Let $Y_{1}$ and $Y_{n}$ be the smallest and largest order statistics, respectively, from a random sample o size $n .$
(a) Use the result of Exercise 141 to determine the joint pdf of $Y_{1}$ and $Y_{n}$ . (Your answer will
include the pdf $f$ and $\mathrm{cdf} F$ of the original random sample.)
(b) Let $W_{1}=Y_{1}$ and $W_{2}=Y_{n}-Y_{1}$ (the latter is the sample range). Use the method of
Sect. 4.6 to obtain the joint pof $W_{1}$ and $W_{2},$ and then derive an expression involving
an integral for the pdf of the sample range.
(c) For the case in which the random sample is from a uniform distribution on $[0,1],$ carry out
the integration of $(b)$ to obtain an explicit formula for the pdf of the sample range. [Hint:
For the Unif[0, 1] distribution, what are $f$ and $F ? ]$

Answer

(a) $g(y_1, y_n) = n(n - 1)[F(y_n) -F(y_1)]^{n - 2}f(y1)f(y-n) $ for $ y_1 < y_n$
(b) $ f(w_1, w_2) = n(n 1)[F(w_1 + w_2) -F(w_1)]^{n - 2}f(w_1)f(w_1 + w2),$
$f _{W_2 }(w_2)=n(n-1)\int_{-\infin}^{\infin}[F(w_1=w_2)-F(w-1)^{n-2}f (w_1)(w_1+w_2)dw_1$ (c) $ n(n- 1)w_{2^{n -2}}(1 -w_2) $ for $ 0 \le w_2\le 1$

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Video Transcript

right. The probability that the absolute value of Y n minus theta is greater than or equal to epsilon. The same thing as the probability of theta minus Y N. To be greater than or equal to. Excellent. This is because sata is the larger one of the two, and so the absolute value is the larger minus the smaller quantity. This is the same thing, manipulating this inequality, the same thing as saying that y 10 is less than or equal to fader minus steps alone. And this is the same thing, saying that the probability That just anyone of the excess say X one falls below three to minus epsilon, raised to the power and now the excess are uniform so that is sata minus epsilon, lot of divided by sudden raised to the ends. Our This being quantity smaller than one and positive, it approaches zero as angles infinity.

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