Probability Hard!!

X ~ Exponential (kx)
Y ~ Exponential (ky)
fX(x) = kx * exp (-kx * x)
fY(y) = ky * exp (-ky * y)
(1):
MZ(t) = E [exp (-z * t)]
= E {exp [-(x - y) * t]}
= E [exp (-x * t) / exp (-y * t)]
= E [exp (-x * t)] / E [exp (-y * t)]
coz X & Y are independent
= MX(t) / MY(t)
= exp (-kx * t) / exp (-ky * t)
= exp [-(kx - ky) * t]
= exp (-kz * t) where kz = kx - ky
Hence, Z ~ Exponential (kx - ky)
(2):
let z = x / y and w = y
then y = w and x = w * z
J = (dz / dx) * (dw / dy) - (dz / dy) * (dw / dx) = 1 / y = 1 / w
fZ,W(z,w) = fX,Y(x,y) / J
= fX(w * z) * fY(w) * w
coz X & Y are independent
= [kx * exp (-kx * w * z)] * [ky * exp (-ky * w)] * w
= kx * ky * w * exp [-w * (kx * z + ky)]
fZ(z) = integral from 0 to infinite fZ,W(z,w) for w
= kx * ky * integral from 0 to infinite {w * exp [-w * (kx * z + ky)] * dw}
= [kx * ky / (kx * z + ky)2] * integral from 0 to -ve infinite {[-w * (kx * z + ky)] * exp [-w * (kx * z + ky)] * d[-w * (kx * z + ky)]}
= kx * ky / (kx * z + ky)2

2009-04-18 18:07:58 補充：
it is no need to find the double integral

fX,Y(x,y) = fX(x) * fY(y) coz they are independent