 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2023/1/c/i/solution

Direct integration of the uniform distribution yields

E U_{j} = b /2

and

E U_{j}^{2} = b^{2} /3

E W_{n} = \prod_{j = 1}^{n} E U_{j} = (b /2)^{n}, E W_{n}^{2} = (b^{2} /3)^{n} .

Subtracting the square of the expected value gives the variance:

E W_{n} = (b /2)^{n}, Var (W_{n}) = (b^{2} /3)^{n} - (b^{2} /4)^{n} .

Since

W_{n} \geq 0

, Markov's inequality implies, for every

ε > 0

P (W_{n} > ε) \leq \frac{( b /2 ) ^{n}}{ε} ⟶ 0 (0 < b < 2) .

Hence

W_{n} \to 0 in probability for 0 < b < 2.

The first moment suffices: the variance need not tend to zero throughout this range.

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