# Almost sure convergence of moments and convergence of distribution

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 $$\mu_n\sim N(M,\frac{2}{n})$$

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.

• 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. Commented Sep 21, 2019 at 11:20
• @kangkanDc Perhaps then I misunderstood what you are actually asking. How do you define $N(\mu,\sigma^2)$ for random variables $\mu$ and $\sigma^2$?
– saz
Commented Sep 21, 2019 at 11:22
• 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)? Commented Sep 21, 2019 at 19:42
• @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.
– saz
Commented Sep 21, 2019 at 20:06