# Methods of Evaluating Estimators

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3 3 Therefore, E(ˆσ) = E ( ) X1 + + X = E( X 1 ) + + E( X ) = σ So ˆσ is a ubiased estimator for σ. Thus the MSE of ˆσ is equal to its variace, i.e. MSEˆσ = E(ˆσ σ) 2 = V ar(ˆσ) = V ar = V ar( X 1 ) + + V ar( X ) 2 = E( X 2 ) (E( X )) 2 = 2σ2 σ 2 ( ) X1 + + X = V ar( X ) = σ2 The Statistic S 2 : Recall that if X 1,, X come from a ormal distributio with variace σ 2, the the sample variace S 2 is defied as S 2 = (X i X) 2 1 It ca be show that ( 1)S2 σ 2 χ 2 1. From the properties of χ 2 distributio, we have [ ] ( 1)S 2 E = 1 E(S 2 ) = σ 2 σ 2 ad V ar [ ] ( 1)S 2 σ 2 = 2( 1) V ar(s 2 ) = 2σ4 1 Example 2: Let X 1, X 2,, X be i.i.d. from N(µ, σ 2 ) with expected value µ ad variace σ 2, the X is a ubiased estimator for µ, ad S 2 is a ubiased estimator for σ 2. Solutio: We have E( X) ( ) X1 + + X = E Therefore, X is a ubiased estimator. The MSE of X is = E(X 1) + + E(X ) = µ This is because V ar( X) ( ) X1 + + X = V ar MSE X = E( X µ) 2 = V ar( X) = σ2 = V ar(x 1) + + V ar(x ) 2 = σ2

4 4 Similarly, as we showed above, E(S 2 ) = σ 2, S 2 is a ubiased estimator for σ 2, ad the MSE of S 2 is give by MSE S 2 = E(S 2 σ 2 ) = V ar(s 2 ) = 2σ4 1. Although may ubiased estimators are also reasoable from the stadpoit of MSE, be aware that cotrollig bias does ot guaratee that MSE is cotrolled. I particular, it is sometimes the case that a trade-off occurs betwee variace ad bias i such a way that a small icrease i bias ca be traded for a larger decrease i variace, resultig i a improvemet i MSE. Example 3: A alterative estimator for σ 2 of a ormal populatio is the maximum likelihood or method of momet estimator ˆσ 2 = 1 (X i X) 2 = 1 S2 It is straightforward to calculate E( ˆσ ( 1 2 ) = E S2 ) = 1 σ2 so ˆσ 2 is a biased estimator for σ 2. The variace of ˆσ 2 ca also be calculated as V ar( ˆσ ( ) 1 ( 2 1)2 ) = V ar S2 = V ar(s 2 ( 1)2 2σ 4 2( 1)σ4 ) = = Hece the MSE of ˆσ 2 is give by E( ˆσ 2 σ 2 ) 2 = V ar( ˆσ 2 ) + (Bias) 2 = 2( 1)σ4 2 + We thus have (usig the coclusio from Example 2) ( ) 1 2 σ2 σ 2 = 2 1 σ 4 2 MSE σ2 ˆ = 2 1 σ 4 < σ4 = 2σ4 < 2σ4 1 = MSE S 2. This shows that ˆσ 2 has smaller MSE tha S 2. Thus, by tradig off variace for bias, the MSE is improved. The above example does ot imply that S 2 should be abadoed as a estimator of σ 2. The above argumet shows that, o average, ˆσ2 will be closer to σ 2 tha S 2 if MSE is used as a measure. However, ˆσ2 is biased ad will, o the average, uderestimate σ 2. This fact aloe may make us ucomfortable about usig ˆσ 2 as a estimator for σ 2. I geeral, sice MSE is a fuctio of the parameter, there will ot be oe best estimator i terms of MSE. Ofte, the MSE of two estimators will cross each other, that is, for some

5 5 parameter values, oe is better, for other values, the other is better. However, eve this partial iformatio ca sometimes provide guidelies for choosig betwee estimators. Oe way to make the problem of fidig a best estimator tractable is to limit the class of estimators. A popular way of restrictig the class of estimators, is to cosider oly ubiased estimators ad choose the estimator with the lowest variace. If ˆθ 1 ad ˆθ 2 are both ubiased estimators of a parameter θ, that is, E(ˆθ 1 ) = θ ad E(ˆθ 2 ) = θ, the their mea squared errors are equal to their variaces, so we should choose the estimator with the smallest variace. A property of Ubiased estimator: Suppose both A ad B are ubiased estimator for a ukow parameter θ, the the liear combiatio of A ad B: W = aa + (1 a)b, for ay a is also a ubiased estimator. Example 4: This problem is coected with the estimatio of the variace of a ormal distributio with ukow mea from a sample X 1, X 2,, X of i.i.d. ormal radom variables. For what value of ρ does ρ (X i X) 2 have the miimal MSE? Please ote that if ρ = 1 1, we get S2 i example 2; whe ρ = 1, we get ˆσ 2 i example 3. Solutio: As i above examples, we defie The, Let ad let t = ρ( 1) The S 2 = (X i X) 2 1 E(S 2 ) = σ 2 ad Var(S 2 ) = 2σ4 1 e ρ = ρ (X i X) 2 = ρ( 1)S 2 ad We ca Calculate the MSE of e ρ as E(e ρ ) = ρ( 1)E(S 2 ) = ρ( 1)σ 2 = tσ 2 V ar(e ρ ) = ρ 2 ( 1) 2 V ar(s 2 ) = 2t2 1 σ4 MSE(e ρ ) = V ar(e ρ ) + [Bias] 2 = V ar(e ρ ) + [ E(e ρ ) σ 2] 2 = V ar(e ρ ) + (tσ 2 σ 2 ) 2 = V ar(e ρ ) + (t 1) 2 σ 4.

6 6 Plug i the results before, we have where MSE(e ρ ) = f(t) = 2t2 1 σ4 + (t 1) 2 σ 4 = f(t)σ 4 ( ) 2t (t 1)2 = 1 t2 2t + 1 whe t = 1, f(t) achieves its miimal value, which is 2. That is the miimal value of MSE(e ρ ) = 2σ4 +1, with ( 1)ρ = t = 1 +1, i.e. ρ = From the coclusio i example 3, we have It is straightforward to verify that whe ρ = MSE ˆ σ2 = σ 4 < 2σ4 1 = MSE S 2. MSE ˆ σ2 = σ 4 2σ4 + 1 = MSE(e ρ) 2 Efficiecy of a Estimator As we poited out earlier, Fisher iformatio ca be used to boud the variace of a estimator. I this sectio, we will defie some quatity measures for a estimator usig Fisher iformatio. 2.1 Efficiet Estimator Suppose ˆθ = r(x 1,, X ) is a estimator for θ, ad suppose E(ˆθ) = m(θ), a fuctio of θ, the T is a ubiased estimator of m(θ). By iformatio iequality, Var(ˆθ) [m (θ)] 2 I(θ) whe the equality holds, the estimator ˆθ is said to be a efficiet estimator of its expectatio m(θ). Of course, if m(θ) = θ, the T is a ubiased estimator for θ. Example 5: Suppose that X 1,, X form a radom sample from a Beroulli distributio for which the parameter p is ukow. Show that X is a efficiet estimator of p.

8 8 Suppose ˆθ is a efficiet estimator for its expectatio E(ˆθ) = m(θ). Let a statistic T be a liear fuctio of ˆθ, i.e. T = aˆθ + b, where a ad b are costats. The T is a efficiet estimator for E(T ), i.e., a liear fuctio of a efficiet estimator is a efficiet estimator for its expectatio. Proof: We ca see that E(T ) = ae(ˆθ) + b = am(θ) + b, by iformatio iequality Var(T ) a2 [m (θ)] 2. I(θ) We also have Var(T ) = Var(aˆθ + b) = a 2 Var(ˆθ) = a 2 [m (θ)] 2 I(θ), sice ˆθ is a efficiet estimator for m(θ), Var(ˆθ) attais its lower boud. Our computatio shows that the variace of T ca attai its lower boud, which implies that T is a efficiet estimator for E(T ). Now, let us cosider the expoetial family distributio f(x θ) = exp[c(θ)t (x) + d(θ) + S(x)], ad we suppose there is a radom sample X 1,, X from this distributio. We will show that the sufficiet statistic T (X i ) is a efficiet estimator of its expectatio. Clearly, l (X θ) = log f(x i θ) = [c(θ)t (X i ) + d(θ) + S(X i )] = c(θ) T (X i )+d(θ)+ S(X i ), ad l (X θ) = c (θ) T (X i ) + d (θ). Therefore, there is a liear relatio betwee T (X i ) ad l (X θ): T (X i ) = 1 c (θ) l (X θ) d (θ) c (θ). Thus, the sufficiet statistic T (X i ) is a efficiet estimator of its expectatio. Ay liear fuctio of T (X i ) is a sufficiet statistic ad is a efficiet estimator of its expectatio. Specifically, if the MLE of θ is a liear fuctio of sufficiet statistic, the MLE is efficiet estimator of θ. Example 7. Suppose that X 1,, X form a radom sample from a ormal distributio for which the mea µ is kow ad the variace σ 2 is ukow. Costruct a efficiet estimator for σ 2.

9 9 Solutio: Let θ = σ 2 be the ukow variace. The the p.d.f. is f(x θ) = 1 exp { 1 } 2πθ 2θ (x µ)2, which ca be recogized as a member of expoetial family with T (x) = (x µ) 2. So (X i µ) 2 is a efficiet estimator for its expectatio. Sice E[(X i µ) 2 ] = σ 2, E[ (X i µ) 2 ] = σ 2. Therefore, (X i µ) 2 / is a efficiet estimator for σ Efficiecy ad Relative Efficiecy For a estimator ˆθ, if E(ˆθ) = m(θ), the the ratio betwee the CR lower boud ad Var(ˆθ) is called the efficiecy of the estimator ˆθ, deoted as e(ˆθ), i.e. e(ˆθ) = [m (θ)] 2 /[I(θ)]. Var(ˆθ) By the iformatio iequality, we have e(ˆθ) 1 for ay estimator ˆθ. Note: some textbooks or materials defie efficiet estimator ad efficiecy of a estimator oly for ubiased estimator, which is a special case of m(θ) = θ i our defiitios. If a estimator is ubiased ad ad its variace attais the Cramér-Rao lower boud, the it is called the miimum variace ubiased estimator (MVUE). To evaluate a estimator ˆθ, we defied the mea squared error as MSE(ˆθ) = Var(ˆθ) + (E(ˆθ) θ) 2 If the estimator is ubiased, the MSE(ˆθ) = Var(ˆθ). Whe two estimators are both ubiased, compariso of their MSEs reduces to compariso of their variaces. Give two ubiased estimators, ˆθ ad θ, of a parameter θ, the relative efficiecy of ˆθ relative to θ is defied as eff(ˆθ, θ) = Var( θ) Var(ˆθ). Thus, if the efficiecy is smaller tha 1, ˆθ has a larger variace tha θ has. This compariso is most meaigful whe both ˆθ ad θ are ubiased or whe both have the same bias. Frequetly, the variaces of ˆθ ad θ are of the form var(ˆθ) = c 1 ad var( θ) = c 2 where is the sample size. If this is the case, the efficiecy ca be iterpreted as the ratio of sample sizes ecessary to obtai the same variace for both ˆθ ad θ.

10 10 Example 8: Let Y 1,, Y deote a radom sample from the uiform distributio o the iterval (0, θ). Cosider two estimators, ˆθ 1 = 2Ȳ ad ˆθ 2 = + 1 Y (), where Y () = max(y 1,, Y ). Fid the efficiecy of ˆθ 1 relative to ˆθ 2. Solutio: Because each Y i follows a uiform distributio o the iterval (0, θ), µ = E(Y i ) = θ/2, ad σ 2 = Var(Y i ) = θ 2 /12. Therefore, so ˆθ 1 is ubiased. Furthermore E(ˆθ 1 ) = E(2Ȳ ) = 2E(Ȳ ) = 2µ = θ, Var(ˆθ 1 ) = Var(2Ȳ ) = 4Var(Ȳ ) = 4σ2 = θ2 3. To fid the mea ad variace of ˆθ 2, recall that the desity fuctio of Y () is give by Thus, g () (y) = [F Y (y)] 1 ( ) y 1 1 for 0 y θ f Y (y) = θ θ 0 otherwise E(Y () ) = θ y dy = θ θ, it follows that E(ˆθ 2 ) = E{[( + 1)/]Y () } = θ, i.e. ˆθ 2 is a ubiased estimator for θ. therefore, the variace of Y () is E(Y 2 ()) = θ θ 0 y +1 dy = Var(Y () ) = E(Y 2 ()) E(Y () ) 2 = + 2 θ2, ( )2 + 2 θ2 + 1 θ. Thus, ( ) ( Var(ˆθ 2 ) = Var Y () = ) 2 [ ( )2 ] + 2 θ2 + 1 θ = θ 2 ( + 2). Fially, the efficiecy of ˆθ 1 relative to ˆθ 2 is give by eff(ˆθ 1, ˆθ 2 ) = Var(ˆθ 2 ) Var(ˆθ 1 ) = θ2 /[( + 2)] θ 2 /(3) =

11 11 3 Exercises Exercise 1. X, the cosie of the agle at which electros are emitted i muo decay has a desity f(x) = 1 + αx 1 x 1 1 α 1 2 The parameter α is related to polarizatio. Show that E(X) = α. Cosider a estimator 3 for the parameter α, ˆα = 3 X. Computer the variace, the bias, ad the mea square error of this estimator. Exercise 2. Suppose that X 1,, X form a radom sample from a ormal distributio for which the mea µ is ukow ad the variace σ 2 is kow. Show that X is a efficiet estimator of µ. Exercise 3. Suppose that X 1,, X form a radom sample of size from a Poisso distributio with mea λ. Cosider ˆλ 1 = (X 1 + X 2 )/2 ad ˆλ 2 = X. Fid the efficiecy of ˆλ 1 relative to ˆλ 2. Exercise 4. Suppose that Y 1,, Y deote a radom sample of size form a expoetial distributio with desity fuctio give by f(y) = { 1 θ e y/θ fory > 0 0 otherwise Cosider two estimators ˆθ 1 = Y (1), ad ˆθ 2 = Ȳ, where Y (1) = mi(y 1,, Y ). Please show that both ˆθ 1 ad ˆθ 2 are ubiased estimator of θ, fid their MSE, ad fid the efficiecy of ˆθ 1 relative to ˆθ 2.

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