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Levy's continuity theorem states that, for a sequence of random variables $\{X_n\}$ with characteristic functions $\{\varphi_n(t)\}$ and a random variable $X$ with a characteristic function $\varphi(t)$, pointwise convergence $\varphi_n(t)\rightarrow\varphi(t)$ implies convergence in distribution $X_n\xrightarrow{D}X$.

Are any converse results known for this?

Why?

I am interested in the moments of the maximum $M_n=\max (X_1,\ldots,X_n)$ where $X_1,\ldots,X_n$ are i.i.d. and have finite support $[x_l,x_u]$. Thus the distribution of $X_i$ belongs to the maximum domain of attraction of the Weibull distribution, meaning that, for appropriately chosen sequence $c_n$, $c_n^{-1}(M_n-x_u)\xrightarrow{D}W_\alpha$ where $W_\alpha$ is Weibull random variable with distribution function $\Psi_\alpha(x)=1-\exp[-(-x)^\alpha]$, $\alpha>0$. A some kind of a converse to Levy continuity theorem might allow me to use the Weibull characteristic function to obtain the desired moments. However, I would appreciate other suggestions -- perhaps there are more elegant ways to get there.

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    $\begingroup$ Well, the converse of the continuity theorem is true and easy: it follows from the fact that if $X_n \to X$ in distribution then $E[f(X_n)] \to E[f(X)]$ for every bounded continuous $f$. (This is often taken as the definition of convergence in distribution.) Take $f(x) = e^{itx}$ to see that $\varphi_n \to \varphi$ pointwise. If I recall correctly, you can also show that the convergence is uniform on compact sets. If you want convergence of moments, you need something like control over the derivatives of the characteristic functions. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2015 at 15:18
  • $\begingroup$ I see, but now I am confused. The CF for Weibull is $\sum_{n=0}^\infty \frac{(it)^n}{n!}\Gamma(1+n/\alpha)$, obviously differentiable at $t=0$, however, it somehow doesn't seem right that the moments of (normalized) max converge to the moments of Weibull. Can anyone comment? (I can rephrase the question, or write a new question) $\endgroup$
    – Bullmoose
    Commented Mar 31, 2015 at 19:45
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    $\begingroup$ Maybe it helps to consider an easier example. Let $X_n = n^2$ with probability $1/n$ and 0 otherwise, and let $X=0$. Then $X_n \to X$ in probability, hence also in distribution, but $E[X_n] \to \infty \ne E[X]$. The chf of $X_n$ is $\varphi_n(t) = (1-\frac{1}{n}) + \frac{1}{n}e^{itn^2}$. You can check that $\varphi_n \to \varphi = 1$ uniformly on compact sets. Moreover $\varphi_n$ and $\varphi$ are smooth. But $\varphi_n'(0) = i n$ which does not converge to $\varphi'(0) =0$ as $n \to \infty$. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2015 at 20:14
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    $\begingroup$ If you want the $k$th moment of $X_n$ to converge to that of $X$, you need something like $\varphi^{(k)}_n(0) \to \varphi^{(k)}(0)$. This isn't implied by pointwise (or uniform) convergence of the $\varphi_n$ themselves; you have to work harder to prove it (if it is even true at all). It's basically a fancier way of saying that convergence in distribution does not imply convergence in $L^k$, and the latter requires more work. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2015 at 20:16
  • $\begingroup$ Ok, yes, that makes complete sense. Thanks for the example and clarification. I think I might be able to show an almost-sure convergence of the normalized max to Weibull in my problem, but that doesn't help me with convergence in $L^k$, which seems to be the only way to ensure moments converge... $\endgroup$
    – Bullmoose
    Commented Mar 31, 2015 at 20:23

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