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Suppose that $X_1, X_2, \ldots, X_n$ are i.i.d. random variables such that $X_1\sim N(\mu, 0.5)$ and $n = 100$. a) Find a maximum number $c$ such that $$P(X_1\le c+\mu, X_2\le c+\mu, \ldots,X_n\le c+\mu) \le 0.0005$$

b) Using a sample mean $x̄_n$=$(1/n)\sum X_i$, find a confidence interval [$c_L, c_U$] such that $$P(c_L\le \mu\le c_U$)=0.95$$

Attempt:

a) Not sure how to begin.

b) $P(1.96\le (\bar x-\mu)/0.25\le 1.96) = [\bar x_n-1.96(0.25,\bar x_n+1.96(0.25)]$

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  • $\begingroup$ Is $0.5$ the variance or the standard deviation? (Both are used.) For (b), in either case your standard deviation is wrong, it should be $\frac{\sigma}{10}$. $\endgroup$ – André Nicolas Jan 16 '14 at 1:55
  • $\begingroup$ 0.5 is the variance. $\endgroup$ – user112267 Jan 16 '14 at 2:11
  • $\begingroup$ Then you want (for example) $c_U=\bar{x}_n+\frac{1.96\sqrt{0.5}}{10}$. $\endgroup$ – André Nicolas Jan 16 '14 at 2:15
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By definition, $c = \max\left\{k \text{ }| \text{ }X_i \leq k \text{ for all } i = 1, 2, \dots, \right\}$. If $X_i \leq c$ for all $i$, then clearly, $\max\left(X_1, X_2, \dots, X_n\right) \leq c$ as well. Thus, we want to find $c$ such that

$\begin{align}P\left(\max\left(X_1, X_2, \dots, X_n\right) \leq c + \mu\right) &= P\left(\text{all } X_i \leq c + \mu\right) \\ &= P\left(X_1 \leq c + \mu\right) \cdot P\left(X_2 \leq c + \mu \right) \cdots P\left(X_n \leq c + \mu\right) \text{ by independence}\\ &= \left[P\left(X_1 \leq c + \mu\right)\right]^{n} \text{ since all }X_i \text{ are identically distributed} \\ &= \left[P\left(\dfrac{X_1 - \mu}{\sqrt{0.5}} \leq \dfrac{c+\mu - \mu}{\sqrt{0.5}}\right)\right]^{n} \\ &= \left[P\left(Z \leq \dfrac{c}{\sqrt{0.5}}\right)\right]^{n} \text{ where } Z \sim \mathcal{N}\left(0,1\right) \\ &\leq 0.0005\text{.}\end{align}$

Solve the rest by using a normal distribution chart or your inverse CDF.

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  • $\begingroup$ So just to make sure. I take 0.0005^100 = 0.926807482, and use Z table to find Z=1.45 $\le$ c/sqrt(0.5) and c is 1.45*sqrt(0.5)=1.0253 . How did you go from 3rd line of your equation to your 4th line? $\endgroup$ – user112267 Jan 16 '14 at 2:50
  • $\begingroup$ Not quite... remember that for any random variable $X \sim \mathcal{N}\left(\mu, \sigma^{2}\right)$ that $\dfrac{X-\mu}{\sqrt{\sigma^{2}}} \sim \mathcal{N}\left(0,1\right)$. That's where the $Z$ comes from. Now using algebra, you want $\left[P\left(Z \leq \dfrac{c}{\sqrt{0.5}}\right)\right]^{n} \leq 0.0005$, or $\left[P\left(Z \leq \dfrac{c}{\sqrt{0.5}}\right)\right] \leq 0.005^{1/n} = 0.005^{1/100}$. Look up the value $z$ where $P\left(Z \leq z\right) = 0.005^{1/100}$, and set $z = \dfrac{c}{\sqrt{0.5}}$. $\endgroup$ – Clarinetist Jan 18 '14 at 13:38

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