hw10solns

# hw10solns - Solutions to Homework 10 6.1 a. LetZ denote the...

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Unformatted text preview: Solutions to Homework 10 6.1 a. LetZ denote the number of contracts assigned to firm 111, then P(X =O,Y :0) = P(X =0,Y =0,Z = 2) = P(contracts 1 assigned to 111) - P(contracts 2 assigned to HI) 1 1 '3 5 l 9 And P(X =1,Y =0) = P(X =1,Y =0,Z =1) 2 P(contracts l assigned to I) - P(contracts 2 assigned to ID) + P(contracts 1 assigned to H1) ~ P(contracts 2 assigned to 1) Similarly, we can derive the result in the table below b. the marginal distribution forY is 2 4 P(Y =0)=ZP(X =i,Y :0) =3 [:0 2 4 P(Y =1) =ZP(X =i,Y :1):6 i=0 P(Y=2)=ZZ:P(X=i,Y=2)=% i=0 P(X=],Y=1)_ P(X=1lY,=1)= P(Y_1) = 0.5 \Dl-P-IOIN 0.4 0.2 P(X s 0.2,Y s 0.4) = j I f(x, y)dxdy "U3 —00 0.4 0.2 = j jldxdy 0 0 20.08 00 0.3 P(0.1 s X s 0.3,Y > 0.4) = j ] f(x, y)dxdy 0.4 OJ 1 0.3 =Hldxdy 0.4 (H =0.12 6.11 a. forOs x S 1, we have fﬂm=jﬂnw® = :[ldy 0 =1 Otherwise, fX (x) = 0 . here X ~ uniform[0,1] b. forOS ysl, we have fﬂw=Iﬂwa 1 = jldx 0 =1 Otherwise, fy(y)=0. hereY ~ uniform[0,1] 6.21 a. f” (x, y) f (xly)= XIY _ 1, OSxSI,OSySI — 0, otherwise c. from 6.11, we have ) LOSxSl fx (x a 0,0therwise f( )_ LOSysl Y y _ 0,0therwise then f ( )f( )_ 1,03xsl,03ysl X x Y y — 0, otherwise Thus fx (x)fy(y) = fx,y(x, y) X andY are independent. (1. when Y = 0.4 , the conditional density function becomes 1, 03x31 0, otherwise ley(x| y =0.4) ={ Then P(X 2 0.8 I Y = 0.4) = I f“, (x! y = 0.4)dx —lw = I ldx 0.8 = 0.2 6.26 a. Deﬁne set A = {(x, y) :0 S y S x < oo,x < 2, y >1}, A is the shadow area in the figure below, forafixedX =xe(l,2), 1< ny, then A can be equivalently written as A={(x,y):l<x<2,] <ny} The probability of interest is P(X < 2,Y >1) = Hf(x, y)dxdy A Te‘xdydx l 2 l 2 ﬁx — l)e"‘dx l e"1 — 2e"2 0.097 b. Define set B = {(x,y):0\$ y S x < oo,x2 2y} ,3 is the shadow area in the ﬁgure below, forafixedX =xe[0,oo), §Sny, then B can be equivalently written as B={(x,y):0\$x<w,§\$y£x} The probability of interest is P(X 2 2Y) = “foe, y)dxdy B e'xdydx /2 X e‘xdx i 2 oo 0 w 0 l l 2 c. For a fixed X = x e [0,oo),0 S y S x , then the marginal density function for the total waiting time X is no) = jf(x,y)dy =xe xe[0,oo) d. For a fixedY = y e [0, 00), y S x < oo , then the marginal density function forY is my) = Jf(x,y)dx = o]e""dx —V =e', ye[0,°0) the conditional density function of X given Y = y is _ fx_y(x,y) “ my) _ ey—x, 0Sny<oo _ {0, otherwise fXIy(XI y) 147 P<X)_—71___} y>§ 5' = 1“ PCXé=21-_-I 7);?) .1 P[7>.5_ -: [L— S cl“ 0-2.5 ZSKO'ZKO‘J. .F i" \$.03 “=3 go-S. x): Ae-Xl‘hDo-zsC T/ WIL/ P(X?’%)y=025> = H ...
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## This note was uploaded on 06/05/2011 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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hw10solns - Solutions to Homework 10 6.1 a. LetZ denote the...

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