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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?

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up vote 1 down vote accepted

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 $\dfrac{e^{it}-1}{it}$.

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