Given a sequence of i.i.d standard normal random variables, ${X_n}$, define $F_n(X)$ as the CDF of the mean of $X_1,\ldots,X_n$. What is $\lim_{n \rightarrow \infty} F_n(X).$
This problem is presented in the textbook before the central limit theorem and so I'm not sure what tools to approach it with. Any hints?
Best Answer
For any $n$, the mean of $n$ i.i.d. $N(0,1)$ variables is $\sim N(0,1/n)$, which can be derived from basic properties of Gaussians - no need of CLT for this.
So intuitively in the limit, the mean becomes constant $0$. Let $Y$ denote this constant r.v. (i.e. $Y=0$ surely) and $Y$ would have this CDF:
$$F_Y(x) = 0, 1 ~~~~\text{depending on}~~~ x < 0, \ge 0 ~~~~\text{respectively}$$
However, there is (I think) a technicality. The question is not asking for $F_Y$. Instead it asks for the limit of $F_n(x)$, and if you look at the formula for CDF of Gaussians, you can see that $F_n(0) = 1/2$ for all $n$. So the limit must be the same way. I.e. I think the question is asking for the limit of a sequence of CDF functions $F_n$, and the limit is:
$$F_\infty(x) = 0, \frac12,1 ~~~~\text{depending on}~~~ x < 0, = 0, > 0 ~~~~\text{respectively}$$
Note that $F_\infty(x)$ is not a proper CDF any more, because it is not right-continuous at $x=0$. However, in my interpretation of the wording of the question, it is asking for the limit of a sequence of functions $F_n$, each of which is a CDF, but it does not care that the limit itself is not a CDF.
Just my interpretation. :)