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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 \DlPIOIN 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“ 02.5 ZSKO'ZKO‘J. .F i" $.03
“=3 goS. x):
AeXl‘hDozsC
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.
 Spring '08
 Staff
 Probability

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