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Imagine that I have a sequence of random variables $\mu_n$. For each realization of $\mu_n$ , I construct a random distributed RV $X_n$ such that $X_n\sim N(\mu_n,E)$. Suppose I am given that $\mu_n \xrightarrow{as}\mu$ where $\mu$ is a scalar. Can i say anything about the asymptotic distribution of $X_n$.

my attempt :

It seems to me that $F_n(x)$ will almost surely converge to $F(x)$ where $F(x)$ is the CDF associated with $N(\mu,E)$. This would also hold for for the characteristic function. Does that imply that i can make some statement about the asymptotic distribution of $X_n$. Does it help if i know that $\mu_n\sim N(M,\frac{2}{n})$

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First of all: Since mean and CDF are deterministic objects. It doesn't make sense to talk about "almost sure" convergence.

If $X_n$ are Gaussian random variables with mean $\mu_n$ and variance $\sigma_n^2$ such that $\mu_n$ and $\sigma_n^2$ converge as $n \to \infty$, say to $\mu$ and $\sigma^2$, then $X_n$ converges in distribution to a Gaussian random variable $X$ with mean $\mu$ and variance $\sigma^2$.

In the case which you mentioned at the very end of your question, i.e. $\mu_n=M$ is constant and $\sigma_n^2 = 2/n$, then $\mu=M$ and $\sigma^2=\lim_{n \to \infty} 2/n=0$, which means that the limit $X$ has mean $M$ and variance $0$, i.e. $X$ is constant almost surely.

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  • $\begingroup$ I understand that talking about distributions with parameters which are random variables is a bit loose. Is there a more precise way of formulating that? I was thinking of compound distributions. In my case these arise in the context of bayesian update of a paramter. The posterior distribution parameters, looking at it from the first period, are themselves random variables. $\endgroup$
    – kangkan Dc
    Commented Sep 21, 2019 at 11:20
  • $\begingroup$ @kangkanDc Perhaps then I misunderstood what you are actually asking. How do you define $N(\mu,\sigma^2)$ for random variables $\mu$ and $\sigma^2$? $\endgroup$
    – saz
    Commented Sep 21, 2019 at 11:22
  • $\begingroup$ Hi Saz. I am not sure I am clear or if this make sense. But here is the experiment. Suppose you have a parameter M about which you have a normal prior. And you have a signal of M, lets say x_t=M+noise. Then the posterior estimate of M will be a normal distribution. This posterior have determinate perimeters. But lets say I am an observer at the beginning who is watching some one run this experiment. I know that after a period n, the person will have some posterior that will depend of <x_n> (sample mean of x). But i <x_n> will be random variable to me. what can i say about F_n(x)? $\endgroup$
    – kangkan Dc
    Commented Sep 21, 2019 at 19:42
  • $\begingroup$ @kangkanDc I don't know much about this. You might want to take a look at the Glivenko-Cantelli theorem which says that the empirical distribution function converges to the cdf... sounds to me as you are looking for a result of this form. $\endgroup$
    – saz
    Commented Sep 21, 2019 at 20:06

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