Review Questions, Chapters 8, 9. f(y) = 0, elsewhere. F (y) = f Y(1) = n ( e y/θ) n 1 1 θ e y/θ = n θ e yn


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1 Stat 366 Lab 2 Solutios (September 2, 2006) page TA: Yury Petracheko, CAB 484, yuryp/ Review Questios, Chapters 8, Suppose that Y, Y 2,..., Y deote a radom sample of size from a populatio with a expoetial distributio whose desity is give by (/θ)e y/θ, y > 0 If Y () = mi(y, Y 2,..., Y ) deotes the smallestorder statistic, show that ˆθ = Y () is a ubiased estimator for θ ad fid MSE(ˆθ). Solutio. Let s fid the distributio fuctio of Y : e y/θ, y > 0 F (y) = Now we ca use the formula F Y() (y) = [ F (y) ] or fy() = ( F (y) ) f(y) to fid the the desity fuctio for Y () : for y > 0, f Y() = ( e y/θ) θ e y/θ = θ e y θ. We ca recogize this desity fuctio to be the desity of the expoetial distributio with parameter θ /, Y () Exp ( θ ). Kowig the distributio of Y () allows us to compute the expectatio of ˆθ = Y () : E[ˆθ] = E[Y () ] = θ = θ. So, E[ˆθ] = θ, ad ˆθ is a ubiased estimator of θ. To fid MSE(ˆθ), use the formula MSE(ˆθ) = V [ˆθ] + ( B(ˆθ) ) 2. Sice the estimator is ubiased, its bias B(ˆθ) equals zero. For the variace, remember that Y () is expoetial. We have MSE(ˆθ) = V [ˆθ] + 0 = 2 V [ Y () ] = 2 θ2 2 = θ2.
2 Stat 366 Lab 2 Solutios (September 2, 2006) page Suppose that Y, Y 2,..., Y deote a radom sample of size from a expoetial distributio with desity fuctio give by (/θ)e y/θ, y > 0 I Exercise 8.5 we determied that ˆθ = Y () is a ubiased estimator of θ with MSE(ˆθ)= θ 2. Cosider the estimator ˆθ 2 = Ȳ, ad fid the efficiecy of ˆθ relative to ˆθ 2. Solutio. First compute the variace of ˆθ 2 : [ ] V [ˆθ 2 ] = V [Ȳ ] = V Y + + Y = V [Y Y ] = ( V [Y ] + + V [Y 2 ] ) ) θ 2 = = ( θ θ 2 2 }{{} times 2 = θ2. To fid the relative efficiecy, we eed to fid the ratio of two variaces: eff(ˆθ, ˆθ 2 ) = V (ˆθ 2 ) V (ˆθ ) = θ2 θ 2 =. We coclude that ˆθ 2 is preferable to ˆθ. 9.6 Let Y, Y 2,..., Y deote a radom sample from the probability desity fuctio (θ + )y θ, 0 < y < ; θ > Fid a estimator for θ by the method of momets. Solutio. Let s fid the first momet of this distributio: µ = The method of momets implies 0 (θ + ) y θ+ dy = (θ + ) yθ+2 θ + 2 = θ + 0 θ + 2. Ȳ = ˆθ + ˆθ + 2 ˆθ = 2Ȳ Ȳ.
3 Stat 366 Lab 2 Solutios (September 2, 2006) page Suppose that Y, Y 2,..., Y deote a radom sample from the Poisso distributio with mea λ. (a) Fid the maximumlikelihood estimator ˆλ for λ. (d) What is the MLE for P (Y = 0) = e λ? Solutio. Let s defie the likelihood fuctio L(λ y, y 2,..., y ): p(y i ) = λ y i e λ y i! = λ y i e λ y. i! The problem ow is to fid the maximum value of this fuctio of λ. Let s make a simplifyig trasformatio: ( ) l y i l λ λ Differetiatio with respect to λ yields: l(y i!). d dx l λ y i = 0. Solvig this equatio: The latter is the MLE for λ. λ = y i, or ˆλ = Y i = Ȳ. To aswer (b), recall the ivariace priciple for MLEs: if t(ˆθ) is a oetooe fuctio, the t(θ) = t(ˆθ). I our case t(λ) = e λ, so ê λ = e ˆλ = e Ȳ. 9.75a Suppose that Y, Y 2,..., Y costitute a radom sample from a uiform distributio with probability desity fuctio 2θ +, 0 y 2θ + 0, elsewhere. Obtai the maximumlikelihood estimator of θ.
4 Stat 366 Lab 2 Solutios (September 2, 2006) page 4 Solutio. This is a somewhat differet problem from the previous oe because the support of the desity fuctio depeds o θ. Recall the idicator fuctio I(A). It is equal to oe whe A is true, ad zero if A is false. We ca write the likelihood fuctio i the followig way: f(y i ) = 2θ + I(0 y i 2θ + ) = (2θ + ) I(0 y i 2θ + ). We ca simplify this eve further if we ote that the product of idicator is ozero oly whe all of the uderlyig coditios fulfill. That is, all y i are less that 2θ + ad positive. Notice that this statemet is equivalet to the followig: 0 y () ad y () 2θ +. (We use order statistics y () = mi(y,..., y ) ad y () = max(y,..., y ).) We have (2θ + ) I(0 y ()) I(y () 2θ + ). Now look at the first part of the likelihood fuctio L, (2θ + ). Notice that this is a decreasig (ad cotiuous) fuctio of θ. If we wat to maximize L, we should choose the value of θ as small as possible. Notice that if 2θ + is smaller tha y (), the the value of L(θ) is zero. So, the miimum of 2θ + is y (). This gives the miimum value for θ ad maximizes the likelihood L(θ). We coclude (provided at least oe observatio i the sample is positive) ( Y () = 2ˆθ + ˆθ = Y() ) Let Y, Y 2,..., Y deote a radom sample from the probability desity fuctio (θ + )y θ, 0 < y < ; θ > Fid the maximumlikelihood estimator for θ. Compare your aswer to the method of momets estimator foud i Exercise 9.6. Solutio. Defie the likelihood fuctio: ( ) θ. (θ + )yi θ = (θ + ) y i Take the logarithms: l l(θ + ) + θ l y i.
5 Stat 366 Lab 2 Solutios (September 2, 2006) page 5 Fid critical poits: so ad fially d dθ l θ + + l y i = 0, θ = l y, i ˆθ = l Y. i This is quite differet from the method of momets estimator foud i Exercise 9.6.
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