 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-39/1/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

E [s^{Y_{j}} ∣ T_{j}] = exp [- \frac{j θ}{2} (1 - s) T_{j}] .

E [s^{Y_{j}}] = \frac{( 2 j )}{( 2 j ) + ( j θ /2 ) ( 1 - s )} = \frac{j - 1}{j - 1 + θ ( 1 - s )} = \frac{p _{j}}{1 - ( 1 - p _{j} ) s}, p_{j} = \frac{j - 1}{j - 1 + θ} .

P (Y_{j} = k) = p_{j} (1 - p_{j})^{k}, k = 0, 1, \dots .

θ = 0

, this reduces to

Y_{j} = 0

 New to topics? Read the docs here!