 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/m/2/a/solution

For

0 \leq z \leq 1

P (M_{n} \leq z) = \prod_{j = 1}^{n} P (X_{j} \leq z) = z^{n} .

Since

M_{n} \leq 1

, for

0 < ε < 1

P (∣ M_{n} - 1∣ > ε) = (1 - ε)^{n} ⟶ 0.

For

ε \geq 1

M_{n}

to one.

 New to topics? Read the docs here!