6

Let $f_n$ denote a sequence of PDFs, and $F_n$ denote the corresponding sequence of CDFs. Given $L^1$ convergence of the PDFs to some PDF $f$,

$$\int_\mathbb{R} |f_n(x) -f(x)| dx \rightarrow 0$$

does this imply $L^2$ convergence of the corresponding CDFs to the corresponding CDF $F$,

$$\int_\mathbb{R} \left(F_n(x) - F(x)\right)^2 dx \rightarrow 0$$

Jimmy R.
  • 35,868
user3825755
  • 265

1 Answers1

3

First note that $0\leqslant F(x),F_n(x)\leqslant 1$ for all $n,x$ so $\sup_{n,x}|F_n(x)-F(x)|=2$. It follows that $$\sup_{n,x}(F_n(x)-F(x))^2\leqslant 2|F_n(x)-F(x)|.$$ From $L^1$ convergence of $f_n$, continuity of $F_n, F$, and nonnegativity of $f_n,f$, we have for each $t\in\mathbb R$ \begin{align} |F_n(t)-F(t)| &= \left| \int_{-\infty}^t f_n(x)\ \mathsf dx -\int_{-\infty}^t f(x)\ \mathsf dx\right|\\ &\leqslant \int_{-\infty}^t |f_n(x)-f(x)|\ \mathsf dx\\ &\leqslant \int_{\mathbb R} |f_n(x)-f(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0, \end{align} and hence $F_n$ converges in distribution to $F$. In fact, as $F$ and $F_n$ are bounded, continuous, and monotone, it follows from Pólya's extension to Dini's theorem (cf. *Problems and Theorems in Analysis, I, p. 270) that $F_n$ converges uniformly to $F$ on any compact subset of $\mathbb R$. Since $F$ is a CDF, we may extend it to a continuous function $\overline F$ on the extended real line $[-\infty,\infty]$ by $\overline F(-\infty)=0$, $\overline F(\infty)=1$ (and similarly for $F_n$, and therefore $F_n$ converges uniformly to $F$ on $[-\infty,\infty]$. Since $F$ and $F_n$ are uniformly continuous, given $\varepsilon>0$ we may choose $N$ so that $n\geqslant N$ implies $$\sup_{x\in\mathbb R}|F_n(x)-F(x)|<\varepsilon. $$ Then $$\int_{\mathbb R} (F_n(x)-F(x))^2\ \mathsf dx\leqslant 2 \int_{\mathbb R}|F_n(x)-F(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0$$ so that $F_n$ converges to $F$ in $L^2$.

Math1000
  • 36,983
  • 1
    The convergence of $|F_n-F|\to{0}$ is uniform; thus, the limit and the integral can be exchanged by the dominated convergence theorem (by definition of uniformity $|F_n-F|\leq{g_n}\to{0}$). – Vossler Mar 31 '16 at 14:24
  • @Vossler Thanks for the hint! – Math1000 Mar 31 '16 at 15:19
  • Thanks for this detailed answer! So the basic idea is to show that the convergence of $|F_n - F| \rightarrow 0$ is uniform in order to exchange the integral and the limit? Regarding the accepted answer here, wouldn't the set over which is integrated have to be finite? – user3825755 Mar 31 '16 at 16:01
  • 1
    Wow. It's been really long since I last saw such a complete, hardcore $\varepsilon-\delta$-style proof. Not sure it was necessary here, but +1 for the patience! – Vossler Mar 31 '16 at 18:02
  • @user3825755 Indeed, $F_n$ converges to $F$ uniformly on any compact set, but $|F_n-F|$ need not have compact support. So we need a way to bound the tails - perhaps there is a more elegant method, but I believe my approach is correct. – Math1000 Mar 31 '16 at 20:00
  • @Math1000 Sorry but I still am confused about the very last step. So we have that $|F_n - F| < \varepsilon$ everywhere, even uniformly on compact sets. But how does the convergence of $\int_\mathbb{R} |F_n - F| dx$ then follow? The dominated convergence theorem would require an integrable upper bound, which the constant $\varepsilon$ is not, right? – user3825755 Apr 01 '16 at 09:40
  • The interchange in order of limit operations is justified by uniform convergence of $F_n$, not by dominated converge. – Math1000 Apr 01 '16 at 09:52
  • @Math1000 Could you potentially point me to a suitable reference? For example here, the statement is only given for integrals over compact sets. – user3825755 Apr 01 '16 at 11:19
  • @user3825755 I found a different approach (with much fewer Greek letters ;) ). – Math1000 Apr 01 '16 at 13:45
  • 1
    @Math1000 thanks for your efforts! So convergence of the integral then follows because $[-\infty, \infty]$ is compact and $|F_n - F|$ converges uniformly on this closed interval? – user3825755 Apr 01 '16 at 13:58
  • Yes, that is the idea. – Math1000 Apr 01 '16 at 14:40
  • 1
    @user3825755 Just a minor clarification in response to Math1000's earlier comment: "The interchange in order of limit operations is justified by uniform convergence, not by dominated convergence": the uniform convergence criterion for exchanging integral and limit is actually a consequence of the DCT. – Vossler Apr 01 '16 at 18:07
  • 1
    Sorry that this has taken a while to reply. Isn't the critical requirement for uniform convergence to imply the desired convergence that the measure of the space over which is integrated is finite? For "normal" compact subsets of $\mathbb{R}$, this is of course so, but the measure of $[-\infty,\infty]$ is no longer finite. e.g. see answers in this question – user3825755 Apr 06 '16 at 11:59
  • @Math1000 In addition to my comment from earlier, also the wikipedia page on uniform convergence indicates that the measure of the integration space is important (although no specific results are given for the Lebesgue integral) – user3825755 Apr 06 '16 at 14:59