Does uniform convergence of distribution functions imply convergence of moments?

Question

Suppose $X_n \sim F_n$, $X \sim F$ with $\Vert F_n - F\Vert_\infty \rightarrow 0$.

Using $EX = EX^+-EX^-$, and the integral tail formula for expectation, I have that \begin{align*} |EX_n - EX| =&\ \left|\int_0^\infty P(X^+_n > x)\ dx-\int_0^\infty P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)\ dx- \int_0^\infty P(X^-_n > x)\ dx \right|\\ =&\ \left|\int_0^\infty P(X^+_n > x)-P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)- P(X^-_n > x)\ dx \right|\\ =&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx +\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\ \leq&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx\right| +\left|\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\ \rightarrow 0 \end{align*} since cdfs of $X_n^\pm$ also converge uniformly to cdf of $X^\pm$, and since, the cdfs are themselves bounded between 0 and 1, so we can take their difference to be very small outside some interval and bound them inside the interval using uniform continuity.

But how can I generalize this to higher order moments?

Answer 1

I think you are trying to find a link to your limit objects, given immense info you have on the distribution of an RVs sequence. However, your attempt this time is also wrong.
Taking $X_n$ and $X$ such that: $$\mathcal{L}(X_n)= \left(1-\frac{1}{n} \right) \delta_0+\frac{1}{n}\delta_{n^2}$$ $$ \text{and } \mathcal{L}(X)= \delta_0$$ and you see, this time, we don't even have the convergence of the first moment, while we have the uniform convergence for CDFs.
So what is the problem here? Also, what is the problem with your above "solution"? That is you did not give a proper treatment to the tails of these random variables. More precise, even if their difference is uniformly small. There integral can always explode because you are integrating on a domain of measure infinity.

To sum up : In order for you to have to desired results, you need to have a control on the tails of $F_n$
And in fact, this is also enough, as given a suitable control, you can imply the uniform integrability of $|X_n|^p$
For example:: $\exists \epsilon> 0$ such that : $$ \lim_{x \rightarrow +\infty} e^{\epsilon x}\sup_n\Vert F_n(x) -1\Vert =0 $$ (together with another limit for the minus infinity tail)

Remark :This condition is closely related to the notion "Exponential tension", which has demonstrated its importance in theory of Large deviation.
P/s: I forgot to put minus one into the bracket, last time.

Answer 2

This is not true as the following example suggests. Let $\mathsf{P}(X_n=e^n)=n^{-1}=1-\mathsf{P}(X_n=0)$ and $X=0$. In this case, $$ \|F_{X_n}-F_X\|_{\infty}=n^{-1}\to 0. $$ However, for each $r>0$, $$ \mathsf{E}X_n^r=\frac{e^{rn}}{n}\to\infty. $$