 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/11/b/solution

Normalize the Rayleigh distribution by substituting

s = x^{2} / (2 σ^{2})

1 = A \int_{0}^{\infty} x e^{- x^{2} / (2 σ^{2})} d x = A σ^{2}, A = σ^{- 2} .

For its sketch, the density starts at zero, is positive for

x > 0

, and tends to zero at infinity. Differentiation gives

f^{'} (x) = \frac{e ^{- x^{2} / (2 σ^{2})}}{σ ^{2}} (1 - \frac{x ^{2}}{σ ^{2}}) .

It therefore has a single maximum at

x = σ

e^{- 1/2} / σ

E [X] = \frac{1}{σ ^{2}} \int_{0}^{\infty} x^{2} e^{- x^{2} / (2 σ^{2})} d x .

Integrating by parts with

x d x e^{- x^{2} / (2 σ^{2})} = - σ^{2} d (e^{- x^{2} / (2 σ^{2})})

makes the boundary term vanish and leaves

\int_{0}^{\infty} e^{- x^{2} / (2 σ^{2})} d x

E [X] = σ \frac{π}{2} .

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