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I'm trying to prove this: Given $\{\mu_n\}$ and $\{\sigma_n\}$ sequences of real numbers such that $\mu_n \rightarrow \mu$ and $\sigma_n \rightarrow \sigma$, if $X_n \sim N(\mu_n, \sigma_n^2)$ and $X \sim N(\mu_n, \sigma_n^2)$ then $X_n \overset{D}{\rightarrow} X$.

I have a couple of questions:

1º Since $X_n \rightarrow X \implies X_n \overset{D}{\rightarrow} X$ wouldn't be easier to show that given $\epsilon > 0$ $\lim P\{| X_n -X |< \epsilon \} = 1$?.

2º And to prove the convergence in distribution, knowing that $\mu_n \rightarrow \mu$ and $\sigma_n \rightarrow \sigma$ then $F(X_n) = \displaystyle\int_{-\infty}^{+\infty}\displaystyle\frac{1}{\sqrt{2\pi\sigma_n^2}}e^{-\frac{(x-\mu_n)^2}{2\sigma_n^2}}dx \rightarrow \displaystyle\int_{-\infty}^{+\infty}\displaystyle\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx = F(X)$ and that's it?

balaya
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2 Answers2

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Just what you mean by $X_n\to X$ is not clear. One could take it to mean almost sure covergence, but then you write $\lim_n P(|X_n-X|<\varepsilon)=1$, and that's about convergence in probability. However, the information given is not enough to justify the conclusion that $X_n\to X$ in probability, let alone almost surely.

It is nonsense to write $$ F(X_n) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma_n}} e^{-(x-\mu_n)^2/(2\sigma_n^2)}\,dx. $$ One could write $$ F_{X_n}(x)=\int_{-\infty}^x \frac{1}{\sqrt{2\pi\sigma_n}} e^{-(w-\mu_n)^2/(2\sigma_n^2)}\,dw $$ and show that that approaches the relevant limit, and that would indeed suffice.

Michael Hardy
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  • Thanks!, and $X_n \rightarrow X$ should be $X_n \overset{P}{\rightarrow X}$. I see now that it would be impossible to prove that, thanks!. – balaya Jul 08 '13 at 14:12
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Assuming you meant that $X~N(\mu,\sigma)$:

1) Is $X_n\rightarrow X$ supposed to mean convergence in probability? If yes then it doesn't make sense to ask for convergence in probability because that is inherently tied to the nature of the probability space that $X_n$ and $X$ are defined on. For example if $X_n$ and $X$ are independent for all $n$ then it's not hard to see you cannot possibly have convergence in probability. That's why you need weak convergence (convergence in distribution) which is the cumulative distribution function converging.

2) Assuming you meant to integrate the Gaussian PDF from $-\infty$ to $z$ (which defines $F_n(z)$ then yes you just need to show convergence for each $z$ to $F(z)$ using something like dominated convergence theorem.

Alex R.
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