STAT 418HW11 SolutionsSections 6.1-6.21, 3, 8, 9, 10; TE 11.Outcome {x, y} (a) f(x,y) Outcome {x, y} (a) f(x,y){1,2} P({1,1}) = 1/36 {5,7} P({5,2} ∪ {2,5}) = 2/36{2,3} P({1,2} ∪ {2,1}) = 2/36 {5,8} P({5,3} ∪ {3,5}) = 2/36{2,4} P({2,2}) = 1/36 {5,9} P({5,4} ∪ {4,5}) = 2/36{3,4} P({3,1} ∪ {1,3}) = 2/36 {5,10} P({5,5}) = 1/36{3,5} P({3,2} ∪ {2,3}) = 2/36 {6,7} P({6,1} ∪ {1,6}) = 2/36{3,6} P({3,3}) = 1/36 {6,8} P({6,2} ∪ {2,6}) = 2/36{4,5} P({4,1} ∪ {1,4}) = 2/36 {6,9} P({6,3} ∪ {3,6}) = 2/36{4,6} P({4,2} ∪ {2,4}) = 2/36 {6,10} P({6,4} ∪ {4,6}) = 2/36{4,7} P({4,3} ∪ {3,4}) = 2/36 {6,11} P({6,5} ∪ {5,6}) = 2/36{4,8} P({4,4}) = 1/36 {6,12} P({6,6}) = 1/36{5,6} P({5,1} ∪ {1,5}) = 2/36Outcome {x, y} (b) f(x,y) (c) f(x,y){1,1} P({1,1}) = 1/36 P({1,1}) = 1/36{1,2} P({1,2}) = 1/36 P({1,2} ∪ {2,1}) = 2/36{1,3} P({1,3}) = 1/36 P({1,3} ∪ {3,1}) = 2/36{1,4} P({1,4}) = 1/36 P({1,4} ∪ {4,1}) = 2/36{1,5} P({1,5}) = 1/36 P({1,5} ∪ {4,1}) = 2/36{1,6} P({1,6}) = 1/36 P({1,6} ∪ {4,1}) = 2/36{2,2} P({2,2} ∪ P({2,1}) = 2/36 P({2,2}) = 1/36{2,3} P({2,3}) = 1/36 P({2,3} ∪ {3,2}) = 2/36{2,4} P({2,4}) = 1/36 P({2,4} ∪ {4,2}) = 2/36{2,5} P({2,5}) = 1/36 P({2,5} ∪ {5,2}) = 2/36{2,6} P({2,6}) = 1/36 P({2,6} ∪ {6,2}) = 2/36{3,3} P({3,3} ∪ {3,2} ∪ {3,1}) = 3/36 P({3,3}) = 1/36{3,4} P({3,4}) = 1/36 P({3,4} ∪ {4,3}) = 2/36{3,5} P({3,5}) = 1/36 P({3,5} ∪ {5,3}) = 2/36{3,6} P({3,6}) = 1/36 P({3,6} ∪ {6,3}) = 2/36{4,4} P({4,4} ∪ {4,3} ∪ {4,2} ∪ {4,1}) = 4/36 P({4,4}) = 1/36{4,5} P({4,5}) = 1/36 P({4,5} ∪ {5,4}) = 2/36{4,6} P({4,6}) = 1/36 P({4,6} ∪ {6,4}) = 2/36{5,5} P({5,5} ∪ {5,4} ∪ {5,3} ∪ {5,2} ∪ {5,1}) = 5/36 P({5,5}) = 1/36{5,6} P({5,6}) = 1/36 P({1,6} ∪ {6,5}) = 2/36{6,6} P({6,6} ∪ {6,5} ∪ {6,4} ∪ {6,3} ∪ {6,2} ∪ {6,1}) = 6/36 P({6,6}) = 1/3613.a.{y1, y2} f(y1, y2){0,0}0@1131A0@1331A=1526{0,1}0@111A0@1121A0@1331A=526{1,0}0@111A0@1121A0@1331A=526{1,1}0@221A0@1111A0@1331A=126b.{y1, y2, y3} f(y1, y2, y3) {y1, y2, y3} f(y1, y2, y3){0,0,0}0@1031A0@1331A=60143{0,1,1}0@221A0@1011A0@1331A=5143{0,0,1}0@111A0@1021A0@1331A=45286{1,0,1}0@221A0@1011A0@1331A=5143{0,1,0}0@111A0@1021A0@1331A=45286{1,1,0}0@221A0@1011A0@1331A=5143{1,0,0}0@111A0@1021A0@1331A=45286{1,1,1}0@331A0@1331A=12868.a.fY(y) = cZy−y(y2− x2)e−ydx =43cy3e−y, 0 < y < ∞Z∞0fY(y)dy = 1 ⇒43cΓ(4) =43c3! = 1 ⇒ c =18.2b.fY(y) =43cy3e−y=16y3e−y, 0 < y < ∞fX(x) =18Z∞|x|(y2− x2)e−ydy =18Z∞|x|y2e−ydy −18Z∞|x|x2e−ydy =14e−|x|(1 + |x|), −∞ < x < ∞becauseZ∞|x|y2e−ydy = −y2e−y+ 2Z∞|x|ye−ydy = −y2e−y+ 2(−ye−y+Z∞|x|e−ydy) = −y2e−y− 2ye−y− 2e−yc.E[X] =Z∞−∞xf(x)dx =Z∞−∞x14e−|x|(1 + |x|)dx = 0,because x14e−|x|(1 + |x|) is an odd function.9.a. It is a joint density function.f(x, y) =67x2+xy2> 0, 0 < x < 1, 0 < y < 2Z20Z1067x2+xy2dxdy =Z206713+y4dy =67y3+y28|20= 1.b.fX(x) =67Z20x2+xy2dy =67(2x2+ x), 0 < x < 1.c.P (X > Y ) =67Z10Zx0x2+xy2dydx =1556.d.P (Y > .5|X < .5) =P (Y > .5, X < .5)P (X < .5)=R2.5R.50x2+xy2dxdyR.50(2x2+ x)dx= .8625.e.E(X) =Z1067x2x2+ xdx =67x42+x33|10=57.f.E(Y ) =Z20Z1067yx2+xy2dxdy =Z2067y13+y4dy =87.310.a.P (X < Y ) =Z∞0Zy0e−(x+y)dxdy=Z∞0e−y[−e−x|y0]dy=Z∞0e−y(1 − e−y)dy= [−e−y+12e−2y]|∞0= 0 + 1 + 0 −12=12.b.P (X < a) =Za0e−xdx= −e−x|a0= 1 − e−a.TE1.P (X ≤ a2, Y ≤ b2) = P (a1< X ≤ a2, b1< Y ≤ b2)+ P (X ≤ a1, b1< Y ≤ b2)+ P (a1< X ≤ a2, Y ≤ b1)+ P (X ≤ a1, Y ≤ b1).The right hand side is the union of 4 mutually exclusive events. Also,P (X ≤ a1, Y ≤ b2) = P (X ≤ a1, b1< Y ≤ b2) + P (X ≤ a1, Y ≤ b1),and similarly,P (X ≤ a2, Y ≤ b1) = P (a1≤ X ≤ a2, Y ≤ b1) + P (X ≤ a1, Y ≤ b1).Therefore,F (a2, b2) = P (a1< X ≤ a2, b1< Y ≤ b2) + F (a1, b2) − F (a1, b1) + F (a2, b1) − F (a1, b1) + F (a1, b1)P (a1< X ≤ a2, b1< Y ≤ b2) = F (a2, b2) + F (a1, b1) − F (a1, b2) − F (a2,