 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/16/c/iii/solution

E [X] = \frac{1}{2} \int_{- \infty}^{\infty} (μ + 2 s t) sech^{2} t d t = μ + s \int_{- \infty}^{\infty} t sech^{2} t d t .

The remaining integrand is an odd function. Its integral is absolutely convergent, since

sech^{2} t \leq 4 e^{- 2∣ t ∣}

and

∣ t ∣ e^{- 2∣ t ∣}

is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. Consequently

E [X] = μ .

Equivalently, the logistic distribution is symmetric about

μ

, and its exponentially decaying tails ensure that its expected value exists.

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