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Unformatted text preview: ECE 310, Spring 2005, HW 10 solutions
Problem E9.1 1 EX =
0 x fX (x) dx =
0 1 0 x 2x dx =
1 2x3 2 = 3 0 3 1 E(X 2 ) = Therefore, x2 2x dx = 1 2x4 = 4 0 2 1 1 4  = 2 9 18 VAR(X) = E(X 2 )  (E X) = Problem E9.2 2 EX =
0 xfX (x) dx =
1  2 2 dx =  2 x x =2
1 E(X 2 ) = Therefore, Problem E9.3 x2 fX (x) dx =
1 2 dx = 2 ln x x =
1 VAR(X) = E(X 2 )  (E X)2 = i=2 EX =
i=0 i pi = 0 0.2 + 1 0.5 + 2 0.3 = 1.1
2 VAR(X) = E (XE X)2 =
i=0 pi (iE X)2 = .2(1.1)2 +.5(0.1)2 +.3(0.9)2 = .49 Problem E9.13 Using properties (E4) and (E5): 1 1 1 E( X + Y ) = E X + E Y = 2 2 2 Problem E9.15
1 1 x dx + 2 = 2.25
0 pX (x) =
y=0 pX,Y (x, y) = pX,Y (x, y = 0) + pX,Y (x, y = 1)
1 1 E(XY ) =
x=0 y=0 xypX,Y (x, y) = 1 0.4 = 0.4 Problem E9.21 COV(aX + b,cY + d) = E (aX + b)(cY + d)  E(aX + b)E(cY + d) = E acXY + adX + bcY + bd  (acE(X)E(Y ) + adE(X) + bcE(Y ) + bd) = acE(XY )  acE(X)E(Y ) = acCOV(X, Y ) 1 Problem E9.22 a)
=0 E(P ) = E(V I) = E(V 2 /R) = b) E(V 2 ) VAR(V )  (E(V ))2 = = 4k T W R R E(V1 + V2 )2 = E(V1 + V2 + 2V1 V2 ) = E(V12 ) + E(V22 ) + 2E(V1 V2 )=0
2 2 = E(V12 ) + E(V22 ) + 2(COV(V1 , V2 )  E(V1 ) E(V2 ) ) = 1 + 2 =0 =0 =0 = 4k T W (R1 + R2 ) Problem E9.28 a) As it is shown in Lemma 9.12.1, b + Ef (X) = EY . Hence: b = EY  2EX = 3  8 = 5 b) Using the derivations from page 273, we obtain: 1 1 1 E(Y  X  1)2 = VAR(Y ) + ( )2 VAR(X) + (1  EY + EX)2  COV(X, Y ) 2 2 2 1 1 2 = 1 + + (1  3 + 2) + = 2 2 2 c) E(XY ) = COV(X, Y ) + (EX)(EY ) = 0.5 + 4 3 = 11.5 EX 2 = VAR(X) + (EX)2 = 2 + 42 = 18 2 ...
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This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HAAS
 Trigraph, dx

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