Stat 700 HW2 Solutions, 9/25/09

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1 Stat 700 HW2 Solutions, 9/25/09 (1). By the spectral theorem, B = k λ j v j v j, where v j are an orthonormal basis of eigenvectors of B with corresponding eigenvalues λ j. Now, since λ j v j = Bv j = B 2 v j = B(λ j v j ) = λ 2 j v j and λ j 0, it follows that λ j = 0 or 1 for all j = 1,..., k. Moreover, r = rank(b) is the number of nonzero eigenvalues λ j, and after renumbering if necessary, there is no loss in generality in assuming λ 1 = = λ r = 1, while λ r+1 = = λ k = 0. Now let Z be a k-vector defined by Z j = v jw for j = 1,..., k, and observe that Z is a linear transformation of W and therefore multivariate normal, but var(z j ) = v jbv j = 0 for j > r, and cov(z i, Z j ) = v ibv j = δ ij for i, j = 1,..., k. Thus, k r r W W = W v j v jw = W v j v jw = Zj 2 χ 2 r since we have seen that Z j, 1 j r, are iid N (0, 1). (2). As in class, Z A 1/2 Y N (0, I n n ), so that R 2 = Z Z χ 2 n, so by univariate change-of-variable, f R (r) = 2 (n 2)/2 r n 1 /2 Γ(n/2) e r2 r > 0 Now it is easy to see that Z/R = Z/(Z Z) 1/2 is independent of R and uniformly distributed on the surface of the unit n-sphere {z R n : z z = 1}, since for any orthogonal matrix U, (Z/R, R) has the same distribution as (UZ/(Z U UZ) 1/2, (Z U UZ) 1/2 ) = (UZ/R, R). This argument can be used to give a rigorous proof both of independence and uniformity of Z/R on the surface of the unit sphere, because orthogonal matrices are transitive on the surface of the unit sphere, i.e. one can be found to carry any unit vector v into any other fixed vector w. For a complete proof in this problem, we need to find a n 1-dimensional coordinate system for the unit sphere so that there will be a density and so that we can use the Jacobian change-of-variable formula and also give a density for the image A 1/2 Z/R = Y/R of a uniform random unit-vector in R n under the linear operator A. Define: 1

2 ω j z j = arccos( zj ), 1 j n 2 z2 n { arccos(zn 1 / zn zn) 2 if z n > 0 ω n 1 = 2π arccos(z n 1 / zn zn) 2 if z n 0 Then the inverse mapping from the spherical coordinates (r, ω 1,..., ω n 1 ) back to z is given by z j = { r j 1 i=1 sin(ω i) cos(ω j ) if 1 j n 1 r n i=1 sin(ω i) if j = n It is easy to see that the Jacobian matrix for this inverse mapping involves r not at all in the first row and as a multiplicative factor in each of the other rows, so that the corresponsing absolute Jacobian determinant is r n 1 K(ω) for a function K of ω = (ω 1,..., ω n 1 ). It follows immediately by the change-of-variable formula for densities that f R,ω (r, ω) = C r n 1 e r2 /2 K(ω) which immediately provides the independence of R and ω and (after adjustment of constants) the separate densities of R and ω. (3). Bickel-Doksum, # (a). This one is obviously not identifiable, because the multivariate normal data with mean-vector (µ + α 1,..., µ + α p ) and variance = σ 2 I p p is left unchanged if µ is replaced by µ + c and at the same time all α j are replaced by α j c, where c is any real number. (b). Now we have (only) to show that with α restricted so that p α j = 0, the mapping from ϑ to (ξ 1,..., ξ p+1 ) (µ + α 1,..., µ + α p, σ 2 ) is 1-to-1. But if (µ + α 1,..., µ + α p are given, then under this side-condition it follows that the sum of these p quantities is pµ. Thus, when ϑ satisfies the side-condition, the proof of 1-to-1 is completed by the formulas: µ = (ξ ξ p )/p ξ, α j = ξ j ξ, σ 2 = ξ p+1 (c). Since we observe only Y X N (µ 1 µ 2, σ 2 ), there is no way to tell ϑ = (µ 1 + c, µ 2 + c) from ϑ = (µ 1, µ 2 ) even if σ 2 is known. (d). This one also is nonidentifiable, since the parameters θ and θ = 2

3 (α + c1 p, λ c1 b, ν, σ 2 ) give rise to exactly the same data-distribution. (e). But here the parameter is identifiable, since p b b p ν = E( X ij ), ν + α i = E( X ij ), ν + λ j = E( X ij ) i=1 (4). Bickel-Doksum, # (a) Unif(0, ϑ) cannot be regular, since the set where the density is positive is (0, ϑ). (b) The set where the probability mass function is positive is {1, 2,..., ϑ}, which depends on ϑ, and therefore the model is not regular. (c) This model, called truncated normal, is not regular because it is of mixed type, neither continuous nor discrete. (d) The shifted discrete values fall in the set {.1 + ϑ,.2 + ϑ,...,.9 + ϑ} where the probability mass function is positive; again not regular. (5). Bickel-Doksum, # (a) The model is X N (µ, σ 2 ) F, Y 2µ+ X N (2µ+ µ, σ 2 ) = N (µ+, σ 2 ) G( ) F ( ). (b) If F ( ) F 0 ( µ), with F 0 continuous but not normal, and we define Y = 2µ+ X, then P (Y y) = P (X 2µ+ y) = 1 F 0 ( (y µ)) which implies that if F 0 is symmetric about 0 (so that X is symmetric about its mean µ), and G = F Y = F X ( ). P (Y µ z) = 1 F 0 ( z) = F 0 (z ) (c) Assume that Y = X + δ(x) for some function δ( ) which is continuous and strictly increasing, and also that G(y) = F Y (y) = F X (y ). Let H( ) be defined as the unique inverse of the function x + δ(x), i.e., let x = H(z) be the solution to x + δ(x) = z. Then H is strictly increasing, and F (y ) = G(y) = P (X + δ(x) y) = P (X H(y)) = F (H(y)) implies that H(y) y for all points y which are points of increase of F (defined as points u such that for all ɛ > 0, F (u + ɛ) > F (u ɛ)), i.e., for all such y, y + δ(y ) = y, or δ(y ) =. We conclude δ( ) if F has no flat places, i.e., if every point is a point of increase. (6). Bickel-Doksum, # Here the parameters are the probability vectors (p(j), 0 j N) and (r(j), 0 j N), and the probability law immediately determines the family of joint probability masses π 1 (j) = P (T = j C) = P (Y = j, T C) 3 i=1

4 π 0 (j) = P (C = j T ) = P (Y = j, C < T ) where Y = min(t, C), and note also that P (Y = j) = π 1 (j) + π 0 (j). Our job is to show that these functions π i, π 0 uniquely determine the vectors p, r. Note first that P (Y j) = P (T j, C j) = p(k) r(k) P (Y = j, T C) = P (T = j, C = k) = p(j) Putting these facts together gives r(k) P (Y = j, T C Y j) = P (Y = j, T C) P (Y j) = p(j) N r(k) N p(k) N r(k) / N = p(j) p(k) Thus the functions π 1, π 0 uniquely determine G(j) = p(j)/ N p(k) from which we find p(0), p(1),..., p(n) by induction. From this and the formula for P (Y j) we find also (for all j) that N r(j) is uniquely determined, and therefore r(j) is also. This completes the proof of identifiability. (7). Bickel-Doksum, # The posterior probability mass function for ϑ given X = k is, for each ϑ = j/4, j = 1, 2, 3, (1/3) ϑ (1 ϑ) k 3 (1/3) (j/4) (1 = ϑ (1 ϑ) k 3 j/4)k (j/4) (1 j/4)k which for k = 2 has the three values (9/20, 8/20, 3/20). (b) The most probable value for k = 2 is ϑ = 1/4. For general k, the most probable value is 3/4 if k = 0, 1/2 if k = 1, and 1/4 if k 2. (c) When the prior is β(r, s), the posterior (now a continuous density) is easily checked to be proportional to t r 1 (1 t) s 1 t (1 t) k, i.e., is β(r + 1, s + k). 4

5 (8) Optional Problem. > v1 = c(1,-1,0)/sqrt(2) > v2 = c(1,1,1)/sqrt(3) orthonormal pair of vectors > sigma = 8*outer(v1,v1)+3*outer(v2,v2) [,1] [,2] [,3] [1,] [2,] [3,] So Sigma^(1/2) = > M = sqrt(8)*outer(v1,v1)+sqrt(3)*outer(v2,v2) [,1] [,2] [,3] [1,] [2,] [3,] The hyperplane of values of Y is the set of vectors t(c((1,2,3)) + column space of R3 = (1,2,3) + a*(1,-1,0) + b*(1,1,1) a,b any real values = (1+a+b, 2-a+b, 3+b) and we want to know (with Z1s = W1 and Z2s=W2 indep N(0,1)) P((1,2,3) + (1,-1,0)*W1*sqrt(8/2) + (1,1,1)*W2*sqrt(3/3) > 0) Inequalities are: 1+2*W1+W2 > 0, 2-2*W1+W2 > 0, 3+W2 > 0 and the last one is implies by the first two, so the probability is P(-1-W2 < 2*W1 < 2+W2 ) = Numerical bivariate ntegral > integrate(function(w) dnorm(w)*(pnorm(1+w/2)-pnorm(-(1+w)/2)), -1.5,Inf)$val ### -1.5 because want 2+W2 > -1-W2 [1] > zmat = mvrnorm(10000,1:3,sigma) Check by simulation > dim(zmat) [1] > mean(apply(zmat>0,1,prod)) [1] ### OK!! 5