Past exam of the options I course of the University of Oxford / 2010 / M / M1

Table of contents
- a M1
  - Solution a
- b M1
  - i b
    - Solution i
  - ii b
    - Solution ii
  - iii b
    - Solution iii
  - iv b
    - Solution iv
- c M1
  - Solution c

a

Solution

For rate

λ > 0

M_{X} (t) = \int_{0}^{\infty} λ e^{- (λ - t) x} d x = \frac{λ}{λ - t}, t < λ .

(1)

This function is finite near zero, so its third derivative there equals the third moment. Therefore

E (X^{3}) = M_{X}^{(3)} (0) = \frac{6}{λ ^{3}} .

(2)

b

 0  0

i

 0  0

Solution

 0  0

Take

Y = a X_{1} + b

. Directly,

E (e^{t Y}) = e^{b t} E (e^{a t X_{1}}) = e^{b t} M_{1} (a t) .

(1)

The formula holds wherever the moment-generating function on the right is finite.

ii

 0  0

Solution

 0  0

Take

Y = X_{1} + X_{2}

. Independence factors the expectation:

E (e^{t Y}) = E (e^{t X_{1}}) E (e^{t X_{2}}) = M_{1} (t) M_{2} (t) .

(1)

iii

 0  0

Solution

 0  0

Let

X_{1}^{'}

be an independent copy of

X_{1}

and take

Y = X_{1} + X_{1}^{'}

. Then

M_{Y} (t) = M_{1} (t)^{2} .

(1)

Independence of the copy is essential. The variable

2 X_{1}

would instead have moment-generating function

M_{1} (2 t)

iv

 0  0

Solution

 0  0

Take

Y = U X_{1}

, where

U

is uniform on

(0, 1)

and independent of

X_{1}

. Conditional expectation gives

E (e^{t U X_{1}}) = \int_{0}^{1} E (e^{u t X_{1}}) d u = \int_{0}^{1} M_{1} (u t) d u .

(1)

For

t

in a sufficiently small neighborhood of zero, finiteness follows from finiteness of the original moment-generating function in such a neighborhood.

This uses the moment-generating function of an independent product.

c

 0  0

Solution

 0  0

Rewrite the function as

M_{Y} (t) = \frac{1}{2} \frac{1}{1 - t} + \frac{1}{2} \frac{2}{2 - t} .

(1)

These are equal mixture weights on the moment-generating functions of exponential distributions of rates

1

and

2

. Moment-generating functions finite near zero determine the distribution, so

Y

has that mixture law. Consequently

P (Y \leq 1) = \frac{1}{2} (1 - e^{- 1}) + \frac{1}{2} (1 - e^{- 2}) = 1 - \frac{e ^{- 1} + e ^{- 2}}{2} .

(2)

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