Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 9F (Probability)

Because $U$ and $V$ are bounded, their exponential series may be integrated term by term for every real $t$. Thus their moment-generating functions satisfyand similarly for $V$. Equality of every moment gives equality term by term, so

Normalization of the Gaussian distribution givesCompleting the square,Thereforeand hence

The normalizing constant satisfiesFor an integer $k$,Translation by the integer $k$ merely permutes the summation indices, so the final sum is $B^{-1}$. ConsequentlyThe last equality uses the moment-generating function of the standard normal variable from part (b).

No. A standard counterexample comes from the log-normal distribution. Letand, for a fixed $0<|\epsilon |\le 1$, letThis is a nonnegative density distinct from $f$. For every nonnegative integer $n$, substituting $z=\log x$ makes the difference of the $n$th moments proportional toIt is the imaginary part ofwhich vanishes because its phase is $2\pi n$. The case $n=0$ also proves that $f_{\epsilon }$ is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.