total before it has been established.

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