Past exam of the probability course of the University of Oxford / 2021 / 2 / c / i

Solution

First justify that the moments being used exist. For every

r > 0

E e^{r ∣ Y ∣} \leq E e^{r Y} + E e^{- r Y} = M_{Y} (r) + M_{Y} (- r) < \infty.

The bound

∣ y ∣^{k} \leq k! r^{- k} e^{r ∣ y ∣}

gives

E ∣ Y ∣^{k} < \infty

for all

k

Y

Y

and

- Y

have the same probability distribution, so for odd

k

E Y^{k} = E (- Y)^{k} = - E Y^{k},

E Y^{k} = 0

Conversely, suppose every odd moment vanishes. For fixed

t \in R

e^{t Y}

are bounded by

e^{∣ t ∣∣ Y ∣}

, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:

M_{Y} (t) = \sum_{k = 0}^{\infty} \frac{t ^{k} E Y ^{k}}{k !} = \sum_{j = 0}^{\infty} \frac{t ^{2 j} E Y ^{2 j}}{( 2 j )!} = M_{Y} (- t) = M_{- Y} (t) .

Y

and

- Y

. Hence

Y has a symmetric distribution ⟺ E Y^{k} = 0 for every odd k .

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