Topic 9: Sampling Distributions of Estimators


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1 Topic 9: Samplig Distributios of Estimators Course 003, 2018 Page 0
2 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be derived from the joit distributio of X 1... X. It is called the samplig distributio because it is based o the joit distributio of the radom sample. Give a samplig distributio, we ca make appropriate tradeoffs betwee sample size ad precisio of our estimator sice samplig distributios o sample size. obtai iterval estimates rather tha poit estimates after we have a sample a iterval estimate is a radom iterval such that the true parameter lies withi this iterval with a give probability (say 95%). choose betwee to estimators we ca, for istace, calculate the measquared error of the estimator, E θ [( ˆθ θ) 2 ] usig the distributio of ˆθ. Page 1
3 Examples: Applicatio: sample size ad precisio 1. What if X i N(θ, 4), ad we wat E( X θ) 2.1? This is simply the variace of X, ad we kow X N(θ, 4/). 4.1 if Cosider a radom sample of size from a Uiform distributio o [0, θ], ad the statistic U = max{x 1,..., X }. The CDF of U is give by: 0 if u 0 ( ) F(X) = uθ if 0 < u < θ 1 if u θ We ca ow use this to see how large our sample must be if we wat a certai level of precisio i our estimate for θ. Suppose we wat the probability that our estimate lies withi.1θ for ay level of θ to be bigger tha 0.95: Pr( U θ.1θ) = Pr(θ U.1θ) = Pr(U.9θ) = 1 F(.9θ) = We wat this to be bigger tha 0.95, or With the LHS decreasig i, we choose log(.05) log(.9) = Our miimum sample size is therefore 29. Page 2
4 Joit distributio of sample mea ad sample variace For a radom sample from a ormal distributio, we kow that the M.L.E.s are the sample mea ad the sample variace 1 (X i X ) 2. We kow that X N(µ, σ2 ) ad ( X i µ σ )2 χ 2 ( sum of squares of stadard ormals) If we replace the populatio mea µ with the sample mea X, the resultig sum of squares, has a χ 2 1 distributio, which is idepedet of the distributio of X. This is stated formally below: Theorem: If X 1,... X form a radom sample from a ormal distributio with mea µ ad variace σ 2, the the sample mea X ad the sample variace 1 (X i X ) 2 are idepedet radom variables ad X N(µ, σ2 ) (X i X ) 2 σ 2 χ 2 1 Note: This is oly for ormal samples. Page 3
5 Applicatio: mea ad variace estimates We have a ormal radom sample ad would like the M.L.E.s of the mea ad stadard deviatio to be withi oefifth of a stadard deviatio of the respective parameters, µ ad σ with some threshold probability. Suppose we wat to choose a sample size such that Pr( X µ 1 5 σ) 2 1 If we use Chebyshev s iequality, we get this probability is greater tha 25, so settig this equal to 1 2, we have = 50 Usig the exact distributio of X, Pr( X µ 1 5 σ) = Pr( X µ σ 1 5 ) Sice we ow have a stadard ormal r.v., we kow Pr(Z >.68) =.25, so we eed the smallest greater tha (.68 5) 2 = 11.6, so = 12 (Stata 14: ivormal(.75)=.6745) Now if we wat to determie so that Pr[( X µ 1 5 σ ad ( ˆσ σ 5 1 σ] 1 2 By the previous theorem, X ad ˆσ are idepedet, so the LHS is the product p 1 p 2 = Pr( X µ 1 5 σ)pr( ˆσ σ 1 5 σ) p 1 = Pr( Z 5 ) = 1 2 (1 Φ( 5 ). p 2 = Pr(.8σ < ˆσ < 1.2σ) = Pr(.64 < ˆσ2 σ 2 Sice V = ˆσ2 σ 2 < 1.44) χ 2 1, we ca search over values of to fid oe that gives us a product of probabilities equal to 1 2. For = 21, p 1 =.64 p 2 =.79 so p 1 p 2 =.5. display chi2(20, 30.24)chi2(20, 13.44) (sice 21*.64=13.44 ad 21*1.44=30.24) Page 4
6 The tdistributio Let Z N(0, 1), let Y χ 2 v, ad let Z ad Y be idepedet radom variables. The X = Z Yv t v The p.d.f of the tdistributio is give by: f(x; v) = v+1 Γ( 2 ) (1 Γ( v 2 ) + x2 ) ( v+1 2 ) πv v Features of the tdistributio: Oe ca see from the above desity fuctio that the tdesity is symmetric with a maximum value at x = 0. The shape of the desity is similar to that of the stadard ormal (bellshaped) but with fatter tails. Page 5
7 Relatio to radom ormal samples RESULT 1: Defie S 2 = (X i X ) 2 The radom variable U = (X µ) S 2 1 t 1 (X µ) σ Proof: We kow that N(0, 1) ad that S2 σ 2 χ 2 1. Dividig the first radom variable by the square root of the secod, divided by its degrees of freedom, the σ i the umerator ad deomiator cacels to obtai U. Implicatio: We caot make statemets about X µ usig the ormal distributio if σ 2 is ukow. This result allows us to use its estimate ˆσ 2 = (X i X ) 2 / sice (X µ) ˆσ/ 1 t 1 RESULT 2 As, U Z N(0, 1) To see why: ad 1 U ca be writte as is close to 1. 1 (X µ) ˆσ t 1. As gets large ˆσ gets very close to σ F 1 (.55) =.129 for t 10,.127 for t 20 ad.126 for the stadard ormal distributio. The differeces betwee these values icreases for higher values of their distributio fuctios (why?) Page 6
8 Cofidece itervals for the mea Give σ 2, let us see how we ca obtai a iterval estimate for µ, i.e. a iterval which is likely to cotai µ with a prespecified probability. ( ) Sice (X µ) σ/ N(0, 1), Pr 2 < (X µ) σ/ < 2 =.955 But this evet is equivalet to the evets 2σ < X µ < 2σ ad X 2σ < µ < X + 2σ With kow σ, each of the radom variables X 2σ ad X + 2σ are statistics. Therefore, we have derived a radom iterval withi which the populatio parameter lies with probability.955, i.e. ( Pr X 2σ < µ < X + 2σ ) =.955 = γ Notice that there are may itervals for the same γ, this is the shortest oe. Now, give our sample, our statistics take particular values ad the resultig iterval either cotais or does ot cotai µ. We ca therefore o loger talk about the probability that it cotais µ because the experimet has already bee performed. We say that (x 2σ < µ < x + 2σ ) is a 95.5% cofidece iterval for µ. Alteratively, we may say that µ lies i the above iterval with cofidece γ or that the above iterval is a cofidece iterval for µ with cofidece coefficiet γ Page 7
9 Cofidece Itervals for meas..examples Example 1: X 1,..., X forms a radom sample from a ormal distributio with ukow µ ad σ 2 = 10. x is foud to be with = 40. A 80% cofidece iterval for the mea µ 10 is give by ( ), ) or (6.523, 7.805). The cofidece coefficiet. is.8 (stata 14: display ivormal(.9) Example 2: Let X deote the sample mea of a radom sample of size 25 from a σ distributio with variace 100 ad mea µ. I this case, = 2 ad, makig use of the cetral limit theorem the followig statemet is approximately true: ( Pr 1.96 < (X µ) 2 ) ( ) < 1.96 =.95 or Pr X 3.92 < µ < X =.95 If the sample mea is give by x = 67.53, a approximate 95% cofidece iterval for the sample mea is give by (63.61, 71.45). Example 3: Suppose we are iterested i a cofidece iterval for the mea of a ormal distributio but do ot kow σ 2. We kow that (X µ) ˆσ/ 1 t 1 ad ca use the tdistributio with ( 1) degrees of freedom to costruct our iterval estimate. With = 10, x = 3.22, ˆσ = 1.17, a 95% cofidece iterval is give by (3.22 (2.262)(1.17)/ 9, (2.262)(1.17)/ 9) = (2.34, 4.10) (display ivt(9,.975) gives you 2.262) Page 8
10 Cofidece Itervals for differeces i meas Let X 1,..., X ad Y 1,..., Y m deote idepedet ormal radom samples. X i N(µ 1, σ 2 ) ad Y i N(µ 2, σ 2 ) respectively. Sample meas ad variaces are X, Ȳ, ˆσ2 1, ˆσ2 2. We kow (usig previous results) that: X ad Ȳ are ormally ad idepedetly distributed with meas µ 1 ad µ 2 ad variaces σ2 σ2 ad m ( X Ȳm) N(µ 1 µ 2, σ2 ˆσ 2 1 σ 2 χ 2 1 ad m ˆσ2 2 σ 2 + σ2 m ) so ( X Ȳm) (µ 1 µ 2 ) N(0, 1) σ 2 + σ2 m χ 2 m 1, so their sum ( ˆσ2 1 + m ˆσ2 2 )/σ2 χ 2 +m 2. Therefore U = ( X Ȳm) (µ 1 µ 2 ) ( ) t +m 2 ˆσ 2 1 +m ˆσ2 2 1 (+m 2) + m 1 Deote the deomiator of U by R. Page 9 Suppose we wat a 95% cofidece iterval for the differece i the meas: ( ) Usig the above tdistributio, we fid a umber b for which Pr b < X < b =.95 The radom iterval ( X Ȳ) br, ( X Ȳ) + br will ow cotai the true differece i meas with 95% probability. A cofidece iterval is ow based o sample values, ( x ȳ m ) ad correspodig sample variaces. Based o the CLT, we ca use the same procedure eve whe our samples are ot ormal.
Topic 9: Sampling Distributions of Estimators
Topic 9: Samplig Distributios of Estimators Course 003, 2018 Page 0 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be
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