Suppose I have a simple uniform continuous "unit" distribution X:
$$\begin{align*}
\forall y \in \mathbb{R} \implies \\
y < 0 : & P(X < y) = 0 \\
y \in [0,1] : & P(X < y) = y \\
y > 1 : & P(X < y) = 1 \\
\end{align*}$$
Let $Y_n$ be a random variable equal to the mean of $n$ independent variables with a distribution of X.
Let $Z_n$ be a random variable defined as $Y_n$ normalized. That is:
$$\begin{align*}
Z_n = \frac{Y_n – mean(Y_n)}{stddev(Y_n)}
\end{align*}$$
Is it correct to say that as $i \rightarrow \infty$, $Z_i$ approaches the standard normal distribution by the central limit thereom ?
If so, then is there some way we can derive/calculate the formula for the normal distribution based on the formula for X above? Perhaps using some calculus or whatever? How would this be done?
Best Answer
As in the typical proof of the central limit theorem, you can show that the limit of the characteristic function is $$e^{-t^2/2}$$ and that this is the characteristic function of a standard normal distribution. Finding a characteristic function and inverting one both use calculus.
Incidentally, the mean of the sum of $n$ of your $[0,1]$ independent uniform random variables is $\frac{n}{2}$ and the standard deviation is $\sqrt{\frac{n}{12}}$. The characteristic function of a $[0,1]$ uniform distribution is $\dfrac{e^{it}-1}{it}$.