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Exercise :

Let $X_1, X_2, \dots, X_{1200}$ be independent and uniformly distributed in the interval $(-1/2,1/2)$.

If : $S=X_1 + X_2 + \dots + X_{1200}$, estimate the probabilities $P\{S>20\}$ and $P\{|S| > 20\}$.

Sorry for not providing an attempt but I'm at a loss on how to start here. I would really appreciate a thorough explanation.

All I can say is that we can observe that :

$$X_1, X_2, \dots, X_{1200} \to U(-1/2,1/2)$$

which means that :

$$f_{X_1}(x_1)=f_{X_2}(x_2)=\dots=f_{X_{1200}}(x_{1200})= 1$$

Please, assist me with this problem as it's an exam question that I am trying to grasp for tomorrow's semester exams.

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  • $\begingroup$ $S$ follows the uniform sum distribution. $\endgroup$ Feb 13, 2018 at 15:17
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    $\begingroup$ Hint: By the Central Limit Theorem, $S$ has an approximately Normal distribution. You will need to compute its mean and standard deviation. $\endgroup$
    – awkward
    Feb 13, 2018 at 15:18
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    $\begingroup$ Second hint: By a symmetry argument you also have $$P(S > 20) = P(S < -20)$$ $\endgroup$
    – Gono
    Feb 13, 2018 at 15:20
  • $\begingroup$ Well, I guess $E[X_i] = 0$ and $V[X_i] = 1/12$ since the distribution is uniform. I can proceed with CLT then, which I was missing as a clue, thanks ! $\endgroup$
    – Rebellos
    Feb 13, 2018 at 15:25

1 Answer 1

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From the Central limit theorem:

$$ S_{1200} =\sum^{1200}_{i=1}X_i \sim N(n\mu,n\sigma^2) $$

here $S_{1200} \sim N(0, 100)$

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  • $\begingroup$ Verified the answ. (+1) In this notation $\sigma^2 = 100,\, \sigma = 10.$ In R statistical software 1 - pnorm(20, 0, 10) returns 0.02275013, so $P(S>20)\approx 0.02275.$ and, by symmetry about $0,\,$ $P(|S|>20)$ is twice that. // $S$ is very close to normal; even for $n=12$ (vs 1200) the approx is quite good. $\endgroup$
    – BruceET
    Feb 13, 2018 at 17:26

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