1 Solution to Problem 6.40

Transcription

1 1 Soluto to Problem 6.40 (a We wll wrte T τ (X 1,...,X where the X s are..d. wth PDF f(x µ, σ 1 ( x µ σ g, σ where the locato parameter µ s ay real umber ad the scale parameter σ s > 0. Lettg Z X µ σ we see that the Z are..d. wth PDF g(z. Thus, T 1 /T τ 1 (X 1,...,X τ (X 1,...,X τ 1 (σz 1 + µ,...,σz + µ τ (σz 1 + µ,...,σz + µ στ 1 (Z 1,...,Z στ (Z 1,...,Z τ 1 (Z 1,...,Z τ (Z 1,...,Z. Sce the dstrbuto of the Z do t deped o the parameters (µ, σ, t follows that T 1 /T s acllary. (b Trval. Soluto to Problem 7. (a Assumg α s kow, we have the log lkelhood, log L(β X α log β β 1 X + C, where C s a costat that does t deped o β. Takg dervatves ad settg equal to 0 gves α β + 1 X β 0 1

2 ad the soluto s ˆβ(α α 1 X. It s easy to check that the secod dervatve of the log lkelhood s egatve, so ths gves us the maxmum. (b The log lkelhood wth both α ad β ukow s log L (α, β X log Γ(α + α log X α log β β 1 X + C. Substtutg the value of β that maxmzes the lkelhood for fxed α gves a cocetrated lkelhood: log L ( α, ˆβ(α X log Γ(α + α log X + α log α + α log X α. + C { log Γ(α + α [ log X + L 1 + log α } + C where C does t deped o α ad L 1 log X. Therefore, we ca maxmze log L ( α, ˆβ(α X over α to fd the MLE for α, the plug ths to ˆβ(α to obta the MLE for β,.e. ˆβ ˆβ(ˆα. 3 Soluto to Problem 7.6 (a The lkelhood s so L(θ x 1,...x θx I [θ, (x θ I [θ, (m x g(τ(x 1,...,x,θ T m X, x h(x 1,...,x

3 s a suffcet statstc. (b To fd the MLE, t suffces to maxmze the g(t, θ gve above,.e. to maxmze L 1 (θ T θ I [θ, (T θ I (0,T (θ. (Oe ca easly check that I [θ, (T I (0,T (θ for ay 0 < θ T. The maxmum s clearly T at θ T,.e. the MLE s ˆθ T m X. (c Presumably, we should compute the mea µ(θ E[X θ ad the solve for θ the equato µ(θ X. OK, so bldly followg the recpe, E[X θ θ xθ 1 x dx θ x 1 dx. θ Oh, I get t. It s a trck questo. The method of momets estmator does t exst, f we follow the usual recpe. Of course, we could compute γ α (θ E[X α θ for ay α < 1, the set ths equal to 1 X α ad solve for θ: Ths yelds a estmate γ α (θ θ x +α dx θ ˆθ ( 1 α X α 1/α. θα 1 α. Obvously dfferet αs wll gve dfferet values. Whch oe should we choose? 4 Soluto to Problem 7.9 The lkelhood ca be wrtte L(θ x θ I [T, (θ, where T max X s the suffcet statstc. The MLE s clearly ˆθ MLE T. 3

4 To compute the mea ad varace of ths, we eed the PDF of T whch s Hece, I partcular, f T (t θ θ t 1 I [0,θ (t. θ E[T θ θ t 1+ dt 0 + θ. E[T θ Var[T θ + 1 θ ( + θ + 1 θ ( + 1 ( + θ. (1 Fally, we ca compute the Mea Squared Error usg equato (7.3.1, p. 330: MSE ( θ, ˆθ MLE Bas ( θ, ˆθ MLE + Var[ˆθMLE θ ( 1 θ ( + 1 ( + θ ( + 1( + θ. Obvously, µ(θ E[X θ θ/. So the Method of Momets estmator s ˆθ MME X. Now X s always a ubased estmator of the mea, so E [ θ ˆθMME θ,.e., ˆθ MME s a ubased estmator of θ. It s easy to check that Var[X θ θ /1, 4

5 e.g. apply (1 above wth 1. Hece, Var [ ˆθMME θ 4Var [ X θ θ /(3. Sce ˆθ MME s a ubased estmator of θ, t follows that MSE ( θ, ˆθ MME θ /(3. Clearly, both MSEs go to 0 as, but the MSE for the MLE goes to 0 faster. Oe ca check that MSE ( θ, ˆθ MME MSE ( θ, ˆθMLE 3 + 3( + 1( + ( 1( 3( + 1( +, whch s 0 for 1 ad, but postve for all >. Hece, the MLE s ever worse tha the MME terms of MSE, ad for ay for ay sample sze 3 s strctly better terms of MSE, so we would prefer the MLE over the MME. Note: Oe may actually do better by correctg for the bas the MLE;.e., the estmator ˆθ UMV UE + 1 T, may actually be better tha the MLE terms of MSE. 5 Soluto to Problem 7.19 (a Note that the Y are depedet wth Y N(βx, σ. Thus, the jot PDF s f ( y 1,...,y β, σ [ (π / (σ / exp 1 (y σ βx [ (σ / exp β x σ exp c(β,σ 1 σ y w 1 (β,σ τ 1 (y 1,...,y 5 + β x σ y (π / w (β,σ h(y 1,...,y τ (y 1,...,y.

6 Clearly ths s a expoetal famly ad the -D suffcet statstc s ( (T 1, T Y, x Y. (b The MLE for β ca be obtaed by mmzg R(β (Y βx. It s also kow as the Least Squares Estmator. Takg d dβ the ormal equato x (Y βx 0. ad settg equal to 0 gves The soluto s We have ˆβ E [ ˆβ β, σ x Y. x x E[Y β, σ x x (βx x x β x β. Ths shows that ˆβ s a ubased estmator of β. (c Note that ˆβ s a lear combato of depedet ormal RVs, so by a theorem stated class, ˆβ has a ormal dstrbuto. We ve already computed the mea, so we eed oly compute the varace. Var [ ˆβ β, σ Var[ x Y ( x x Var[Y ( x 6 x σ ( x σ. x

7 Note that a aalyss of uts (e.g., suppose the Y s are klograms ad the x s are meters does t show ay problem wth ths result. 6 Soluto to Problem 7.0 (a E [ Y x E[Y x βx x β, so the estmator s ubased. (b Var [ Y x Var[Y ( x σ ( x Now, we clam that x 1 ( x, whch shows that the MLE from the prevous exercse has MSE o larger tha ths estmator. Thk of a radom varable X obtaed by pckg oe of hte x s at radom. Ths s essetally the uform dstrbuto o {x 1,...,x }, except f a value appears more tha oce the set, the t gets probablty proportoal to the umber of tmes t appears. Clearly Var[X E[X (E[X 1 ( 1 x x 0. Ths proves our clam. I fact, we have strct equalty uless the x s are all equal, whch case the MLE ad the estmator ths problem are fact equal to Ȳ /x where x s the commo value of the x s. 7