Asked by amelie
In a study of particulate pollution in air samples over a smokestack, X1 represents the amount of pollant per sample when a cleaning device is not operating, and X2 represents the amount per sample when the cleaning device is operating. Assume that (X1, X2) has a joint probability density function.
f(X1,X2) =[ 1 for 0<=X1<=2 0<=X2<=1
2 X2<=X1
[ 0 ELSEWHERE
The random variable Y=X1-X2 represents the amount by which the weight of emitted pollutant can be reduced by using the cleaning device.
a) Find E(Y) AND V(Y)
I have solved the E(Y)already.But I
have problems with the variance.
Work
=========
a) E(Y)= 1
E(X1)=2 E(X2)=1
V(Y)= V(X1-X2)= V(X1)-V(X2)
V(X1)= E(X1 ^2) - E(X1) ^2
= ç(FROM 0 TO 2) X1^2 dX1 - 4
= -1 1/3
V(X2)= ...
= ç(FROM 0 TO 1) 2X2 ^2 dX2 - 1
= -1/3
ANS= -1, WRONG
CORRECT ANS= 1/6
i though i was doing the work correctly until i got negative variances, can you help me figure out what am i going wrong.
f(X1,X2) =[ 1 for 0<=X1<=2 0<=X2<=1
2 X2<=X1
[ 0 ELSEWHERE
The random variable Y=X1-X2 represents the amount by which the weight of emitted pollutant can be reduced by using the cleaning device.
a) Find E(Y) AND V(Y)
I have solved the E(Y)already.But I
have problems with the variance.
Work
=========
a) E(Y)= 1
E(X1)=2 E(X2)=1
V(Y)= V(X1-X2)= V(X1)-V(X2)
V(X1)= E(X1 ^2) - E(X1) ^2
= ç(FROM 0 TO 2) X1^2 dX1 - 4
= -1 1/3
V(X2)= ...
= ç(FROM 0 TO 1) 2X2 ^2 dX2 - 1
= -1/3
ANS= -1, WRONG
CORRECT ANS= 1/6
i though i was doing the work correctly until i got negative variances, can you help me figure out what am i going wrong.
Answers
Answered by
sami
E[y]=3/2
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