Given $X_1,...,X_n \sim Uni(0,1)$. Let $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$ be the sample mean.

Also have been given: $E(\bar{X}_n)=\frac{1}{2}$, $Var(\bar{X}_n)=\frac{1}{12n}$

According to CLT: $\bar{X}_n \sim N(E(\bar{X}_n), Var(\bar{X}_n))$

Then why is $$P(|\bar{X}_n- \frac{1}{2}|≥0.1)=2*P(\bar{X}_n≥0.6)$$

I know that

$$|\bar{X}_n- \frac{1}{2}|≥0.1$$ $$= \bar{X}_n- \frac{1}{2}≥0.1 \text{ or } \frac{1}{2}-\bar{X}_n ≥ 0.1$$ $$= \bar{X}_n ≥ 0.6 \text{ or } \bar{X}_n ≤ 0.4$$

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    $\begingroup$ This is definately not true in general, you're argument shows that you need to be able to conclude that $Pr(\bar X_n \geq 0.6) = Pr(\bar X_n \leq 0.4)$. Perhaps you know something about $X_n$? Perhaps some symmetry can help you? $\endgroup$ – Matthew Drury Mar 25 '16 at 16:23
  • $\begingroup$ This is clearly false if $X_n=0$, a constant random variable. $\endgroup$ – Alex R. Mar 25 '16 at 16:24
  • $\begingroup$ Can you prove that the pdf of $\bar X_n$ is symmetric about some point? Do you see what this would get you? $\endgroup$ – Matthew Drury Mar 25 '16 at 16:30
  • $\begingroup$ Why all the negatives? :( $\endgroup$ – mavavilj Mar 26 '16 at 5:46

This is not true in general. However, if $\bar{X}_n$ is symmetric about $\tfrac12$, so that $$ \Pr(\bar X_n > \tfrac12 + y) = \Pr(\bar X_n < \tfrac12 - y) $$ then it does hold: \begin{align} \Pr\left( \lvert \bar X_n - \tfrac12 \rvert \ge \varepsilon \right) &= \Pr\left( \bar X_n \ge \tfrac12 + \varepsilon \right) + \Pr\left( \bar X_n \le \tfrac12 - \varepsilon \right) \\&= 2 \Pr\left( \bar X_n \ge \tfrac12 + \varepsilon \right) \end{align} (where the first equality holds because the events $( \bar X_n \ge \tfrac12 + \varepsilon )$ and $( \bar X_n \le \tfrac12 - \varepsilon )$ are disjoint, and the second by the assumed symmetry property).

This is the case if $\bar X_n \sim \mathcal N(\tfrac12, \sigma^2)$. Note, though, that the CLT only says this holds asymptotically, not for any particular $n$. But in your case of $X_n = \frac1n X_i$, $X_i \stackrel{\text{iid}}{\sim} \mathrm{Unif}(0, 1)$, note that each $X_i$ has this symmetry property about $\tfrac12$, and so $\bar X_n$ does as well.

  • $\begingroup$ So does one use the symmetry property of the individual $X_i$s to reason about the symmetry of $\bar{X}_n$ or is the symmetry of $\bar{X}_n$ found by the CLT? $\endgroup$ – mavavilj Mar 25 '16 at 18:26
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    $\begingroup$ @mavavilj The first one; the CLT again only shows the symmetry asymptotically. It's possible to show that the mean of symmetric variables is again symmetric; this should be intuitive, but a proof requires a little bit of effort. $\endgroup$ – Dougal Mar 25 '16 at 18:48
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    $\begingroup$ @mavavilj Definitely a worthwhile effort to convince yourself of that fact. $\endgroup$ – Matthew Drury Mar 25 '16 at 19:05

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