求助 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, ∞