chapter 1

Chapter 6: Point Estimation27.a. f ( x x ; α,β )1( x x ... x )α− 1 −Σxi/ β1 2 ne,...,n=, so the log likelihood isnαnβ Γ∑( α )i( α −1) ln( x ) − − nαln( β ) − nlnΓ( α )∑∑iβ0 yields ln ( ) − n ln ( β ) − n Γ( α ) = 0x. Equating both∑dnαx iand = = 02dαβ βdifficult system of equations to solve.∑x id d anddα dβ, a veryb. From the second equation in a, = nα⇒ x = αβ = µ , so the mle of µ isµˆ = X .βx ito28.⎛ x⎝ θ2n2a. ⎜ exp[ − x / 2θ] ⎟...⎜ exp[ − x / 2θ] ⎟ = ( x ... x )2[ − x / 2θ]⎞ ⎛ x⎞ exp Σ1 i1n1 nn⎠⎝ θnatural log of the likelihood function is ln ( x ... x ) − nln( θ )2n Σx derivative wrt θ and equating to 0 gives − + i= 0θ 2θ222Σx θ = i. The mle is therefore ˆΣX θ = i2n2nestimator suggested in Exercise 15.b. For x > 0 the cdf of X if F( x θ ) = P( X ≤ x)i⎠n, soθ2Σxi−2θΣx n = i2. The. Taking the2θ and, which is identical to the unbiased2⎡− x ⎤; is equal to 1−exp ⎢ ⎥⎣ 2θ ⎦this to .5 and solving for x gives the median in terms of θ :that ln (.5)12( 1.38630θ ˆ) .2⎡− x ⎤.5 = exp ⎢ ⎥⎣ 2θ ⎦2− x= , so x = µ ~ = 1. 38630 . The mle of µ ~is therefore2θ. Equatingimplies214