 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/6/a/ii/solution

Expand the square and use linearity of expectation, together with

E [X] = μ

and

E [1] = 1

var (X) = E [X^{2} - 2 μ X + μ^{2}] = E [X^{2}] - 2 μ E [X] + μ^{2} = E [X^{2}] - (E [X])^{2} .

This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

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