 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/m/1/a/solution

Z^{2} \geq a^{2} 1_{{∣ Z ∣ \geq a}}

. Taking expectations and dividing by

a^{2} > 0

P (∣ Z ∣ \geq a) \leq E Z^{2} / a^{2} .

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