 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/18/b/ii/solution

When the relevant moments are finite, the mean and variance are

μ = E [X] = \int_{- \infty}^{\infty} x f (x) d x, σ^{2} = E [(X - μ)^{2}] = \int_{- \infty}^{\infty} (x - μ)^{2} f (x) d x .

Equivalently,

σ^{2} = \int x^{2} f (x) d x - μ^{2}

. The mean requires an integrable absolute first moment; finite second moment guarantees both quantities are finite.

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