Part IB Statistics. Theorems with proof. Based on lectures by D. Spiegelhalter Notes taken by Dexter Chua. Lent 2015

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1 Part IB Statistics Theorems with proof Based on lectures by D. Spiegelhalter Notes taken by Dexter Chua Lent 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine. Estimation Review of distribution and density functions, parametric families. Examples: binomial, Poisson, gamma. Sufficiency, minimal sufficiency, the Rao-Blackwell theorem. Maximum likelihood estimation. Confidence intervals. Use of prior distributions and Bayesian inference. [5] Hypothesis testing Simple examples of hypothesis testing, null and alternative hypothesis, critical region, size, power, type I and type II errors, Neyman-Pearson lemma. Significance level of outcome. Uniformly most powerful tests. Likelihood ratio, and use of generalised likelihood ratio to construct test statistics for composite hypotheses. Examples, including t-tests and F -tests. Relationship with confidence intervals. Goodness-of-fit tests and contingency tables. [4] Linear models Derivation and joint distribution of maximum likelihood estimators, least squares, Gauss-Markov theorem. Testing hypotheses, geometric interpretation. Examples, including simple linear regression and one-way analysis of variance. Use of software. [7] 1

2 Contents IB Statistics (Theorems with proof) Contents 0 Introduction 3 1 Estimation Estimators Mean squared error Sufficiency Likelihood Confidence intervals Bayesian estimation Hypothesis testing Simple hypotheses Composite hypotheses Tests of goodness-of-fit and independence Goodness-of-fit of a fully-specified null distribution Pearson s Chi-squared test Testing independence in contingency tables Tests of homogeneity, and connections to confidence intervals Tests of homogeneity Confidence intervals and hypothesis tests Multivariate normal theory Multivariate normal distribution Normal random samples Student s t-distribution Linear models Linear models Simple linear regression Linear models with normal assumptions The F distribution Inference for β Simple linear regression Expected response at x Hypothesis testing Hypothesis testing Simple linear regression One way analysis of variance with equal numbers in each group

3 0 Introduction IB Statistics (Theorems with proof) 0 Introduction 3

4 1 Estimation IB Statistics (Theorems with proof) 1 Estimation 1.1 Estimators 1.2 Mean squared error 1.3 Sufficiency Theorem (The factorization criterion). T is sufficient for θ if and only if for some functions g and h. f X (x θ) = g(t (x), θ)h(x) Proof. We first prove the discrete case. Suppose f X (x θ) = g(t (x), θ)h(x). If T (x) = t, then f X T =t (x) = P θ(x = x, T (X) = t) P θ (T = t) = g(t (x), θ)h(x) g(t (y), θ)h(y) {y:t (y)=t} g(t, θ)h(x) = g(t, θ) h(y) = h(x) h(y) which doesn t depend on θ. So T is sufficient. The continuous case is similar. If f X (x θ) = g(t (x), θ)h(x), and T (x) = t, then f X T =t (x) = = y:t (y)=t g(t (x), θ)h(x) g(t (y), θ)h(y) dy g(t, θ)h(x) g(t, θ) h(y) dy = h(x), h(y) dy which does not depend on θ. Now suppose T is sufficient so that the conditional distribution of X T = t does not depend on θ. Then P θ (X = x) = P θ (X = x, T = T (x)) = P θ (X = x T = T (x))p θ (T = T (x)). The first factor does not depend on θ by assumption; call it h(x). Let the second factor be g(t, θ), and so we have the required factorisation. Theorem. Suppose T = T (X) is a statistic that satisfies f X (x; θ) f X (y; θ) Then T is minimal sufficient for θ. does not depend on θ if and only if T (x) = T (y). 4

5 1 Estimation IB Statistics (Theorems with proof) Proof. First we have to show sufficiency. We will use the factorization criterion to do so. Firstly, for each possible t, pick a favorite x t such that T (x t ) = t. Now let x X N and let T (x) = t. So T (x) = T (x t ). By the hypothesis, f X (x;θ) f X (x t:θ) does not depend on θ. Let this be h(x). Let g(t, θ) = f X(x t, θ). Then f X (x; θ) = f X (x t ; θ) f X(x; θ) = g(t, θ)h(x). f X (x t ; θ) So T is sufficient for θ. To show that this is minimal, suppose that S(X) is also sufficient. By the factorization criterion, there exist functions g S and h S such that Now suppose that S(x) = S(y). Then f X (x; θ) = g S (S(x), θ)h S (x). f X (x; θ) f X (y; θ) = g S(S(x), θ)h S (x) g S (S(y), θ)h S (y) = h S(x) h S (y). This means that the ratio f X(x;θ) f X (y;θ) does not depend on θ. By the hypothesis, this implies that T (x) = T (y). So we know that S(x) = S(y) implies T (x) = T (y). So T is a function of S. So T is minimal sufficient. Theorem (Rao-Blackwell Theorem). Let T be a sufficient statistic for θ and let θ be an estimator for θ with E( θ 2 ) < for all θ. Let ˆθ(x) = E[ θ(x) T (X) = T (x)]. Then for all θ, E[(ˆθ θ) 2 ] E[( θ θ) 2 ]. The inequality is strict unless θ is a function of T. Proof. By the conditional expectation formula, we have E(ˆθ) = E[E( θ T )] = E( θ). So they have the same bias. By the conditional variance formula, var( θ) = E[var( θ T )] + var[e( θ T )] = E[var( θ T )] + var(ˆθ). Hence var( θ) var(ˆθ). So mse( θ) mse(ˆθ), with equality only if var( θ T ) = Likelihood 1.5 Confidence intervals 1.6 Bayesian estimation 5

6 2 Hypothesis testing IB Statistics (Theorems with proof) 2 Hypothesis testing 2.1 Simple hypotheses Lemma (Neyman-Pearson lemma). Suppose H 0 : f = f 0, H 1 : f = f 1, where f 0 and f 1 are continuous densities that are nonzero on the same regions. Then among all tests of size less than or equal to α, the test with the largest power is the likelihood ratio test of size α. Proof. Under the likelihood ratio test, our critical region is { C = x : f } 1(x) f 0 (x) > k, where k is chosen such that α = P(reject H 0 H 0 ) = P(X C H 0 ) = C f 0(x) dx. The probability of Type II error is given by β = P(X C f 1 ) = f 1 (x) dx. Let C be the critical region of any other test with size less than or equal to α. Let α = P(X C f 0 ) and β = P(X C f 1 ). We want to show β β. We know α α, ie f 0 (x) dx f 0 (x) dx. C C Also, on C, we have f 1 (x) > kf 0 (x), while on C we have f 1 (x) kf 0 (x). So f 1 (x) dx k f 0 (x) dx C C C C f 1 (x) dx k f 0 (x) dx. C C C C Hence β β = f 1 (x) dx f 1 (x) dx C C = f 1 (x) dx + f 1 (x) dx C C C C f 1 (x) dx f 1 (x) dx C C C C = f 1 (x) dx f 1 (x) dx C C C C k f 0 (x) dx k f 0 (x) dx C C C C { } = k f 0 (x) dx + f 0 (x) dx C C C C { } k f 0 (x) dx + f 0 (x) dx C C C C = k(α α) 0. C 6

7 2 Hypothesis testing IB Statistics (Theorems with proof) C C C C C (f 1 kf 0) β /H 1 C C C (f 1 kf 0) C C α /H 0 β/h 1 α/h Composite hypotheses Theorem (Generalized likelihood ratio theorem). Suppose Θ 0 Θ 1 and Θ 1 Θ 0 = p. Let X = (X 1,, X n ) with all X i iid. Then if H 0 is true, as n, 2 log Λ X (H 0 : H 1 ) χ 2 p. If H 0 is not true, then 2 log Λ tends to be larger. We reject H 0 if 2 log Λ > c, where c = χ 2 p(α) for a test of approximately size α. 2.3 Tests of goodness-of-fit and independence Goodness-of-fit of a fully-specified null distribution Pearson s Chi-squared test Testing independence in contingency tables 2.4 Tests of homogeneity, and connections to confidence intervals Tests of homogeneity Confidence intervals and hypothesis tests Theorem. (i) Suppose that for every θ 0 Θ there is a size α test of H 0 : θ = θ 0. Denote the acceptance region by A(θ 0 ). Then the set I(X) = {θ : X A(θ)} is a 100(1 α)% confidence set for θ. (ii) Suppose I(X) is a 100(1 α)% confidence set for θ. Then A(θ 0 ) = {X : θ 0 I(X)} is an acceptance region for a size α test of H 0 : θ = θ 0. Proof. First note that θ 0 I(X) iff X A(θ 0 ). For (i), since the test is size α, we have P(accept H 0 H 0 is true) = P(X A ( θ 0 ) θ = θ 0 ) = 1 α. 7

8 2 Hypothesis testing IB Statistics (Theorems with proof) And so P(θ 0 I(X) θ = θ 0 ) = P(X A(θ 0 ) θ = θ 0 ) = 1 α. For (ii), since I(X) is a 100(1 α)% confidence set, we have P (θ 0 I(X) θ = θ 0 ) = 1 α. So P(X A(θ 0 ) θ = θ 0 ) = P(θ I(X) θ = θ 0 ) = 1 α. 2.5 Multivariate normal theory Multivariate normal distribution Proposition. (i) If X N n (µ, Σ), and A is an m n matrix, then AX N m (Aµ, AΣA T ). (ii) If X N n (0, σ 2 I), then Proof. X 2 σ 2 = XT X σ 2 = X 2 i σ 2 χ2 n. Instead of writing X 2 /σ 2 χ 2 n, we often just say X 2 σ 2 χ 2 n. (i) See example sheet 3. (ii) Immediate from definition of χ 2 n. Proposition. Let X N n (µ, Σ). We split X up into two parts: X = where X i is a n i 1 column vector and n 1 + n 2 = n. Similarly write ( ) ( ) µ1 Σ11 Σ µ =, Σ = 12, µ 2 Σ 21 Σ 22 where Σ ij is an n i n j matrix. Then (i) X i N ni (µ i, Σ ii ) (ii) X 1 and X 2 are independent iff Σ 12 = 0. Proof. (i) See example sheet 3. (ii) Note that by symmetry of Σ, Σ 12 = 0 if and only if Σ 21 = 0. From ( ), M X (t) = exp(t T µ+ 1 2 tt Σt) for each t R n. We write t = ( X1 X 2 ), Then the mgf is equal to M X (t) = exp (t T1 µ 1 + t T2 Σ 11 t tt2 Σ 22 t tt1 Σ 12 t ) tt2 Σ 21 t 1. From (i), we know that M Xi (t i ) = exp(t T i µ i tt i Σ iit i ). So M X (t) = M X1 (t 1 )M X2 (t 2 ) for all t if and only if Σ 12 = 0. ( t1 t 2 ). 8

9 2 Hypothesis testing IB Statistics (Theorems with proof) Proposition. When Σ is a positive definite, then X has pdf ( ) n 1 exp 2π f X (x; µ, Σ) = 1 Σ Normal random samples [ 1 2 (x µ)t Σ 1 (x µ) Theorem (Joint distribution of X and SXX ). Suppose X 1,, X n are iid N(µ, σ 2 ) and X = 1 n Xi, and S XX = (X i X) 2. Then (i) X N(µ, σ 2 /n) (ii) S XX /σ 2 χ 2 n 1. (iii) X and SXX are independent. Proof. We can write the joint density as X N n (µ, σ 2 I), where µ = (µ, µ,, µ). Let A be an n n orthogonal matrix with the first row all 1/ n (the other rows are not important). One possible such matrix is A = 1 n 1 n 1 n 1 1 n n n(n 1) n(n 1) n(n 1) n(n 1) Now define Y = AX. Then We have Y N n (Aµ, Aσ 2 IA T ) = N n (Aµ, σ 2 I). Aµ = ( nµ, 0,, 0) T. ]. (n 1) n(n 1) So Y 1 N( nµ, σ 2 ) and Y i N(0, σ 2 ) for i = 2,, n. Also, Y 1,, Y n are independent, since the covariance matrix is every non-diagonal term 0. But from the definition of A, we have Y 1 = 1 n n i=1 X i = n X. So n X N( nµ, σ 2 ), or X N(µ, σ 2 /n). Also Y Yn 2 = Y T Y Y1 2 = X T A T AX Y1 2 = X T X n X 2 n = Xi 2 n X 2 = i=1 n (X i X) 2 i=1 = S XX. So S XX = Y Y 2 n σ 2 χ 2 n 1. Finally, since Y 1 and Y 2,, Y n are independent, so are X and S XX. 9

10 2 Hypothesis testing IB Statistics (Theorems with proof) 2.6 Student s t-distribution Proposition. If k > 1, then E k (T ) = 0. If k > 2, then var k (T ) = k k 2. If k = 2, then var k (T ) =. In all other cases, the values are undefined. In particular, the k = 1 case, this is known as the Cauchy distribution, and has undefined mean and variance. 10

11 3 Linear models IB Statistics (Theorems with proof) 3 Linear models 3.1 Linear models Proposition. The least squares estimator satisfies 3.2 Simple linear regression X T X ˆβ = X T Y. (3) Theorem (Gauss Markov theorem). In a full rank linear model, let ˆβ be the least squares estimator of β and let β be any other unbiased estimator for β which is linear in the Y i s. Then var(t T ˆβ) var(t T β ). for all t R p. We say that ˆβ is the best linear unbiased estimator of β (BLUE). Proof. Since β is linear in the Y i s, β = AY for some p n matrix A. Since β is an unbiased estimator, we must have E[β ] = β. However, since β = AY, E[β ] = AE[Y] = AXβ. So we must have β = AXβ. Since this holds for any β, we must have AX = I p. Now cov(β ) = E[(β β)(β β) T ] Since AXβ = β, this is equal to = E[(AY β)(ay β) T ] = E[(AXβ + Aε β)(axβ + Aε β) T ] = E[Aε(Aε) T ] = A(σ 2 I)A T = σ 2 AA T. Now let β ˆβ = (A (X T X) 1 X T )Y = BY, for some B. Then BX = AX (X T X 1 )X T X = I p I p = 0. By definition, we have AY = BY + (X T X) 1 X T Y, and this is true for all Y. So A = B + (X T X) 1 X T. Hence cov(β ) = σ 2 AA T = σ 2 (B + (X T X) 1 X T )(B + (X T X) 1 X T ) T = σ 2 (BB T + (X T X) 1 ) = σ 2 BB T + cov(ˆβ). Note that in the second line, the cross-terms disappear since BX = 0. So for any t R p, we have var(t T β ) = t T cov(β )t = t T cov(ˆβ)t + t T BB T tσ 2 = var(t T ˆβ) + σ 2 B T t 2 var(t T ˆβ). 11

12 3 Linear models IB Statistics (Theorems with proof) Taking t = (0,, 1, 0,, 0) T with a 1 in the ith position, we have var( ˆβ i ) var(β i ). 3.3 Linear models with normal assumptions Proposition. Under normal assumptions the maximum likelihood estimator for a linear model is ˆβ = (X T X) 1 X T Y, which is the same as the least squares estimator. Lemma. (i) If Z N n (0, σ 2 I) and A is n n, symmetric, idempotent with rank r, then Z T AZ σ 2 χ 2 r. (ii) For a symmetric idempotent matrix A, rank(a) = tr(a). Proof. (i) Since A is idempotent, A 2 = A by definition. So eigenvalues of A are either 0 or 1 (since λx = Ax = A 2 x = λ 2 x). (ii) Since A is also symmetric, it is diagonalizable. So there exists an orthogonal Q such that Λ = Q T AQ = diag(λ 1,, λ n ) = diag(1,, 1, 0,, 0) with r copies of 1 and n r copies of 0. Let W = Q T Z. So Z = QW. Then W N n (0, σ 2 I), since cov(w) = Q T σ 2 IQ = σ 2 I. Then Z T AZ = W T Q T AQW = W T ΛW = rank(a) = rank(λ) = tr(λ) = tr(q T AQ) = tr(aq T Q) = tr A Theorem. For the normal linear model Y N n (Xβ, σ 2 I), (i) ˆβ N p (β, σ 2 (X T X) 1 ) (ii) RSS σ 2 χ 2 n p, and so ˆσ 2 σ2 n χ2 n p. (iii) ˆβ and ˆσ 2 are independent. Proof. r wi 2 χ 2 r. i=1 12

13 3 Linear models IB Statistics (Theorems with proof) We have ˆβ = (X T X) 1 X T Y. Call this CY for later use. Then ˆβ has a normal distribution with mean and covariance So (X T X) 1 X T (Xβ) = β (X T X) 1 X T (σ 2 I)[(X T X) 1 X T ] T = σ 2 (X T X) 1. ˆβ N p (β, σ 2 (X T X) 1 ). Our previous lemma says that Z T AZ σ 2 χ 2 r. So we pick our Z and A so that Z T AZ = RSS, and r, the degrees of freedom of A, is n p. Let Z = Y Xβ and A = (I n P ), where P = X(X T X) 1 X T. We first check that the conditions of the lemma hold: Since Y N n (Xβ, σ 2 I), Z = Y Xβ N n (0, σ 2 I). Since P is idempotent, I n P also is (check!). We also have rank(i n P ) = tr(i n P ) = n p. Therefore the conditions of the lemma hold. To get the final useful result, we want to show that the RSS is indeed Z T AZ. We simplify the expressions of RSS and Z T AZ and show that they are equal: Z T AZ = (Y Xβ) T (I n P )(Y Xβ) = Y T (I n P )Y. Noting the fact that (I n P )X = 0. Writing R = Y Ŷ = (I n P )Y, we have RSS = R T R = Y T (I n P )Y, using the symmetry and idempotence of I n P. Hence RSS = Z T AZ σ 2 χ 2 n p. Then Let V = ˆσ 2 = RSS n ( ) ˆβ = DY, where D = R σ2 n χ2 n p. ( C I n P Since Y is multivariate, V is multivariate with ) is a (p + n) n matrix. cov(v ) = Dσ 2 ID T ( ) = σ 2 CC T C(I n P ) T (I n P )C T (I n P )(I n P ) T ( ) = σ 2 CC T C(I n P ) (I n P )C T (I n P ) ( ) = σ 2 CC T 0 0 I n P Using C(I n P ) = 0 (since (X T X) 1 X T (I n P ) = 0 since (I n P )X = 0 check!). Hence ˆβ and R are independent since the off-diagonal covariant terms are 0. So ˆβ and RSS = R T R are independent. So ˆβ and ˆσ 2 are independent. 13

14 3 Linear models IB Statistics (Theorems with proof) 3.4 The F distribution Proposition. If X F m,n, then 1/X F n,m. 3.5 Inference for β 3.6 Simple linear regression 3.7 Expected response at x 3.8 Hypothesis testing Hypothesis testing Lemma. Suppose Z N n (0, σ 2 I n ), and A 1 and A 2 are symmetric, idempotent n n matrices with A 1 A 2 = 0 (i.e. they are orthogonal). Then Z T A 1 Z and Z T A 2 Z are independent. Proof. Let X i = A i Z, i = 1, 2 and ( ) W1 W = = W 2 Then W N 2n (( 0 0 ) ( A1 A 2 ) Z. ( )), σ 2 A1 0 0 A 2 since the off diagonal matrices are σ 2 A T 1 A 2 = A 1 A 2 = 0. So W 1 and W 2 are independent, which implies and are independent W T 1 W 1 = Z T A T 1 A 1 Z = Z T A 1 A 1 Z = Z T A 1 Z W T 2 W 2 = Z T A T 2 A 2 Z = Z T A 2 A 2 Z = Z T A 2 Z Simple linear regression One way analysis of variance with equal numbers in each group 14