$Central Limit Theorem question$

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$Central Limit Theorem question$

Suppose that $\lbrace X_i \rbrace$ are IID continuous random variables with the following common pdf:
$$f(x)=\left\{\begin{matrix} 2-2x & 0\leq x\leq1 \\ 0 & else \end{matrix}\right.$$
Let $Z$ be a standard normal random variable. Use the Central Limit Theorem to find the number $a$ so that
$$\lim_{n\rightarrow \infty}P\left ( \frac{1}{n} \sum_{i=1}^{n} X_i\leq\frac{1}{3}+\frac{2}{\sqrt n} \right ) = P(Z\leq a) $$

Probability

Ian Dumais

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The mean of $X_i$ is $\int _0 ^1 x(2-2x) \,dx=x^2-2x^3/3 \big|_0^1=1/3$. And the mean of $X_i ^2 $ is $$\int _0^1x^2 (2-2x)\,dx=2x^3/3-x^4/2\Big|_0^1=2/3-1/2=1/6$$ So the variance of $X_i$ is $E[X_i^2]-E[X_i]^2=1/6-1/9=1/18$.

Let $\bar{X}_n:={1\over n}\sum_{i=1}^n X_i$. Then by central limit theorem $$\lim_{n\to \infty} P\Big({\sqrt n(\bar{X}_n-1/3)\over 1/18}\le z\Big)=P(Z\le {z})$$So for $z=36$ we get $$\lim_{n\to \infty} P\Big( \bar{X}_n\le {1\over 3} + {2\over\sqrt n}\Big)=\lim_{n\to \infty} P\Big({\sqrt n(\bar{X}_n-1/3)\over 1/18}\le 36\Big)=P(Z\le 36)$$Thus $a=36$.

Martin

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$Central Limit Theorem question$

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