Suppose we have $$S_n = \frac{X_1 + X_2 + ... + X_n}{n}$$ where all the $X_i$ ~ $N(0,1)$.
Then I am asked to find the CDF of $S_n$, and also find the limit of $S_n$ as $n$ approaches infinity.
Now the way I found the CDF was to find out what distribution $S_n$ was, which is easily found as:
$S_n$ ~ $N(0,\frac{1}{n})$. Then it is easily computed from this.
What I am struggling with is the limit factor. If $n$ approaches infinity, $S_n$ ~ $N(0,0)$, which makes no sense. But the textbook says that the limit of $S_n$ is also in fact a CDF. Maybe I have to use the integral definition of the CDF and take the limit of an integral(but that's another problem in itself). I'm not sure if I'm doing something wrong, or if there's something I'm missing. Any help would be appreciated.
It is correct that $S_n\sim\mathcal{N}(0,\frac{1}{n})$, but its exact meaning is that as $n\to\infty$, the pdf (which is a Gaussian function) becomes more condensed at zero. Let's assume $X=S_n$ for simplicity. We need to have $$\int_{-\infty}^{\infty}f_{X}(x)dx=1$$ Hence, we should have $f_{X}(0)\to\infty$ as well.
It follows that $$\lim_{n\to\infty} f_X(x)=\delta(x)$$ So the arithmetic mean of $n$ standard normal variables tends to zero with probability one as $n\to \infty$.