Convergence in distribution ( Two equivalent definitions)

I read that for convergence in distribution it is equivalent to have that either the characteristic functions of the random variables convergence pointwise or we have that $F_{X_n} \rightarrow F_{X}$ pointwise, where $F$(the distribution function) is continuous.

I could not find a proof of this, so I was wondering how hard it is to show? Does anybody here have a (Internet)-reference or could sketch the idea?

• I am not so sure whether you are asking is that or not but probably you can find it in Davar Khoshnevisan's Probability book. Chapter 7. Page 92. Mar 2 '14 at 17:55
• Yeah, actually the question that comes to mind reading this is WHICH are the textbooks you are using? (And pointwise convergence of the characteristic functions does not imply convergence in distribution.)
– Did
Mar 2 '14 at 18:27
• to answer your question, I do not have any book on this currently around here, so I would have to go to the library anyway....ah, so only the converse(convergence in distribution implies pointwise convergence of the characteristic functions) is true?
– user66906
Mar 2 '14 at 18:31
• More precisely, $F_{X_n} \rightarrow F_{X}$ in the points of continuity, Lévy's continuity theorem : en.wikipedia.org/wiki/L%C3%A9vy%E2%80%99s_continuity_theorem Mar 2 '14 at 23:44

Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6).

Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem:

Let $\mu_n$, $1\leq n \leq \infty$ be probability measures with characteristic functions $\phi_n$.

(i) If $\mu_n$ converges weakly to $\mu_\infty$, then $\phi_n(t)$ converges pointwise to $\phi_\infty(t)$.

(ii) If $\phi_n(t)$ converges pointwise to a limit $\phi(t)$ that is continuous at $0$, then the associated sequence of distributions $\mu_n$ is tight and converges weakly to a measure $\mu$ with characteristic function $\phi$.

Statement (i) follows from the Portmanteau Theorem characterization of weakly convergence that uses bounded and continuous functions, by noting that $\mathrm e^{itx}$ is bounded and continuous in $x$.

For (ii), Durrett first proves that sequence of distributions $\mu_n$ is tight (not easy, uses the continuity of $\phi$ at $0$; I'll try to update some details later), which in turn implies the existence of a weakly convergent subsequence. The distribution, call it $\mu$, this subsequence converges weakly to, must have characteristic function $\phi$. Moreover, every subsequence has a further subsequence that converges weakly to $\mu$. Using again the Portmanteau Theorem characterization mentioned above (and a general topological fact: if every subsequence has a further subsequence that converges to some point, then the whole sequence converges to that point) one can show that the whole sequence of distributions $\mu_n$ converges weakly to $\mu$.

Edit: The tightness follows from the observation that: $$\mu_n\left(\{x\;|\;|x|>2u^{-1} \}\right) \leq u^{-1}\int_{-u}^u (1- \phi_n(t))dt,$$ using Fubini's theorem to re-write the integral as: $$u^{-1}\int_{-u}^u (1- \phi_n(t))dt = \int_{-\infty}^\infty \left(1- \frac{\sin(ux)}{ux}\right)\mu_n(dx),$$ and the fact that $|\sin x|\leq |x|$ for all $x$.

Edit2: For relationship of tightness and weak convergence topology see also Prokhorov's Theorem and its corollaries.

• Please try to include the essential parts of the answer here and provide the link to the text for reference. Link-only answers can become invalid if the linked page changes. Mar 3 '14 at 11:40
• @ArthurFischer I edited Durrett's theorem text and some proof ideas. I'll try to clarify later.
– ir7
Mar 3 '14 at 15:39
• @ir7 Could you please elaborate how you got the inequality of your observation? Dec 29 '15 at 17:41