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$X_n$ is a sequence of random variables, and $\{\varphi_n(t)\}$ is the corresponding sequence of characteristic functions. Show that $X_n \stackrel{d}{\longrightarrow} 0$ iff $\{\varphi_n(t)\}$ converges to 1 in some neighbourhood of $t=0$.

The necessity follows immediately from Levy's continuity theorem. I am struggling with showing the sufficiency: it remains to show that $\varphi_n(t) \rightarrow 1$ for all $t \in \mathbb{R}$, but how to do that? Appreciate for any help!

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    $\begingroup$ Hint: what are all the moments? $\endgroup$ – Adam Hughes Jul 31 '14 at 5:05
  • $\begingroup$ Can you clarify which directions you're referring to as "necessity" and "sufficiency"? I'm a little confused. In particular, you don't need Levy's continuity theorem to prove "if $X_n \overset{d}{\to} 0$ then $\varphi_n \to 1$", only the definition of convergence in distribution. $\endgroup$ – Nate Eldredge Jul 31 '14 at 5:50
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Suppose $X_n \overset{d}{\to} 0$. It should either be the definition of weak convergence, or a theorem you know, that $E f(X_n) \to f(0)$ for all bounded continuous $f$. Take $f(x) = e^{itx}$ to conclude $\varphi_n(t) \to 1$ for all $t$. (You don't need Lévy here.)

Suppose $\varphi_n(t) \to 1$ on some neighborhood of 0. Following the proof of the Lévy continuity theorem, conclude that $\{X_n\}$ is tight. Now use the double subsequence trick. If it is not the case that $X_n \overset{d}{\to} 0$, then there is a subsequence $X_{n_k}$ which does not have 0 as a limit point. (This uses the fact that convergence in distribution is metrizable.) By tightness, there is a further subsequence $X_{n_{k_j}}$ which converges in distribution to some random variable $X$ which is not 0. But now show that the chf of $X$ is 1 on a neighborhood of 0. Differentiate under the integral sign to show $EX^2 = 0$, a contradiction.

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  • $\begingroup$ By sufficiency I mean showing that $\varphi(t) \overset{d}{\to} 1$ on some neighbourhood of $t=0$ implies $X_n \overset{d}{\to} 0$. I am confused how do you use continuity theorem to conclude that $\lim_{n \to \infty} \varphi_n(t)$ exists for all $t \in \mathbb{R}$? $\endgroup$ – math major Jul 31 '14 at 6:51
  • $\begingroup$ I guess the usual statement of the continuity theorem has as its hypothesis that $\varphi(t) := \lim_{n \to \infty} \varphi_n(t)$ exists for all $t$ and that $\varphi$ is continuous at 0. However, if you inspect the proof, you should find you only need the fact that the limit exists on a neighborhood of 0 and is continuous at 0. $\endgroup$ – Nate Eldredge Jul 31 '14 at 6:58
  • $\begingroup$ Yes, the statement of continuity theorem in the textbook that I am using is exactly the same as what you mentioned. I have gone through the whole proof again, but I can't see still why existence of $\varphi(t) = \lim_{n \to \infty}\varphi_n(t)$ on a neighbourhood of $t = 0$ implies the existence of $\lim_{n}\varphi_n(t)$ on the real line. Under the special condition that you gave, I can deduce that the distribution functions of $X_n$ are uniformly tight, which by Prokhorov's theorem implies existence of a weakly convergent subsequence, and I am stopped here .. $\endgroup$ – math major Jul 31 '14 at 8:00
  • $\begingroup$ Sorry, I claimed a little more than I actually know how to prove. But see my edit for something that does work. $\endgroup$ – Nate Eldredge Jul 31 '14 at 13:08
  • $\begingroup$ Ah, yes, I got what you mean now! Thanks so much! $\endgroup$ – math major Aug 2 '14 at 4:38

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