Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / M / 1 / b

Take $Y = a X_{1} + b$ . From the definition of a moment-generating function,

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

, for

t

where the expectation is finite.

ii

Take $Y = X_{1} + X_{2}$ . Independence makes the joint exponential expectation factor:

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

Take $Y = X_{1} + X_{1}^{'}$ , where

X_{1}^{'}

is an independent copy of

X_{1}

. The preceding factorization gives

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

. Using

2 X_{1}

would instead give

M_{1} (2 t)

, so an independent copy is required.

Take $Y = U X_{1}$ , where

U

is uniform on

(0, 1)

and independent of

X_{1}

. Conditioning on

U

gives

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

For

t

near zero this is finite because the moment-generating function of

X_{1}

is finite on a neighborhood of zero; Tonelli's theorem justifies the conditioning integral for the nonnegative exponential.

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