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

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence

Let $X$ have the Pareto distribution with probability density function $f_X(x)=3x^{-4}$ for $x\ge 1$, and zero otherwise. Its normalization of a probability distribution and first two moments follow fromSet $Y=X-3/2$. ThenThe probability distribution of $Y$ is not a symmetric distribution: $Y\ge -1/2$ almost surely, whereas $\mathbb{P}(Y>1/2)=\mathbb{P}(X>2)=1/8>0$ and $\mathbb{P}(Y<-1/2)=0$. For $X\ge 3$ we have $|Y|\ge X/2$, soThus $\boxed{Y=X-3/2}$ has every property required.

Take a discrete probability distribution on three points:The probabilities are positive and sum to one. Direct calculation givesandNevertheless it is not a symmetric distribution, since $\mathbb{P}(Y=3)=1/10$ while $\mathbb{P}(Y=-3)=0$. Vanishing of just the first and third moments does not force all odd moments to vanish.

It suffices that the moment-generating function be finite on some open interval containing zero:Equivalently, some two-sided exponential moment exists: $\mathbb{E}{e}^{r|Y|}<\infty$ for some $r>0$. For the forward implication, choose $0<r<\delta$ and use $e^{r|Y|}\le e^{rY}+e^{-rY}$; the converse follows from $e^{tY}\le e^{r|Y|}$ for $|t|\le r$.

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.