# How do you use the central limit theorem to approximate the sample mean when pdf f(x)=2x

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)

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

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