Question

Let > 0 and let X1, X2, ..., Xn be a random sample from the distribution with the probability density function f(x; 1) = 212x3e-dız?, x > 0. a. Find E(X), where k > -4. Enter a formula below. Use * for multiplication, / for divison, ^ for power, lam for \, Gamma for the function, and pi for the mathematical constant 11. For example, lam^k*Gamma(k/2)/pi means ik r(k/2)/ I. Hint 1: Consider u = 1x2 or u = x2. Hint 2: f(a) = $.•°2a-1e-udu, a > 0. Elxk) = A Tries 0/10 b. Obtain the method of moments estimator of , . Enter a formula below. Use m1 for the sample mean X. For example, 23*pi^2/m1 means 23 3 2/X. Your expression should NOT contain any function. Hint 3: T(a) = (a – 1)(a – 1) Hint 4: T () = 5 1 = A Tries 0/10 c. Suppose n=5, and x1=1.26, X2=0.60, X3=0.85, X4=0.53, X5=0.58. Find the method of moments estimator of I. Give your answer to 4 decimal places. À= * Tries 0/5 d. Obtain the maximum likelihood estimator of 1, 1. Enter a formula below. Use m2 for the second moment of X, i.e.

Answer 1

Solutich: Let >0 let X ... has pdf frugt 7 = 21² 213 è la saiso I a) E(X) = 52k frais) au = 2x² 8 x 8tk & 1x² dn = 242 stk 21526 b- be e92 2 able EXK) = Clam 4-(kl2))*Gamme (C+)/22 b) for k=1 F(x) = = . 16-birki -3 VT ET SE = VTT /E(X) = (34) sqrt(piltum)] be population mean &x=texi be sample mean by method of moment 3 I = x i 17 = x² J = 7 7 7 M = (9116 )* (Pi | my ^2) el let nas : 126, 0.60 0.85,0.53,0.58 X=2 Exi = 0.764 X = 6 * 1776 4 12 3 = 3.0275

d) Likelihood function L(x) = π fruid) = 2x2x3 e 27,2 = 22h 43 et 2x² log-likelihood functiry enLCA) = neztahlnx +3 Eina; - 1 2 2 ² diff. w.rot. & it equute to zen. DALCA) = 0 +2 +0 - x = 0 î = 21 m2 es let n=5, 40126, 0.6, 0.85,0.53,058 {x? = 3.2875 1 - 08.28743 (P = 3:0419] f n=5 2:0.95, d. 06, 0.6s, os, s.45 Exit = 1:8773 a = 1(1,8773) to = 5.3268]