 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2005/m/1/solution

M_{X} (t) = E [e^{tX}]

where finite. For independent

X, Y

, independence factors the expectation, giving

M_{X + Y} (t) = M_{X} (t) M_{Y} (t)

wherever both factors are finite.

t < 1

, put

s = (1 - t) x

to obtain

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

Here

Γ (n) = (n - 1)!

, obtained by repeated integration by parts. For

t \geq 1

the defining integral diverges.

Each

Y_{i}

(1 - t)^{- 1}

M_{\sum_{i = 1}^{n} Y_{i}} (t) = (1 - t)^{- n} (t < 1) .

The uniqueness theorem for moment-generating functions applies when the functions are finite on a neighborhood of zero. Hence the sum has the same distribution as $X_{n}$ , a Gamma distribution of shape

n

and rate one, with the displayed probability density function in the source.

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