 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/9/a/i/solution

Markov's inequality: if $X \geq 0$ and $a > 0$ , then $P (X \geq a) \leq E [X] / a$ . Pointwise,

X \geq a 1_{{X \geq a}} .

Taking expected values, using their monotonicity and the identity

E [1_{E}] = P (E)

, gives

E [X] \geq a P (X \geq a)

. Division by the positive threshold proves the inequality. If the expectation is infinite, the extended-valued bound remains true but uninformative.

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