 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2021/2/c/iv/solution

It suffices that the moment-generating function be finite on some open interval containing zero:

M_{Y} (t) < \infty for every ∣ t ∣ < δ, for some δ > 0.

Equivalently, some two-sided exponential moment exists:

E e^{r ∣ Y ∣} < \infty

for some

r > 0

. For the forward implication, choose

0 < r < δ

and use

e^{r ∣ Y ∣} \leq e^{r Y} + e^{- r Y}

; the converse follows from

e^{t Y} \leq e^{r ∣ Y ∣}

for

∣ t ∣ \leq r

Under this weaker hypothesis, all moments exist. For

∣ t ∣ < r

, dominated convergence justifies the Taylor series argument in (i). Vanishing odd moments gives

M_{Y} (t) = M_{- Y} (t)

near zero, and the local uniqueness theorem for moment-generating functions gives the symmetric distribution. The reverse implication follows directly from equality of the two probability distributions.

More generally, it suffices that all moments exist and the probability distribution have moment determinacy:

Y

and

- Y

then have identical moments and hence identical probability distributions. Existence of all moments by itself does not provide that uniqueness.

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