Let $(X_n)$ be a sequence of Gaussian random variables $N(m_n, \sigma_n)$. Let's say there is a random variable X sucht that $X_n$ goes to X in distribution ($n \rightarrow \infty$). In this case, we have: $\mathbb E(e^{itX_n}) \rightarrow \mathbb E(e^{itX})$ Why the ... is the sequence $(m_n)$ (n>=1) bounded? :S Thanks for any help.
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If $X_n$ converges in distribution the means of $X_n$ must converge. It's also true that every convergent sequence is bounded.
Tim
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So, in this case, $( \sigma_n )$ converges also? And what would be the limit of the sequence of $ (m_n)$ ? And, in all these cases, X is a Gaussian variable, correct? – JohnD May 10 '13 at 12:59
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That's correct, because the $X_n$ are all Gaussian (So I can write down the distributions in terms of $m_n$ and $\sigma_n$ The sequence can only converge in distribution if the sequences $m_n$ and $\sigma_n$ converge. – Tim May 10 '13 at 13:19
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I just realized that I didn't get why $(m_n)$ is convergent. – JohnD May 11 '13 at 19:58