 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit/3/solution

Assume

λ > 0

- \infty

x

F_{X} (x) = {0, 1 - e^{- λ x}, x < 0, x \geq 0.

E [X] = \int_{0}^{\infty} x λ e^{- λ x} d x = [- x e^{- λ x}]_{0}^{\infty} + \int_{0}^{\infty} e^{- λ x} d x = \frac{1}{λ} .

Similarly,

E [X^{2}] = [- x^{2} e^{- λ x}]_{0}^{\infty} + 2 \int_{0}^{\infty} x e^{- λ x} d x = \frac{2}{λ ^{2}} .

The endpoint terms vanish because exponential decay dominates each fixed power of

x

. Therefore

E [X] = \frac{1}{λ}, Var (X) = E [X^{2}] - E [X]^{2} = \frac{1}{λ ^{2}} .

The mean has the same units as

X

, while the variance has their square;

λ

has reciprocal units.

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