STAT 418HW12 SolutionsSections 6.2-6.515, 20, 21, 23, 27, 28, 32, 33; TE 915.a. Let A denote the area of region R, then A =R R(x,y)∈Rdydx. Because f(x, y) is a joint pdf,1 =Z Z(x,y)∈Rf(x, y)dydx =Z Z(x,y)∈Rcdydx = cZ Z(x,y)∈Rdydx = cA.Therefore c = 1/A.b.f(x, y) =14, x ∈ (−1, 1), y ∈ (−1, 1)fX(x) =Z1−114dy =14+14=12, x ∈ (−1, 1), X ∼ U (−1, 1)fY(y) =Z1−114dx =14+14=12, y ∈ (−1, 1), Y ∼ U(−1, 1)Because the joint density is the product of the marginal densities, f(x, y) = fX(x)fY(y),X and Y are independent.c.P (X2+ Y2≤ 1) =Z Zx2+y2≤114dydx=14Z Zx2+y2≤1dydx=π124=π4.20.a. Yes.f(x, y) = fX(x)fY(y) = xe−xe−y, x > 0, y > 0.b. No.fX(x) =Z1xf(x, y)dy = 2(1 − x), 0 < x < 1fY(y) =Zy0f(x, y)dx = 2y, 0 < y < 1.121.a. f(x, y) ≥ 0 for all x, y.Z∞−∞f(x, y)dxdy =Z10Z1−y024xy dx dy=Z1012y(1 − y)2dy=Z1012(y − 2y2+ y3)dy= 12(1/2 − 2/3 + 1/4) = 1.b.E(X) =Z10xfX(x)dx=Z10xZ1−x024xy dy dx=Z1012x2(1 − x)2dx = 2/5.c.E(Y ) =Z10yfY(y)dy=Z10yZ1−y024xy dx dy=Z1012y2(1 − y)2dy = 2/5.23.a. Yes.f(x, y) = 6x(1 − x) · 2y = fX(x) · fY(y), 0 < x < 1, 0 < y < 1.b.E(X) =Z106x2(1 − x)dx = 1/2.c.E(Y ) =Z102y2dy = 2/3.d.V ar(X) =Z106x3(1 − x)dx − 1/4 = 1/20.e.V ar(Y ) =Z102y3dy − 4/9 = 1/18.227. Given: The joint pdf of (X, Y ) isf(x, y) = e−y, x ∈ (0, 1), y > 0.a. Let W = X + Y and V = Y . Then X = W − V and Y = V. The Jacobian of thistransformation isabs1 −10 1= 1.Therefore the joint pdf of (W, V ) isf(w, v) = e−v, 0 < w − v < 1, v > 0 ⇒ 0 < w − 1 < v < w.Therefore the marginal pdf of W isfW(w) =Rw0e−vdv, 0 ≤ w ≤ 1Rww−1e−vdv, w > 1=1 − e−w, 0 ≤ w ≤ 1e1−w− e−w, w > 1b. Let W = X/Y and V = Y . Then X = W V and Y = V. The Jacobian of thistransformation isabsv w0 1= v.Therefore the joint pdf of (W, V ) isf(w, v) = ve−v, 0 < wv < 1, v > 0 ⇒ 0 < v < 1/w.Therefore the marginal pdf of W isfW(w) =Z1/w0ve−vdv, w > 0= 1 − e−1/w/w − e−1/w, w > 028.a. Given: The joint pdf of (X1, X2) isf(x1, x2) = λ1λ2e−λ1x1−λ2x2, x1> 0, x2> 0.Let Z = X1/X2and V = X2. Then X1= ZV and X2= V. The Jacobian of thistransformation isabsv z0 1= v.Therefore the joint pdf of (Z, V ) isf(z, v) = λ1λ2ve−v(λ1z+λ2), z > 0, v > 0.Therefore the marginal pdf of Z isfZ(z) =Z∞0λ1λ2ve−v(λ1z+λ2)dv= λ1λ2(λ1z + λ2)−2, z > 0.3b.P (X1< X2) = P (Z < 1)=Z10λ1λ2(λ1z + λ2)−2dzLet u = λ1z + λ2.= λ2Zλ1+λ2λ21u2du= λ2−1uλ1+λ2λ2= 1 −λ2λ1+ λ2=λ1λ1+ λ2.32. Assume that weekly sales are independent.a. Let Z ∼ N (0, 1) and W = X1+ X2denote weekly sales over the next 2 weeks, thenW ∼ N(4400, 2302+ 2302).P (W > 5000) = PW − 4400√105800>5000 − 4400√105800= P (Z > 600/325.27) = P (Z > 1.84) = .0326.b. Let X denote the sales in a week.P (X > 2000) = PX − 2200230>2000 − 2200230= P (Z > −.87) = .8078.Let p = 0.8078. The probability that weekly sales exceeds 2000 in at least 2 of the next3 weeks is P(X > 2000 in all 3 weeks ∪ X > 2000 in exactly 2 of the next weeks)= p3+32p2(1 − p).33. Let Z ∼ N (0, 1), X denote Jill’s score, and Y denote Jack’s score.a.P (X < Y ) = P ((X − Y ) < 0)= PX − Y − (170 − 160)√202+ 152<−(170 − 160)√202+ 152= P (Z < −10/25) = P (Z < −.4) = .3446.b.P (X + Y > 350) = PX + Y − (170 + 160)√202+ 152>350 − (170 + 160)√202+ 152= P (Z > 20/25) = P (Z > .8) = .2119.4TE9. X1, . . . , Xnare independent random variables, each exponentially distributed withparameter λ. Let Y = min(X1, . . . , Xn), and let G(y) denote the CDF of Y . Then, for y > 0,G(y) = P (Y ≤ y)= P (min(X1, . . . , Xn) ≤ y)= 1 − P (min(X1, . . . , Xn) > y)= 1 − P (X1> y ∩ X2> y ∩ . . . ∩ Xn> y)= 1 −nYi=1P (Xi> y), by independence= 1 −nYi=1e−λy, because P (Xi> y) =R∞yλe−λxdx = e−λyfor all i= 1 − e−nλy,and G(y) = 0 for y ≤ 0. The density function of Y , g(y), is thereforeddyG(y) =nλe−nλy, y > 00, y ≤