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) .

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

ii

Solution

Take

Y = X_{1} + X_{2}

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

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} .

Independence of the copy is essential. The variable

2 X_{1}

M_{1} (2 t)

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 .

For

t

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

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