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Let $X_{1},\ldots, X_{n}$ be a collection of independent identically distributed (iid) samples from a population with pdf

$$ f(x) = \begin{cases} 2x,&\text{if } 0 \leq x \leq 1\\ 0,&\text{otherwise} \end{cases} $$

How can I use the central limit theorem (CLT) to find the approximate distribution for the sample mean, $\bar{X}_n = \tfrac{1}{n}(X_1+\ldots+X_n)$, as $n$ approaches infinity?

(the answer is $\mathcal{N}(2/3, 1/18n)$ I just don't know how to get there)

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This is how you can apply the central limit theorem in your case. First, you need to determine the mean and variance of your distribution. Let $X$ be a random variable that follows the given distribution with pdf $f$. Then,

$$ \mathbb{E}[X] = \int_0^1 2x^2 \mathrm{d}x = 2/3 $$

and

$$ \operatorname{Var}[X] = \int_0^1 (x-\tfrac{2}{3})^2 2x^2\mathrm{d}x = \frac{1}{18}. $$

Then, the Lindeberg–Lévy CLT says that since $X_1, X_2, \ldots$ are iid samples, which follow the given distribution (which has finite variance), then

$$ \sqrt{n} (\bar{X}_n - \tfrac{2}{3}) \to \mathcal{N}(0, \tfrac{1}{18}) $$

in distribution, where $\bar{X}_n = \tfrac{1}{n}(X_1 + \ldots + X_n)$ is the sample average.

This means that, for large $n$, the distribution of $\sqrt{n}(\bar{X}_n - 2/3)$ can be approximated by $\mathcal{N}(0, 1/18)$, or, equivalently, the distribution of $\bar{X}_n$ can be approximated by

$$ \bar{X}_n \overset{\text{approx.}}{\sim} \mathcal{N}(\tfrac{2}{3}, \tfrac{1}{18 n}). $$

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