Statistics 351 Probability I Fall 2006 (200630) Final Exam Solutions. θ α β Γ(α)Γ(β) (uv)α 1 (v uv) β 1 exp v }

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1 Statistics 35 Probability I Fall 6 (63 Final Exam Solutions Instructor: Michael Kozdron (a Solving for X and Y gives X UV and Y V UV, so that the Jacobian of this transformation is x x u v J y y v u v u v vu + vu v u By Theorem I, the joint density of (U, V is therefore given by v f U,V (u, v f X,Y (uv, uv uv J provided that < u < and < v < θ α β Γ(αΓ(β (uvα (v uv β exp v } v θ θ α β Γ(αΓ(β uα ( u β v α+β exp Γ(α + β Γ(αΓ(β uα ( u β v } θ θ α β Γ(α + β vα+β exp v } θ (b We recognize that the joint density for U and V can be factored as a product of the densities for U and V, respectively Thus, f U (u Γ(α + β Γ(αΓ(β uα ( u β, < u < which we recognize as the density of a Beta(α, β random variable (a We see that f X,Y (x, y for all x, y, and that Thus, f X,Y f X,Y (x, y dx dy is a legitimate density y dy dx 4x 3 dx x 4 (b We compute f X (x (c We compute f Y (y (d We compute f X,Y (x, y dy f X,Y (x, y dx y y dy 4x 3, < x < y dx y ( y, < y < f X Y y (x f X,Y (x, y f Y (y y y ( y y, y < x <

2 (e We compute f Y Xx (y f X,Y (x, y f X (x y 4x 3 3y x 3, < y < x (f We compute (g We compute E(X xf X (x dx x 4x 3 dx 4 5 E(Y X x yf Y Xx (y dy y 3y x 3 3x4 dy 4x x (h Using properties of conditional expectation (Theorem II, we compute ( 3 E(Y E( E(Y X E 4 X 3 4 E(X (i Solution : Using properties of conditional expectation (Theorem II gives ( E(XY E( E(XY X E(X E(Y X E X 3 4 X 3 4 E(X Since we conclude Solution : By definition, E(XY 3 (a Let E(X x f X (x dx xyf X,Y (x, y dx dy E(XY B ( x 4x 3 dx 4 6 3, xy y dy dx x 4 x4 dx 3 so that Y BX By Theorem V3, Y is MVN with mean ( Bµ and covariance matrix BΛB ( ( x x 5 dx 3 6 ( 3 3 y 3 dy dx

3 3 (b Note that det ( so that ( 3 3 ( 3 4 Thus, we can conclude f Y,Y (y, y π exp ( } 3 3 y + y y + 4y 4 (a We recognize f X (x, y as the density function of a multivariate normal random variable with mean ( µ and covariance matrix Λ where Inverting this matrix gives That is, Λ X N 4 (b The characteristic function of X is ϕ X (t, t exp ( / / / Λ ( 4 ((, ( 4 } (4t + 4t t + t 4 (c Recall that since X (X, Y is multivariate normal, the distribution of Y X x is normal with mean µ y + ρ σy σ x (x µ x and variance σy( ρ where ρ corr(x, Y From (a, we know that µ y µ x, σ y, σ x, and ρ cov(x,y σ xσ y Therefore, µ y + ρ σ y (x µ x + σ x (x x and σ y( ρ so that Y X x N( x, 5 By definition, f X,Y (x, y f Y Xx (yf X (x so that f X,Y (x, y π e (y x e x π π exp ( ( (y x + x } π exp ( ( x xy + y } which we recognize as the density function of a multivariate normal random variable with mean ( µ (continued

4 and covariance matrix Λ where Inverting this matrix gives That is, Thus, we conclude that Y N(, Λ Λ (X, Y N ( ( ((, ( 6 (a In order to find the eigenvalues of Λ, we must find those values of λ such that det(λ λi Therefore, ( 6 λ 5 det(λ λi (6 λ 5 λ λ+36 5 λ λ+ (λ (λ 5 6 λ so that the eigenvalues of Λ are λ and λ 6 (b Since λ, and since λ, (Λ λ I (Λ λ I ( ( we conclude that eigenvectors for λ and λ are ( v and v respectively Therefore, the diagonal matrix is ( λ D diag(λ, λ λ and the orthogonal matrix is since v v C ( v v v v ( ( ( ( ( / / / / 6 (c If Y C X, then by Theorem V3, Y is MVN with mean C µ and covariance matrix C ΛC C ΛC D using our result from (b Hence, we conclude (( ( Y N, 6 (d Since Y is multivariate normal we know from Definition I that Y and Y are each onedimensional normals We also know from Theorem V7 that the components of Y are independent if and only if they are uncorrelated From (c we know that cov(y, Y so that Y and Y are, in fact, independent

5 7 Observe that since f X,Y (x, y x for < x <, < y <, we can immediately conclude that X and Y are independent with f X (x x, < x <, and f Y (y, < y < Therefore, using the law of total probability, P (X < Y < X P (x < Y < x X xf X (x dx P (x < Y < xf X (x dx since P (x < Y < x X x P (x < Y < x by the independence of X and Y Now, P (x < Y < x x f Y (y dy x dy x x so that P (x < Y < xf X (x dx (x x f X (x dx x(x x dx (a Let U X ( X ( and V X (+X ( so that U X M and V X Solving for X ( and X ( we find X ( V U and X ( V + U The Jacobian of this transformation is Since J f X(,X ( (y, y! e y e y, < y < y <, we find from Theorem I that the joint density of (U, V is therefore given by f U,V (u, v f X(,X ( (v u, v + u J! e (v u e (v+u 4e v provided that < u < v < In other words, the joint density of the sample median X M and the sample mean X is f XM,X (u, v 4e v, < u < v < 8 (b The density of the sample median X M is given by f XM (u provided that < u < That is, X M Exp(/ 8 (c The density of the sample mean X is given by u f X (v f XM,X (u, v dv 4e v dv e v e u v provided that < v < That is, X Γ(, / u v f XM,X (u, v du 4e v du 4ve v 9 (a Recall that since X (X, Y is multivariate normal, the distribution of Y X x is normal with mean µ y +ρ σy σ x (x µ x and variance σy( ρ where ρ corr(x, Y Thus, since µ x µ y and σ x σ y, we find that ρ corr(x, Y cov(x, Y and we conclude E(Y X ρx so that cov(x, Y E(Y X cov(x, Y ρx cov(x, Y ρ cov(x, X ρ ρ var(x ρ ρ Hence, X and Y E(Y X are uncorrelated u

6 9 (b Since X (X, Y is multivariate normal, we know from Definition I that any linear combination of the components of X must be a one-dimensional normal In particular, this means that Y ρx Y E(Y X is normal Since X is also normal, and since we know from Theorem V7 that the components of a multivariate normal are uncorrelated if and only if they are independent, we conclude that X and Y E(Y X must be independent (since we showed in (a that they are uncorrelated (a Since X 4 Po(4, we find P (X 4 j 4j j! e 4, j, (b Using the definition of conditional probability and the fact that increments of the Poisson process are independent, we have P (X 4 j X 3 P (X 4 j, X 3 P (X 3 Since X 4 X 3 Po(, we find P (X 4 X 3 j, X 3 P (X 3 P (X 4 X 3 j P (X 3 P (X 3 P (X 4 X 3 j P (X 4 j X 3 P (X 4 X 3 j j (j! e e, j, (j! (c Using the definition of conditional probability and the fact that increments of the Poisson process are independent, we have P (X X 3 P (X, X 3 P (X 3 P (X 3 X, X P (X 3 P (X 3 X P (X P (X 3 Since X 3 X Po(, X 3 Po(3, and X Po(, we find P (X X 3! e! e 3! e 3 3 (d By adding and subtracting X 3, we compute cov(x 3, X 4 cov(x 3, X 4 X 3 + X 3 cov(x 3, X 4 X 3 + cov(x 3, X 3 + var(x 3 using the fact that the increments X 4 X 3 and X 3 are independent Since X 3 Po(3 we know var(x 3 3 so that cov(x 3, X 4 var(x 3 3 (e By adding and subtracting X, we compute E(X 3 X j E(X 3 X +X X j E(X 3 X X j+e(x X j E(X 3 X +j where we have used the facts that E(X 3 X X j E(X 3 X since X 3 X and X are independent, and E(X X j j by taking out what is known (See Theorems II and II Since X 3 X Po( we know E(X 3 X so that E(X 3 X j + j, j,,,

7 (a Let T 8 denote the time after waking at which Keith lights his 8th cigarette Since T 8 Γ(8, 4, we conclude E(T so that he is expected to light his 8th cigarette at noon, namely hours after : am (b The probability that he lights 3 cigarettes or more between noon and : pm is P (X 3 X 3 P (X 3 X < 3 since X 3 X Po(4 P (X 3 X P (X 3 X P (X 3 X 4! e 4 4! e 4 4 3e 4! e 4