作业帮概率统计英文题目 2008
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概率统计英文题目 2008
Solution:
(1) The expected value of Z is equal to the sum of the expected values of X and Y,i.e.,
E(Z)=E(X)+E(Y).
Now E(X) = ∫(0,∞)αe^(-αx)dx = 1/α,and E(Y) = ∫(0,∞)βe^(-βy)dy = 1/β.Therefore,
E(Z) = E(X)+E(Y) = (1/α)+( 1/β) = (α+β)/(αβ).
(2) For Z=X+Y,we can use the following formula for the density function f(z),assuming X and Y are independent of each other
f(z)=∫(-∞,+∞)fx(z-y)fy(y)dy
=∫(0,z){αe^[-α(z-y)}[βe^(-βy)]dy
= αβe^(-αz)∫(0,z)e^[-(β-α)y]dy.
Now we proceed with the condition α≠β,and we get
f(z) = [αβ/(α-β)][e^(-αz)-e^(-βz)],(z≥0).
and with the condition α=β,we get
(3) f(z)=(α^2)ze^(-αz),(z≥0).
Alternatively,we can take the limit β→α from the result in (2) to get the same result for (3).And the result in (1) can also be obtained using f(z) in (2) and (3) and E(Z)=∫(0,∞)zf(z)dz.
(1) The expected value of Z is equal to the sum of the expected values of X and Y,i.e.,
E(Z)=E(X)+E(Y).
Now E(X) = ∫(0,∞)αe^(-αx)dx = 1/α,and E(Y) = ∫(0,∞)βe^(-βy)dy = 1/β.Therefore,
E(Z) = E(X)+E(Y) = (1/α)+( 1/β) = (α+β)/(αβ).
(2) For Z=X+Y,we can use the following formula for the density function f(z),assuming X and Y are independent of each other
f(z)=∫(-∞,+∞)fx(z-y)fy(y)dy
=∫(0,z){αe^[-α(z-y)}[βe^(-βy)]dy
= αβe^(-αz)∫(0,z)e^[-(β-α)y]dy.
Now we proceed with the condition α≠β,and we get
f(z) = [αβ/(α-β)][e^(-αz)-e^(-βz)],(z≥0).
and with the condition α=β,we get
(3) f(z)=(α^2)ze^(-αz),(z≥0).
Alternatively,we can take the limit β→α from the result in (2) to get the same result for (3).And the result in (1) can also be obtained using f(z) in (2) and (3) and E(Z)=∫(0,∞)zf(z)dz.