Econ 325 Notes on Point Estimator and Confidence Interval 1 By Hiro Kasahara

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1 Poit Estimator Eco 325 Notes o Poit Estimator ad Cofidece Iterval 1 By Hiro Kasahara Parameter, Estimator, ad Estimate The ormal probability desity fuctio is fully characterized by two costats: populatio mea µ ad populatio variace. The probability mass fuctio of Beroulli radom variable is fully defied by the populatio fractio of success, p. These costats are called parameters ad we geerally use the Greek letter θ to deote them. We are ofte iterested i kowig the populatio parameter such as populatio mea ad populatio variace. To guess the populatio value of mea ad variace, we use their sample aalogues, i.e., sample mea ad sample variace. A poit estimator of θ is a fuctio of the radom sample, deoted by ˆθ: ˆθ = ˆθX 1, X 2,..., X. Here, the right had side of the equatio provides a mappig from the sample {X 1, X 2,..., X } to real value. Namely, ˆθX 1, X 2,..., X a formula to compute the sample aalog of correspodig populatio parameter e.g., for sample mea X, we have ˆθX 1, X 2,..., X = 1 i=1 X i. The estimator ˆθ is a radom variable because the sample {X 1, X 2,..., X } is radomly draw. Whe we evaluate ˆθX 1, X 2,..., X at the realized sample, the ˆθ is called a estimate. The evaluated value of the fuctio ˆθX 1, X 2,..., X at the realized sample is ot a radom variable ay more rather, it is costat. Ubiasedess A estimator ˆθ is said to be a ubiased estimator of the parameter θ if The bias of a estimator ˆθ is defied as E[ˆθ] = θ. Bias = E[ˆθ] θ. The bias of a ubiased estimator is zero by defiitio. Example 1. The sample mea X = 1 i=1 X i is a ubiased estimator of the populatio mea µ because E[ X] = 1 i=1 E[X i] = 1 i=1 µ = µ. The sample variace s 2 = 1 1 i=1 X i X 2 is a ubiased estimator of the populatio variace. However, the estimator ˆ = 1 i=1 X i X 2 is ot a ubiased estimator of because E[ˆ ] = E[ i=1 X i X 2 ] = 1E[ 1 1 i=1 X i X 2 ] = 1 σ2 <. 1 c Hiroyuki Kasahara. Not to be copied, used, revised, or distributed without explicit permissio of copyright ower. 1

2 Example 2. Give a radom sample of size {X 1, X 2,..., X }, cosider a estimator ˆX = X 1, which oly uses the first observatio while igores all other 1 observatios. This estimator is a ubiased estimator of µ because E[X 1 ] = µ. We ca also cosider a estimator defied by the weighted average of X i s as ˆX = i=1 w ix i, where {w i } i=1 is a sequece of umbers such that i=1 w i = 1. The, this estimator is a ubiased estimator of µ because E[ i=1 w ix i ] = i=1 w iµ = µ. Example 3. The sample fractio ˆp is a ubiased estimator fo the populatio fractio p. This is because the sample fractio is viewed as the sample average of idepedet Beroulli radom variables, i.e., ˆp = X = 1 i=1 X i, where X i = 0 with probability 1 p ad X i = 1 with probability p. Takig the expectatio, E[ˆp] = 1 i=1 E[X i] = 1 i=1 p = p. Efficiecy Cosider the case for = 2 ad let X 1 ad X 2 are radomly sampled from the populatio distributio with mea µ ad variace. Cosider the followig two estimators for µ: X = 1 2 X 1 + X 2 ad X = 1 3 X X 2. Both X ad X are ubiased estimators because E[ X] = 1 2 µ+ 1 2 µ = µ ad E[ X] = 1 3 µ+ 2 3 µ = µ. I fact, we may cosider a estimator of the form give by ax ax 2 for ay fixed value of a ad we ca verify that ax ax 2 is a ubiased estimator because E[aX ax 2 ] = µ. While ubiasedess is a desirable property of estimators, we have multiple ubiased estimators. Which estimators do we wat to choose amog all ubiased estimators? The aswer is: the estimator that has the smallest variace. Ituitively, the smaller the variace is, the closer the realized value of the estimator is to the populatio mea o average. I fact, if the variace of the ubiased estimator is zero, we have populatio mea for every realized value of the estimator. Let ˆθ 1 ad ˆθ 2 be two ubiased estimators. The ˆθ 1 is said to be more efficiet tha ˆθ 2 if V arˆθ 1 < V arˆθ 2. If ˆθ 1 is a ubiased estimator that has the smallest variace amog all ubiased estimators, the ˆθ 1 is said to be the most efficiet, or the miimum variace ubiased estimator. Example. Cosider the case for = 2 ad X 1 ad X 2 are radomly sampled from the populatio distributio with mea µ ad variace. What is the most efficiet ubiased estimator? To aswer this, we cosider the class of ubiased estimator of the form ax ax 2 for ay fixed value a. The variace of ax ax 2 is give by V arax ax 2 = {a a 2 }, where CovX 1, X 2 = 0 by radom samplig. Therefore, we may fid the most efficiet ubiased estimator by miimizig ga = a a 2 = 2a 2 2a + 1 with respect to a. The first order coditio is give by g a = a 2 = 0 so that ga is miimized at a = 1/2. Therefore, 1X X 2 2 = X is the most efficiet ubiased estimator for µ. 2

3 Cosistecy A poit estimator ˆθ is said to be cosistet if ˆθ coverges i probability to θ, i.e., for every ɛ > 0, lim P ˆθ θ < ɛ = 1 see Law of Large Number. Example 5. Suppose that X 1, X 2,..., X are radomly sampled from a populatio with mea µ ad variace. Is X = 1 i=1 X i a cosistet estimator of µ? How about 1 1 i=1 X i ad i=1 X i? Are these two estimators cosistet? The sample variace s 2 = 1 1 i=1 X i X 2 is a cosistet estimator of. Is the estimator ˆ = 1 i=1 X i X 2 a cosistet estimator of? Example 6. Suppose that X 1, X 2,..., X are idepedet Beroulli radom variables, i.e., ˆp = X = 1 i=1 X i, where X i = 0 with probability 1 p ad X i = 1 with probability p. The parameter p is the populatio fractio of idividuals with X i = 1. The sample fractio is defied as ˆp = X = 1 i=1 X i. By the Law of Large Numbers, the sample fractio ˆp is a cosistet estimator of the populatio fractio p. The variace of ˆp is give by V arˆp = V ar 1 i=1 X i = 1 2 V ar X1 + X X = V arx i = V arx i 2 = p1 p, where the last lie follows from V arx i = E[X i p 2 ] = 0 p 2 1 p + 1 p 2 p = p1 p. Because V arˆp = p1 p ivolves the ukow populatio parameter p, we do ot kow the value of V arˆp = p1 p. We ca costruct a estimator for V arˆp = p1 p by replacig p with ˆp, where the latter ca be computed from the sample, as V arˆp = ˆp1 ˆp. V arˆp = ˆp1 ˆp is a cosistet estimator of V arˆp. Cofidece Iterval We may estimate iterval rather tha a poit. The idea of iterval estimatio is to costruct a radom iterval such that the costructed iterval cotais the true parameter θ with a pre-specified probability, 1 α. Such a iterval is called 1 α percet cofidece iterval, where 1 α is called the cofidece level. The cofidece iterval is characterized by the lower limit L ad the upper limit U, both of which is a fuctio of the radom sample X 1,..., X so that P LX 1,..., X θ UX 1,..., X = 1 α. Note that both LX 1,..., X ad UX 1,..., X are radom variables. The case that is large or ˆθ Nθ, V arˆθ with V arˆθ kow. Suppose that a poit estimator ˆθ is approximately ormally distributed with mea θ ad variace V arˆθ, i.e., ˆθ N θ, V arˆθ. Two represetative cases are: 1. X 1, X 2,..., X are radomly sampled from some distributio that is differet from ormal distributio but the sample size is large. I this case, ˆθ is defied as the 3

4 average of radom variable, i.e., ˆθ = X = 1 X i so that we may apply the Cetral Limit Theorem to have ˆθ NE[X i ], V arx i /; for example, the sample mea X Nµ, /. Whe is large, we may essetially treat V arx i as if it is kow ad give by the sample variace. 2. X 1, X 2,..., X are radomly sampled from ormal distributio Nµ, with kow variace. I this case, the sample average is ormally distributed with mea µ ad variace /. I these cases, we may costruct the 95 percet cofidece iterval with [ ] [L, U] = ˆθ 1.96 V arˆθ, ˆθ V arˆθ, so that P ˆθ 1.96 V arˆθ θ ˆθ V arˆθ = I geeral, the cofidece iterval with cofidece level 1 α is costructed as P ˆθ z α/2 V arˆθ θ ˆθ + z α/2 V arˆθ = 1 α, 1 where z α/2 is determied such that P Z z α/2 = α/2 whe Z N0, 1. Here, z α/2 V arˆθ is called as the margi of error. We may cofirm 1 by reformulatig the iequality o the left had side of 1 i terms of a stadardized radom variable ˆθ θ V arˆθ as follows. P ˆθ z α/2 V arˆθ θ ˆθ + z α/2 V arˆθ } } =P {ˆθ θ z α/2 V arˆθ ad {z α/2 V arˆθ ˆθ θ =P ˆθ θ z α/2 ad z α/2 ˆθ θ V arˆθ V arˆθ =P z α/2 ˆθ θ V arˆθ z α/2 where Z = ˆθ θ V arˆθ N0, 1 2 =P z α/2 Z z α/2 = 1 α. Example 7 Cofidece Iterval for populatio mea µ. Give the radom sample X 1,..., X draw from Nµ, ad is kow, the sample average X = 1 i=1 X i is a estimator of µ with V ar X = µ is give by [ [L, U] = = σ. Therefore, 95 percet cofidece iterval for X 1.96 σ, X σ ].

5 If we would like to costruct 90 percet cofidece iterval with α = 0.1, z 0.05 = 1.65 i.e., P Z 1.65 = 0.05 for Z N0, 1. Therefore, 95 percet cofidece iterval for µ is give by [ [L, U] = X 1.65 σ, X σ ]. Example 8 Survey o the U.S. presidetial electio i Florida. The survey was coducted betwee Oct. 20 ad 2, 2016, i Florida after the third ad fial presidetial debate. The survey result shows that, amog 1166 likely registered voters who support either Clito or Trump, there are 602 Clito voters ad 56 Trump voters. What is the 95 precet cofidece iterval for the populatio fractio of Clito voters? Let p be the populatio fractio of Clito voters. Each voter s preferece is a Beroulli radom variable X i with P X i = 1 = p ad P X i = 0 = 1 p, where X i = 1 meas Clito voter while X i = 0 meas Trump voter. The sample average is give by ˆp = The stadard deviatio of ˆp is give by p1 p/, which ca be estimated as ˆp1 ˆp/ = /1166 = The margi of error is, therefore, = The, we may costruct the 95 percet cofidece iterval with [ [L, U] = ˆp 1.96 ˆp1 ˆp/, ˆp ˆp1 ] ˆp/ = [0.88, 0.55]. Therefore, the populatio fractio of Clito voters is betwee 0.88 ad 0.55 with probability 95 percet. Therefore, i this Florida s poll, Clito s lead is withi the margi of error. Example 9 Survey o the U.S. presidetial electio i North Carolia. The survey was coducted betwee November 3 ad 6, 2016, i North Carolia. The survey result shows that, amog 791 likely registered voters who support either Clito or Trump, there are 00 Clito voters ad 391 Trump voters. What is the 95 precet cofidece iterval for the populatio fractio of Clito voters? The kowledge of V arˆθ is required for costructig cofidece iterval as show i 1. Typically, V arˆθ depeds o populatio parameter that is ukow e.g., V ar X = / ad V arˆp = p1 p/ but we ca estimate V arˆθ. The estimator of V ar X ad V arˆp are give by Vˆar X = s 2 / ad Vˆarˆp = ˆp1 ˆp/. Whe V arˆθ is ot kow, we replace V arˆθ with its estimator Vˆarˆθ i costructig cofidece iterval. I the above example of the U.S. presidetial electio Example 8, this is what we did: we replaced V arˆp = p1 p with its estimator ˆp1 ˆp. This is fie as log as the sample size is large because the estimator of V arˆp coverges i probability to V arˆp ad we may essetially treat V arˆp as kow i costructig the cofidece iterval. Whe is small, however, this is ot the case aymore. The radomess of the estimator of V arˆp does ot go away whe is small ad the costructed cofidece iterval i 1 by replacig V arˆθ with its estimator does ot cotai θ with probability 1 α aymore. The case that is small Whe is small, it is geerally difficult to costruct cofidece iterval for two reasos. First, we may ot use the Cetral Limit Theorem to claim that ˆθ is ormally distributed. Secod, replacig V arˆθ with its estimator Vˆarˆθ itroduces a additioal source of radomess. 5

6 I both cases, the stadardized radom variable usig the estimator of V arˆθ ˆθ θ Vˆarˆθ is ot a stadard ormal radom variable, ad therefore the cofidece iterval 1 is ot valid ay more because 1 is costructed uder the assumptio that N0, 1 see ˆθ θ V arˆθ 2. While it is geerally difficult to costruct cofidece iterval, there is oe exceptioal case where we may costruct cofidece iterval usig Studet s t-distributio. Suppose that we have a radom sample X 1, X 2,..., X from Nµ,. I this case, we have X µ s/ Studet s t distributio with 1 degrees of freedom. Therefore, the cofidece iterval for µ with cofidece level 1 α is costructed as s P X t 1,α/2 µ X s + t 1,α/2 = 1 α, 3 where t 1,α/2 is determied such that P T t 1,α/2 = α/2 whe T Studet s t distributio with 1 degrees of freedom. We may cofirm 3 by reformulatig the iequality o the left had side of 3 i terms of a stadardized radom variable X µ s/ as follows. s P X t 1,α/2 µ X s + t 1,α/2 { } { X µ =P s/ t 1,α/2 ad t 1,α/2 X } µ s/ =P t 1,α/2 X µ s/ t 1,α/2 where T = X µ s/ =P t 1,α/2 T t 1,α/2 = 1 α, Studet s t distributio with 1 degrees of freedom i. A few commets. First, we eed the assumptio that X 1, X 2,..., X are draw from the ormal distributio. If X i is a Beroulli radom variable, the X µ s/ does ot follow t-distributio. Secod, as, s 2 p, so that the Studet s t distributio coverges to the stadard ormal distributio as. I fact, at = 31, the critical value for 95 percet cofidece iterval usig t-distributio is give by 2.02 which is close to Cofidece iterval for sample variace Suppose that {X 1, X 2,..., X } is a radom sample from a ormal distributio with E[X i ] = µ ad Var[X i ] =. The, the radom variable 1s2 has a distributio kow as the chisquare distributio with 1 degree of freedom which we deote by χ 2 1, i.e., 1s 2 = χ

7 Let χ 2 1,α/2 ad χ2 1,1 α/2 be the value such that P χ2 1 > χ 2 1,α/2 = α/2 ad P χ2 1 > χ 2 1,1 α/2 = 1 α/2 so that P χ 2 1,1 α/2 < 1s2 < χ 2 1,α/2 = 1 α. The, we may costruct the cofidece iterval for from the sample variace s = 1 1 i=1 X i X 2 as follows. 1 α = P χ 2 1,1 α/2 < 1s2 < χ 2 1,α/2 1 1 = P < < = P χ 2 1,α/2 1s 2 χ 2 1,α/2 1s 2 < < 1s2 χ 2 1,1 α/2 χ 2 1,1 α/2. Therefore, with P L < < U 1 = 1 α L = 1s2 χ 2 1,α/2 ad U = 1s2. χ 2 1,1 α/2 Example 10 Cofidece Iterval ad Hypothesis Testig for Sample Variace. Suppose that you are a plat maager for producig electrical devices operated by a thermostatic cotrol. Accordig to the egieerig specificatios, the stadard deviatio of the temperature at which these cotrols actually operate should ot exceed 2.0 degrees Fahreheit. As a plat maager, you would like to kow how large the populatio stadard deviatio σ is. We assume that the temperature is ormally distributed. Suppose that you radomly sampled 25 of these cotrols, ad the sample variace of operatig temperatures was s 2 = 2.36 degrees Fahreheit. i Compute the 95 percet cofidece iterval for the populatio stadard deviatio σ. ii Test the ull hypothesis H 0 : σ = 2 agaist the alterative hypothesis H 1 : σ > 2 at the sigificace level α = The distributio of 1s2 is give by chi-square distributio with 1 degrees of freedom. Let χ 2 1 be a radom variable distributed by chi-square distributio with 1 degrees of freedom ad let χ 2 1,α be the value such that Prχ 2 1 < χ 1,α = α. The, Prχ 2 1,1 α/2 1s2 χ 2 1,α/2 = 1 α ad, therefore, Pr 1s 2. χ 2 1,α/2 1s2 χ 2 1,1 α/2 Now, whe α/2 = 0.025, chi-square table gives χ 2 2,0.025 = ad χ 2 2,0.975 = so that the lower limit of the 95 percet CI is limit is 1s 2 χ 2 1, s 2 χ 2 1,0.025 = = 1.39 ad the upper = =.567. Therefore, the 95 percet CI for the populatio variace is [1.39,.567] ad, because of the mootoic relatioship betwee the variace ad the stadard deviatio, the 95 percet CI for the populatio stadard deviatio is give by [ 1.39,.567] = [1.200, 2.137]. 7

8 To test the ull hypothesis H 0, a fid the distributio of stadardized radom variable whe H 0 is true, i.e., = 2 2 =, b fid the rejectio regio which is the regio 1s 2 where the radom variable 1s2 is ulikely i.e., with the probability less tha 5 percet to fall ito if H 0 is true, c look at the realized value of 1s2 ad ask if the realized value of 1s2 is a ulikely value to happe if H 0 is true by checkig if 1s2 falls ito the rejectio regio. For a, whe H 0 is true, 1s2 is distributed accordig to the chi-square distributio with the degree of freedom equal to 1 = 2. For b, because H 1 : σ > 2, we cosider oesided test; amely, the very high value of 1s2 is cosidered to be evidece agaist H 0 but ot the low value of 1s2. Uder H 0, Pr 1s 2 χ 2 1,α = Pr 2s 2 χ 2 2,α = 1 α for α = 0.05, where χ 2 2,0.05 = Therefore, the rejectio regio for 2s2 is give by 36.15,, i.e., we reject H 0 if 2s2 > 36.15, or equivaletly, s 2 > 36.15/6 = 6.02 because such a value of s 2 is ulikely to happe if H 0 is true. c The realized value of s 2 is 2.36, which does ot fall i the rejectio regio i.e., 2.36 belogs to the regio which is ot ulikely happe if H 0 is true ad hece there is ot sufficiet evidece agaist H 0. We do ot reject H 0. 8