14.30 Exam 2 Solutions

14.30 Exam 2 Solutions RR 1 A: a) E(aX + bY + c) = aE(X) + bE(Y ) + c Z Z RR RR (aX + bY + c)fX,Y (x, y)dxdy = aXfX,Y (x, y) + bY fX,Y (x, y) + cfX,Y ¤ £R ¤ R (x,£Ry) R R = aX fX,Y (x, y)dy dx + bY fX,Y (x, y)dx dy + c = aXfX (x) + R bY fY (y)dy + c = aE(X) + bE(Y ) + c Q.E.D. b) ¤ RE[E(aY | X)] = aE(Y ) R £R | X)fX (x)dx = ayf RE(aY R R (y£R| x)dy fX (x)dx ¤ a R Y£Rf (y | x)fX (x)dydx = aR Y f (y | x)fX (x)dx dy ¤ a Y fX,Y (x, y)dx dy = a Y fY (y)dy = aE(Y ) Q.E.D. c) First lets show: Cov(X + Y, X − Y ) = V arX − V arY Cov(X + Y, X − Y ) = E [(X + Y )(X − Y )] − E(X + Y )E(X − Y ) EX 2 − EY 2 − E 2 (X) + E 2 (Y ) = EX 2 − E 2 (X) − [EY 2 − E 2 (Y )] = V arX − V arY now, Corr(X + Y, X − Y ) = √ Cov(X+Y,X−Y ) V ar(X+Y ).V ar(X−Y ) we p look at the denominator: p solved the numerator, now lets V ar(X + Y ).V ar(X − Y ) = [V arX q+ V arY + 2Cov(X, Y )] . [V arX + V arY − 2Cov(X, Y )] √ 2 2 2 V ar X + V ar Y + 2V arX.V arY = (V arX + V arY ) = V arX + V arY Q.E.D. 1 B: V arX = EX 2 − E 2 X EX = 0.0.4 + 1.0.6 = 0.6 EX 2 = 02 .0.4 + 12 .0.6 = 0.6 V arX = 0.6 − 0.36 = 0.24 1 Similarly, EY = 0.6P E(Z) = E(XY ) = xyf (x, y) = 0.25 Cov(X, Y ) = E(XY ) − EXEY = 0.25 − 0.36 = −0.11 2 a: X ∼ U (1, 3) 1 fX (x) = 3−1 = 12 Y = −α ln(3X) Now in order to apply the 1-step method, we need to first confirm that this function is monotonic: dY α for 1 < X < 3 dX = − 3X < 0 which is monotone decreasing function in the range of X. Range of y: 1< X < 3 −α ln 9 < y < −α ln 3 where: Y = −α ln(3X) Y X = 13 e− α now 2-step method would be: R 1 e− Yα 1 FY (y) = 1 − FX (x) = 1 − 13 2 dx Y Y = 1 − 16 e− α + 12 = 32 − 16 e− α ¾ ½ 1 −Y α for − α ln 9 < y < −α ln 3 dFY (y) 6α e = f (y) = Y dy 0 otherwise and¡ 1-step ¢method would be: 1 −1 ¯fX −1r ¯(y) = 2 ¯ dr (y) ¯ 1 −Y e α ¯ dy ¯ = 3α ¾ ½ 1 −Y ¢ ¯¯ −1 ¯¯ ¡ α for − α ln 9 < y < −α ln 3 6α e fY (y) = fX r−1 (y) . ¯ dr dy(y) ¯ = 0 otherwise 2 b: x f (x) = β1 e− β for iid X1 and X2 y = max{aX1 , X2 + c} P (Y ≤ y) = P (aX1 ≤ y).P (X2 + c ≤ y) y P (X (Xi2 h≤ y − c) hR1y ≤ a ).P R y−c 1 − x i x − 1 β dx = 0a β e β dx . 0 βe y y−c y−c y y(a+1)−ac = FY (y) (1 − e− aβ ).(1 − e− β ) = 1 − e− β − e− aβ + e− aβ ) ( y(a+1)−ac − − y−c − y aβ β aβ (a+1)e e e Y (y) + aβ − if y (c, ∞) = fY (y) = dFdy β aβ 0 otherwise 2 3 a: XBoston ∼ N (65, 9) XSantiago ∼ N (60, 4) P (XB < 66.5) = P (µ + σZ < 66.5) where Z ∼ N (0, 1) = P (Z < 66.5−65 ) 3 = P (Z < 0.5) = 0.6915 So Probability randomly chosen woman is taller than Alice = (1 − 0.6915) = 0.3085 Number of women in Boston taller than Alice: 0.3085 ∗ 2, 000, 000 = 617, 000 3 b: Distribution of "sum of independent normally distributed variables" is also normal:¡ ¢ 1 2 so, XB + XB + XS1 + XS2 + XS3 ∼ N (µsum , σ 2sum ) ∼ N (2µB + 3µS , 2σ2B + 3σ 2B ) = N (310, 30) 3 c: we want: P (−1 < X − µ < 1) where X is the average height and µ is the population mean = P ( −1 < X−µ < √12 ) √2 √2 n √ n √ n = P ( − 2 n < Z < 2n ) √ = 2P (Z < 2n ) − 1 We want at√least 95% probability: = 2P (Z < 2n ) − 1 ≥ 0.95 √ P (Z < 2n ) ≥ 0.975 √ n 2 ≥ 1.96 n ≥ 16 4 a: X ∼ N (50, 100) P (40 < X < 60) = P ( 40−50 < Z < 60−50 10 10 ) = P (−1 < Z < 1) = 2P (Z < 1) − 1 = 2(0.8413) − 1 = 0.6826 4 b: 3 New Technology results in: X ∼ N (50, σ 2 ), where is σ2 unknown We know: P (40 < X < 60) = 0.95 P ( 40−50 < X−50 < 60−50 σ σ σ ) = 0.95 −10 10 P ( σ < Z < σ ) = 0.95 2P (Z < 10 σ ) − 1 = 0.95 P (Z < 10 σ ) = 0.975 10 = 1.96 σ σ = 5.102 so change in standard deviation is: 10 − 5.102 = 4.898 4 c: Notice that this question points you towards another familiar distribution, that is the binomial distribution. We are given the number of observations,and we calculated the probability of success in part a) of this question. So we need to plug in: n = 555; p = 0.6826; and E(Y ) = np; V ar(Y ) = np(1 − p); so: E(Y ) = 378.843; V ar(Y ) = 120.245 where Y is represented by the binomial distribution. Define event M =255 cathodes out of 555 satisfy customer’s specifications, then: µ ¶ µ ¶ n 555 m n−m P (M = m | n, p) = p (1−p) = 0.6826255 (1−0.6826)555−255 m 255 4 d: A copper cathode L has a price of 32 L2 cents. R 3of law 2 Expected price: 2 L fX (x)dx £ ¤ R = 32 L2 fX (x)dx = 32 E(X 2 ) = 32 V ar(X) + E 2 (X) = 32 [100 + 2500] = 3900 = $39.00 4