2. ## Re: Order statistic

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3. ## Re: Order statistic

Originally Posted by variousbubble
Suppose that $X_1, \ldots, X_5$ are i.i.d. uniform random variables on the interval $(3,6)$. Let $\bar{X}$ be the sample mean, with mean $\mu_{\bar{X}}$ and standard deviation $\sigma_{\bar{X}}$. Find the probability that the first order statistic $Y_1$ and the fifth order statistic $Y_5$ differ from $\mu_{\bar{X}}$ by less than $\sigma_{\bar{X}}$.

$f_X(x)=\frac{1}{b-a}=\frac{1}{6-3} = \frac{1}{3} , \hspace{0.5cm} 3

$F_X(x)=\int^{t=x}_{t=3}\frac{1}{3} dt = \frac{t}{3}\vert^{t=x}_{t=3} = \frac{x}{3}-1 , \hspace{0.5cm} 3

$\mu = \frac{a+b}{2} = \frac{3+6}{2} = \frac{9}{2} = 4.5$

$\sigma^{\color{red}{2}\color{black}} = \frac{(b-a)^2}{12} = \frac{(6-3)^2}{12} = \frac{9}{12} = 0.75 \implies \sigma=\sqrt{0.75}$

$\mu_{\bar{X}} = \mu = 4.5$

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{0.75}}{\sqrt{5}} = \sqrt{0.15} \approx 0.3873$

$P\left(\left[|Y_1-\mu_{\bar{X}}| < \sigma_{\bar{X}}\right] \cap \left[|Y_5-\mu_{\bar{X}}| < \sigma_{\bar{X}}\right] \right) \approx P\left(\left[4.1127 < Y_1 < 4.8873\right] \cap \left[4.1127 < Y_5 < 4.8873\right] \right)$

If both the sample min and sample max of the $X_i$s are bounded by 4.1127 and 4.8873, then all $X_i$s from the sample are bounded by 4.1127 and 4.8873.

Thus,

$P\left(\left[4.1127 < Y_1 < 4.8873\right] \cap \left[4.1127 < Y_5 < 4.8873\right] \right) = P\left(\left[4.1127 < X_1 < 4.8873\right] \cap \left[4.1127 < X_2 < 4.8873\right] \cap \cdots \cap \left[4.1127 < X_5 < 4.8873\right]\right)$

Since the $X_i$s are independent, we can rewrite the right-hand side from above as

$P\left(4.1127 < X_1 < 4.8873\right) \cdot P\left(4.1127 < X_2 < 4.8873\right) \cdot \cdots \cdot P\left(4.1127 < X_5 < 4.8873\right)$ ,

or, equivalently,

$\left[P\left(4.1127 < X_1 < 4.8873\right)\right]^5$,

which is,

$\tiny{\left[F_X(4.8873) - F_X(4.1127)\right]^5 = \left[\left(\frac{4.8873}{3}-1\right) - \left(\frac{4.1127}{3}-1\right)\right]^5 \approx 0.2582^5 \approx 0.00115}$.