# BASIC STATISTICS. f(x 1,x 2,..., x n )=f(x 1 )f(x 2 ) f(x n )= f(x i ) (1)

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3 BASIC STATISTICS 3 We ca the write 9 as (x i a 2 (x i x 2 + ( x a 2 ( Equatio is clearly miimized whe a x. Now cosider part b of theorem. Expad the secod expressio i part b ad simplify (x i x 2 x 2 i 2 x x i + x 2 (2 x 2 i 2 x 2 + x 2 x 2 i x Theorem 2 o expected values ad variaces of sums. Theorem 2. Let X,X 2, X be a radom sample from a populatio ad let g(x be a fuctio such that Eg(X ad Varg(X exist. The followig two items hold. a: E( g(x i (Eg(X b: Var( g(x i (Varg(X Proof. First cosider part a of theorem 2. Write the expected value of the sum as the sum of the expected values ad the ote that Eg(X Eg(X 2...Eg(X i...eg(x because the X i are all from the same distributio. ( E g(x i E(g(X i (Eg(X (3 First cosider part b of theorem 2. Write the defiitio of the variace for a variable z as E(z E(z 2 ad the combie terms i the summatio sig. ( [ ( 2 Var g(x i E g(x i E g(x i ] (4 Now write out the bottom expressio i equatio 4 as follows ( Var g(x i E [g(x E(g(X ] 2 + E [g(x E(g(X ] E [g(x 2 E(g(X 2 ] + E [g(x E(g(X ] E [g(x 3 E(g(X 3 ] + + E [g(x 2 E(g(X 2 ] E [g(x E(g(X ] + E [g(x 2 E(g(X 2 ] 2 (5 + E [g(x 2 E(g(X 2 ] E [g(x 3 E(g(X 3 ] E [g(x E(g(X ] E [g(x E(g(X ] + + E [g(x E(g(X ] 2 Each of the squared terms i the summatio is a variace, i.e., the variace of X i var(x. Specifically E [g(x i E(g(X i ] 2 Varg(X i Varg(X (6

4 4 BASIC STATISTICS The other terms i the summatio i 5 are covariaces of the form E [g(x i E(g(X i ] E [g(x j E(g(X j ] Cov [g(x i,g(x j ] (7 Now we ca use the fact that the X ad X j i the sample X,X 2,,X are idepedet to assert that each of the covariaces i the sum i 5 is zero. We ca the rewrite 5 as ( Var g(x i E [g(x E(g(X ] 2 + E [g(x 2 E(g(X 2 ] E [g(x E(g(X ] 2 Var(g(X + Var(g(X 2 + Var(g(X 3 + (8 Varg(X i Varg(X Varg(X 2.3. Theorem 3 o expected values of sample statistics. Theorem 3. Let X,X 2, X be a radom sample from a populatio with mea µ ad variace σ 2 <. The a: E X µ b: Var X σ2 c: ES 2 σ 2 Proof of part a. I theorem 2 let g(x g(x i Xi. This implies that Eg(X i µ The we ca write ( ( E X E X i E X i (EX µ (9 Proof of part b. I theorem 2 let g(x g(x i Xi. This implies that Varg(X i σ2 The we ca write ( ( Var X Var X i Var X 2 i (V arx σ2 (20 2 Proof of part c. As i part b of theorem, write S 2 as a fuctio of the sum of square of X i mius times the mea of X i squared ad the simplify ( [ ] ES 2 E Xi 2 X 2 ( EX 2 E X 2 ( ( σ (σ 2 + µ µ2 σ 2 The last lie follows from the defiitio of a radom variable, i.e., (2

6 6 BASIC STATISTICS Estimatio Usig the Method of Momets. I geeral µ r will be a kow fuctio of the parameters θ,θ 2, θ k of the distributio of Y, that is µ r g r (θ,θ 2, θ k. Now let y,y 2,,y be a radom sample from the desity f( ; θ,θ 2, θ k. Form the K equatios µ g (θ,θ 2, θ k ˆµ µ 2 g 2 (θ,θ 2, θ k ˆµ 2. µ K g K (θ,θ 2, θ k ˆµ K y i yi 2 (27 The estimators of θ,θ 2, θ k, based o the method of momets, are obtaied by solvig the system of equatios for the K parameter estimates ˆθ, ˆθ 2, θ ˆ K. This priciple of estimatio is based upo the covetio of pickig the estimators of θı i such a maer that the correspodig populatio (theoretical momets are equal to the sample momets. These estimators are cosistet uder fairly geeral regularity coditios, but are ot geerally efficiet. Method of momets estimators may also ot be uique. y K i Example usig desity fuctio f(y (p +y p. Cosider a desity fuctio give by f(y (p +y p 0 y (28 0 otherwise Let Y,Y 2, Y deote a radom sample from the give populatio. Express the first momet of Y as a fuctio of the parameters. E(Y 0 0 yf(y dy y (p +y p dy y p+ (p +dy (29 yp+2 (p + (p +2 p + p +2 The set this expressio of the parameters equal to the first sample momet ad solve for p. 0

7 BASIC STATISTICS 7 µ E(Y p + p +2 p + p +2 y i ȳ (30 p +(p +2ȳ pȳ +2ȳ p pȳ 2ȳ p( ȳ 2ȳ ˆp 2ȳ ȳ Example usig the Normal Distributio. Let Y,Y 2, Y deote a radom sample from a ormal distributio with mea µ ad variace σ 2. Let (θ,θ 2 (µ, σ 2. Remember that µ µ ad σ 2 E[Y 2 ] E 2 [Y ]µ 2 (µ 2. µ E(Y µ (3 µ 2 E(Y 2 σ 2 + E 2 [Y ]σ 2 + µ 2 Now set the first populatio momet equal to its sample aalogue to obtai µ ˆµ ȳ Now set the secod populatio momet equal to its sample aalogue σ 2 + µ 2 y i ȳ (32 y 2 i σ 2 yi 2 µ 2 (33 σ yi 2 µ2 Now replace µ i equatio 33 with its estimator from equatio 32 to obtai ˆσ yi 2 ȳ2 (34 ˆσ (y i ȳ 2 This is, of course, from the sample stadard deviatio defied i equatios 4 ad 5.

8 8 BASIC STATISTICS Example usig the Gamma Distributio. Let X,X 2, X deote a radom sample from a gamma distributio with parameters θ ad α. The desity fuctio is give by f(x; θ, α θ α Γ(α xα e x θ 0 x< (35 0 otherwise Fid the first momet of the gamma distributio by itegratig as follows E(X 0 θ α Γ(α x θ α Γ(α xα e x θ dx (36 If we multiply equatio 36 by θ +α Γ( + α we obtai 0 x (+α e x θ dx E(X θ+α Γ( + α θ α Γ(α 0 θ +α Γ( + α x(+α e x θ dx (37 The itegrad of equatio 37 is a gamma desity with parameters θ ad α This itegrad will itegrate to oe so that we obtai the expressio i frot of the itegral sig as the E(X. E(X θ+α Γ( + α θ α Γ(α θ Γ( + α Γ(α (38 The gamma fuctio has the property that Γ(t (t Γ(t or Γ(v +vγ(v. Replacig Γ( + α with α Γ(α i equatio 38, we obtai θ Γ( + α E(X Γ(α θαγ(α Γ(α θα (39 We ca fid the secod momet by fidig E(X 2. To do this we multiply the gamma desity i equatio 36 by x 2 istead of x. Carryig out the computatio we obtai E(X 2 0 θ α Γ(α x 2 θ α Γ(α xα e x θ dx (40 0 x (2+α e x θ dx

9 BASIC STATISTICS 9 If we the multiply 40 by θ 2+α Γ(2 + α we obtai E(X 2 θ2+α Γ(2 + α θ α Γ(α 0 θ2+α Γ(2 + α θ α Γ(α θ2 (α + Γ( + α Γ(α θ2 α(α + Γ(α Γ(α θ 2 α(α + θ 2+α Γ(2 + α x(2+α e x θ dx Now set the first populatio momet equal to the sample aalogue to obtai θα x i x (42 ˆα x θ Now set the secod populatio momet equal to its sample aalogue (4 θ 2 α ( α + x 2 i θ 2 θ 2 x2 i α( α + x2 i ( x (( x θ θ + θ 2 ( x 2 θ 2 x2 i x 2 + x θ ( + x θ x 2 i x θ x 2 i x 2 θ x2 i x 2 x (x i x 2 x ( Method of least squares estimatio. Cosider the situatio i which the Y i from the radom sample ca be writte i the form Y i β + ɛ i ˆβ + e i (44 where E(ɛ i 0 ad Var(ɛ i σ 2 for all i. This is equivalet to statig that the populatio from which y i is draw has a mea of β ad a variace of σ 2.

10 0 BASIC STATISTICS The least squares estimator of β is obtaied by miimizig the sum of squares errors, SSE, defied by SSE e 2 i (y i ˆβ 2 (45 The idea is to pick the value of ˆβ to estimate β which miimizes SSE. Pictorially we select the value of ˆβ which miimizes the sum of squares of the vertical deviatios i figure. FIGURE. Least Squares Estimatio The solutio is obtaied by fidig the value of β that miimizes equatio 45. SSE β 2 (y i ˆβ( 0 ˆβ y i ȳ This method chooses values of the parameters of the uderlyig distributio, θ, such that the distace betwee the elemets of the radom sample ad predicted values are miimized. i (46

11 BASIC STATISTICS 3.3. Method of maximum likelihood estimatio (MLE. Least squares is idepedet of a specificatio of a desity fuctio for the paret populatio. Now assume that y i f ( ; θ (θ,..., θ K i. ( Motivatio for the MLE method. If a radom variable Y has a probability desity fuctio f( ; θ characterized by the parameters θ (θ,...,θ k, the the maximum likelihood estimators (MLE of θ,...,θ k are the values of these parameters which would have most likely geerated the give sample Theoretical developmet of the MLE method. The joit desity of a radom sample y,y 2,..., y is give by L g (y,,y ; θ f(y ; θ f (y 2 ; θ f(y 3 ; θ f(y ; θ. Give that we have a radom sample, the joit desity is just the product of the margial desity fuctios. This is referred to as the likelihood fuctio. The MLE of the θ i are the θ i which maximize the likelihood fuctio. The ecessary coditios for a optimum are: L 0i, 2,..., k (48 θ i This gives k equatios i k ukows to solve for the k parameters θ,..., θ k. I may istaces it will be coveiet to maximize l l L rather tha L give that the log of a product is the sum of the logs Example. Let the radom variable X i be distributed as a ormal N(µ,σ 2 so that its desity is give by f(x i ; µ, σ 2 Its likelihood fuctio is give by 2 πσ 2 e 2 ( x i µ σ 2 (49 L Π f(x i ; µ, σ 2 f(x f(x 2 f(x ( e 2 ( x i µ σ 2 e 2 ( x µ σ 2 2πσ 2 ( 2πσ 2 e 2 σ 2 (xi µ2 (50 l L l 2 l(2 πσ 2 2 σ 2 (x i µ 2 The MLE of µ ad σ 2 are obtaied by takig the partial derivatives of equatio 50

12 2 BASIC STATISTICS l µ ˆσ 2 l σ 2 2 (x i µ 0 ˆµ x i [ ] 2π 2πˆσ 2 2ˆσ 2 (2ˆσ 2 2 ( ( (ˆσ (x i ˆµ 2 x (x i ˆµ 2 0 ˆσ (x 2 i ˆµ 2 (5 ˆσ 2 (x i ˆµ 2 ˆσ 2 (x i x 2 ( s 2 The MLE of σ 2 is equal to the sample variace ad ot S 2 ; hece, the MLE is ot ubiased as ca be see from equatio 2. The MLE of µ is the sample mea Example 2 - Poisso. The radom variable X i is distributed as a Poisso if the desity of X i is give by e λ λ x i x i x! i a o egative iteger f(x i ; λ 0 otherwise mea (X λ Var (X λ The likelihood fuctio is give by [ e λ ] [ λ x e λ ] λ x L x! x! (52 e λ λ Σ xi π x i! (53 l L l λ + x i l λ l ( π x i! To obtai a MLE of λ, differetiate l with respect to λ:

13 BASIC STATISTICS 3 l λ + x i λ 0 (54 ˆλ x i x Example 3. Cosider the desity fuctio f ( y (p +y p 0 y 0 otherwise (55 The likelihood fuctio is give by L Π (p + yp i l L l l [(p + y p i ] (56 (l (p + +p l y i To obtai the MLE estimator differetiate 56 with respect to p l p ( p + + l y i 0 ˆp + ( l y i ˆp + ( l y i (57 ˆp + l y i ˆp l y i Example 4. Cosider the desity fuctio The likelihood fuctio is give by f(y p yi ( p yi 0 p (58

14 4 BASIC STATISTICS L Π p yi ( p yi p i yi l L l ( p i yi ( y i l p + y i l ( p To obtai the MLE estimator differetiate 59 with respect to p where we assume that 0 < p <. i (59 l p y i p ( i y i p 0 y i p ( i y i p y i p y i p p y i (60 y i p y i ˆp 3.4. Priciple of Best Liear Ubiased Estimatio (BLUE Priciple of Best Liear Ubiased Estimatio. Start with some desired properties ad deduce a estimator satisfyig them. For example suppose that we wat the estimator to be liear i the observed radom variables. This meas that if the observatios are y,...,y, a estimator of θ must satisfy where the a i are to be determied. ˆθ a i y i ( Some required properties of the estimator (arbitrary. : E(ˆθ θ (ubiased 2: Var(ˆθ VAR( θ(miimum variace where θ is ay other liear combiatio of the y i that also produces a ubiased estimator Example. Let Y,Y 2,...,Y deote a radom sample draw from a populatio havig a mea µ ad variace σ 2. Now derive the best liear ubiased estimator (BLUE of µ. Let the proposed estimator be deoted by ˆθ. It is liear so we ca write it as follows. ˆθ a i y i (62 If the estimator is to be ubiased, there will be restrictios o the a i. Specifically

15 BASIC STATISTICS 5 Now cosider the variace of ˆθ. Ubiasedess [ ] Var( ˆθ Var a i y i E(ˆθ E µ > ( a i y i a i E(y i a i µ a i a i a 2 i Var(y i+σσ i j a i a j Cov(y i y j a 2 i σ2 because the covariace betwee y i ad y j (i j is equal to zero due to the fact that the y s are draw from a radom sample. (63 (64 subject to the co- The problem of obtaiig a BLUE of µ becomes that of miimizig strait i a i. This is doe by settig up a Lagragia i a2 i The ecessary coditios for a optimum are L(a, λ a 2 i λ( a i (65 L 2a λ 0 a... L 2a λ 0 a L λ a i +0 The first equatios imply that a a 2 a 3...a so that the last equatio implies that (66

16 6 BASIC STATISTICS a i 0 a i 0 a i (67 a i ˆθ a i y i Note that equal weights are assiged to each observatio. y i ȳ 4. FINITE SAMPLE PROPERTIES OF ESTIMATORS 4.. Itroductio to sample properties of estimators. I sectio 3 we discussed alterative methods of estimatig the ukow parameters i a model. I order to compare the estimatig techiques we will discuss some criteria which are frequetly used i such a compariso. Let θ deote a ukow parameter ad let ˆθ ad θ be alterative estimators. Now defie the bias, variace ad mea squared error of ˆθ as Bias (ˆθ E (ˆθ θ Var(ˆθ E MSE (ˆθ E (ˆθ E (ˆθ 2 (ˆθ θ 2 Var(ˆθ + ( Bias The result o mea squared error ca be see as follows 2 MSE(θ E (ˆθ θ (ˆθ 2 E (ˆθ E (ˆθ + E θ ( ( 2 E ((ˆθ E (ˆθ + E ˆθ θ 2 [ (ˆθ E (ˆθ E (ˆθ +2 E θ (ˆθ 2 (ˆθ] E (ˆθ E + 2 ( 2 (ˆθ E (ˆθ + E (ˆθ θ sice E (ˆθ E (ˆθ E (ˆθ Var + ( 2 Bias(ˆθ ( 2 E (ˆθ θ 0 (68 ( Specific properties of estimators.

17 BASIC STATISTICS Ubiasedess. ˆθ is said to be a ubiased estimator of θ if E (ˆθ θ. I figure 2, ˆθ is a ubiased estimator of θ, while θ is a biased estimator. f Θ FIGURE 2. Ubiased Estimator f Θ f Θ Θ Θ Miimum variace. ˆθ is said to be a miimum variace estimator of θ if Var(ˆθ Var( θ (70 where θ is ay other estimator of θ. This criterio has its disadvatages as ca be see by otig that ˆθ costat has zero variace ad yet completely igores ay sample iformatio that we may have. I figure 3, θ has a lower variace tha ˆθ Mea squared error efficiet. ˆθ is said to be a MSE efficiet estimator of θ if MSE(ˆθ MSE( θ (7 where θ is ay other estimator of θ. This criterio takes ito accout both the variace ad bias of the estimator uder cosideratio. Figure 4 shows three alterative estimators of θ Best liear ubiased estimators. ˆθ is the best liear ubiased estimator (BLUE of θ if

18 8 BASIC STATISTICS FIGURE 3. Estimators with the Same Mea but Differet Variaces f Θ f Θ f Θ Θ ˆθ a i y i liear E(ˆθ θ ubiased (72 Var(ˆθ Var( θ where θ is ay other liear ubiased estimator of θ. For the class of ubiased estimators of ˆθ, the efficiet estimators will also be miimum variace estimators Example. Let X,X 2,...,X deote a radom sample draw from a populatio havig a populatio mea equal to µ ad a populatio variace equal to σ 2. The sample mea (estimator of µ is calculated by the formula X i X ad is a ubiased estimator of µ from theorem 3 ad equatio 9. (73 Two possible estimates of the populatio variace are ˆσ 2 S 2 (X i X 2 (X i X 2 We have show previously i theorem 3 ad equatio 2 that ˆσ 2 is a biased estimator of σ 2 ; whereas S 2 is a ubiased estimator of σ 2. Note also that

19 BASIC STATISTICS 9 FIGURE 4. Three Alterative Estimators f Θ f Θ f Θ f Θ Θ ( ˆσ 2 S 2 E (ˆσ ( 2 E ( S 2 ( σ 2 Also from theorem 3 ad equatio 20, we have that Var ( X σ2 (74 (75 Now cosider the mea square error of the two estimators X ad S 2 where X,X 2,...X are a radom sample from a ormal populatio with a mea of µ ad a variace of σ 2.

20 20 BASIC STATISTICS E ( 2 ( σ X µ Var X 2 E ( S 2 σ 2 2 ( Var S 2 2 (76 σ4 The variace of S 2 was derived i the lecture o sample momets. The variace of ˆσ 2 is easily computed give the variace of S 2. Specifically, (( Varˆσ 2 Var s 2 ( 2 Var ( S 2 ( ( σ 4 2( σ4 We ca compute the MSE of ˆσ 2 usig equatios 68, 74, ad 77 as follows 2 MSE ˆσ 2 E (ˆσ 2 σ 2 2( σ ( σ4 2 + σ 4 ( 2( 2 + [( ] 2 σ 2 σ 2 ( 2 ( σ 4 2 ( ( 2 σ 4 + σ ( 2 2+ σ ( 2 σ (78 Now compare the MSE s of S 2 ad ˆσ 2. ( ( 2 2 MSEˆσ 2 σ 4 < σ 4 2 MSE S 2 (79 So ˆσ 2 is a biased estimator of S 2 but has lower mea square error.

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