Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 12F (Probability)

Table of contents
- a 12F
  - Solution a
- b 12F
  - Solution b
- c 12F
  - Solution c

a

Solution

Direct integration of the gamma density gives, for

t < 1

,

M_{S} (t) = \frac{1}{( n - 1 )!} \int_{0}^{\infty} s^{n - 1} e^{- (1 - t) s} d s = (1 - t)^{- n} .

b

Solution

For

0 < x, y < 1

a = x^{- 1/ n} - 1

and

b = y^{- 1/ n} - 1

. Conditional on

S

,

P (X_{1} \leq x, X_{2} \leq y ∣ S) = e^{- (a + b) S} .

Using part (a),

F_{X_{1}, X_{2}} (x, y) = (x^{- 1/ n} + y^{- 1/ n} - 1)^{- n} .

Differentiation gives

f_{X_{1}, X_{2}} (x, y) = \frac{n + 1}{n} \frac{x ^{- 1 - 1/ n} y ^{- 1 - 1/ n}}{( x ^{- 1/ n} + y ^{- 1/ n} - 1 ) ^{n + 2}} .

Putting

y = 1

yields

F_{X_{1}} (x) = x

, so

X_{1}

is uniform on

(0, 1)

.

c

Solution

Because

u \mapsto (1 + u / S)^{- n}

decreases,

X_{i} < X_{i + 1}

exactly when

E_{i} > E_{i + 1}

. Therefore

K \geq i

exactly when

E_{1} \leq \dots \leq E_{i}

, an event of probability

1/ i!

independent of

S

. The tail-sum rearrangement gives

E \sum_{i = 1}^{K} S^{i} = \sum_{i = 1}^{n} E [S^{i}] P (K \geq i) .

Since

E [S^{i}] = (n + i - 1)! / (n - 1)!

, the

i

th term is

(i n + i - 1)

, proving the formula.

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