Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove this conclusion but have some problems with one of the steps.

Assume $X_1,\ldots,X_n,\ldots$ is a sequence of Gaussian random variables, converging almost surely to $X$, prove that $X$ is Gaussian.

We use characteristics function here. Since $|\phi_{X_n}(t)|\leq 1$, by dominated convergent theorem, we have for any $t$

$$ \lim_{n\rightarrow\infty}e^{it\mu_n-t^2\sigma_n^2/2}=\lim_{n\rightarrow \infty}\phi_{X_n}(t) = \lim_{n\rightarrow \infty}\mathbb{E}\left[e^{itX_n}\right] = \mathbb{E}\left[e^{itX}\right] = \phi_X(t) $$

this is the step that I cannot figure out: $e^{it\mu_n-t^2\sigma_n^2/2}$ converges for any $t$ if and only if $\mu_n$ and $\sigma_n$ converges.

Let $\mu=\lim_n \mu_n$, and $\sigma=\lim_n\sigma_n$, then $\phi_X(t)=e^{it\mu-t^2\sigma^2/2}$, which proves that $X$ is a Gaussian random variable.

Why can we get that $\mu_n$ and $\sigma_n$ converge? This looks intuitive for me, but I cannot make a rigorous prove.

share|improve this question
    
try using the continuous mapping theorem which states that if $x_n \to x, g(x_n) \to g(x)$ –  jay-sun Nov 8 '12 at 0:44
    
@jay , Thanks for your reply. With the property of continuous mapping, I can get that if $\mu_n\rightarrow \mu$ and $\sigma_n\rightarrow \sigma$ then $\exp(it\mu_n-t^2\sigma_n^2/2)\rightarrow \exp(it\mu-t^2\sigma^2/2)$. But I don't know how to get the reverse statement. –  pluskid Nov 8 '12 at 0:56
    
Apply to all possible limits of subsequences. Also, try to heavily use that the exponent $it\mu_n-t^2\sigma_n^2/2$ converges for all $t$. –  Berci Nov 8 '12 at 0:58
    
@Berci, Can you give a little more hint here? I'm kind of stuck to infer that converge for all $t$ means what. –  pluskid Nov 8 '12 at 1:29

1 Answer 1

up vote 5 down vote accepted
  • First, we note that the sequence $\{\sigma_n\}$ and $\{\mu_n\}$ has to be bounded. It's a consequence of what was done in this thread, as we have in particular convergence in law. What we use is the following:

If $(X_n)_n$ is a sequence of random variables converging in distribution to $X$, then for each $\varepsilon$, there is $R$ such that for each $n$, $\mathbb P(|X_n|\geqslant R)\lt \varepsilon$ (tightness).

To see that, we assume that $X_n$ and $X$ are non-negative (considering their absolute values). Let $F_n$, $F$ the cumulative distribution function of $X_n$, $X$. Take $t$ such that $F(t)\gt 1-\varepsilon$ and $t$ is a continuity point of $F$. Then $F_n(t)\gt 1-\varepsilon$ for $n\geqslant N$ for some $N$. And a finite collection of random variables is tight.

  • Now, fix an arbitrary strictly increasing sequence $\{n_k\}$. We extract further sub-sequences of $\{\sigma_{n_k}\}$ and $\{\mu_{n_k}\}$, which converge respectively to $\sigma$ and $\mu$. Taking the modulus, we can see that $e^{-\sigma^2/2}=|\varphi_X(1)|$, so $\sigma$ is uniquely determined.
  • We have $e^{it\mu}=\varphi_X(t)e^{t\sigma^2/2}$ for all $t\in\Bbb R$, so $\mu$ is also completely determined.
share|improve this answer
    
Thanks Davide, that clarified all my puzzles! –  pluskid Nov 9 '12 at 3:59
    
why does convergence in law imply tightness??? –  Andy Teich Jan 15 at 19:17
    
@AndyTeich See edit. –  Davide Giraudo Jan 15 at 20:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.