 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-2/3f/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

u = t^{c}

1 = A_{c} \int_{0}^{\infty} t^{c - 1} e^{- t^{c}} d t = \frac{A _{c}}{c} \int_{0}^{\infty} e^{- u} d u, A_{c} = c .

S_{c} (v) = P (X_{c} \geq v) = \int_{v}^{\infty} c t^{c - 1} e^{- t^{c}} d t = e^{- v^{c}}, v \geq 0.

Since the distribution is continuous, the choices

>

and

\geq

have the same probabilities. The conditional probability is consequently

P (X_{c} \geq s + t ∣ X_{c} \geq t) = \frac{S _{c} ( s + t )}{S _{c} ( t )} = e^{- [(s + t)^{c} - t^{c}]} .

For

s, t > 0

and

c > 1

, the supplied inequality, applied to

s / t

, gives

(s + t)^{c} > t^{c} + s^{c}

. Taking exponentials yields

P (X_{c} \geq s + t ∣ X_{c} \geq t) < e^{- s^{c}} = P (X_{c} \geq s) .

This expresses the increasing hazard function

h_{c} (t) = c t^{c - 1}

: surviving to a later age shortens the remaining lifetime in this conditional distribution. At

c = 1

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