Question

Problem 2.1. Let Y1, ...,Yn be a random sample from a uniform distribution on the interval [0 – 1,20 + 1]. a. Find the density function of X = Y;-0 (note that Yi ~ Uf0 - 1,20 + 1]). b. Find the density function of Y(n) = max{Y;, i = 1,...,} c. Find a moment estimator of . d. Use the following data to obtain a moment estimate for 4: 11.72 12.81 12.09 13.47 12.37.

Answer 1

Solution a unit ohm Let Y..... yo be a random Sample from distribution on the interval L 9-1, 29+] (26 +1) - (0-1) 20+1-Oto f (y) – öte 0-1 ye 20+ otherwise function of x = 0 a) To find the density Y,-0.2% Y, = 2x+ 0 to differentiate wort Now x on both sides then we get dx1 = 2 V, 70(0-1, 20+] f(4.) = 672 Range of x when = 20 +1, x=0$ cs Scany =824, X=Ž

. Therefore 12 Now g(x) le port (Probability densite function) of X 3(x) = f(.). glej tong * (2) Car 36x) = 2 <x< Therefore **[3,47] Here g(x) is pdf (Probability density function) since S 30) de = 1 b) To find the density function of Ycm = max í vi, i = 1, .. You is hth onder statistic Yon = max { vi , i=1,2,.... } Now the Pdf (Probability density function) of wth ander Statistic is given by fr (V) = f (Yem) En [F (y)]**+() ma - n Here CamScanner

Not the x is given as 3) F(x) cumulative density function (cdo ) of y- (0-1) (20+1)-(0-1) y-ot = 20+1 -0+1 y-ot! 0 +2 y-ot! t' (ym) -n 1 x ot 2 where 0-1 Yan) <20+ flyn) = n. n. (yo+1) (C) To find a moment estimator of a | Yop..... Yn → u 0-1, 20+] Now, | E(X)= H 0-1+20+1 E(N=/= - u = 3012 > Now sample moments is given as Tau W Now we have to find the moment coqueting e Theth equations of ☺ & © CamScanner estimator then by

30- Theretele moment estimatel of ois et d) Given data a moment estimator 16.72, 12.81, 12009,13.4 712.37 By using given dot a then to obtain for o Now to find the X 11-72+12.817 12.09+ 13.47712.37 x S F = 62.46 7 = 12.492 We know that the X value in equation in then Now to substitute we get o 2 (12492) 24.984 3 O^ = 8.3.28 Therefore the moment estimator for o is 98.328 inScanner