Question

4 Another Paradox in continuous conditional probability Let X, Y be two independent exponential random variables; i.e. with cdf: 1 ~ e^-x 2 0 Fx(x) = Fy(y) = otherwise. Let a > 0 and b > 0. With X and Y, we cook up the following random variables: Z = (X - a)/Y and U = X - aY (a) Show that P[Z < b] = 1 - e^(-b) for b > 0 and provide the pdf of Z for strictly positive values (density for negative values is not requested here). (b) Compute the joint density P(y, z) (y, z) for z > 0 (Formulae for z < 0 are slightly more complicated and not asked). Conclude that the conditional density p(y|z) for z > 0 is given by: p(y|z) = y(z + 1)e^(-y(z+1)) for y > 0.

          4 Another Paradox in continuous conditional probability Let X, Y be two independent exponential random variables; i.e. with cdf: 1 ~ e^-x 2 0 Fx(x) = Fy(y) = otherwise.
Let a > 0 and b > 0. With X and Y, we cook up the following random variables:
Z = (X - a)/Y and U = X - aY
(a) Show that P[Z < b] = 1 - e^(-b) for b > 0 and provide the pdf of Z for strictly positive values (density for negative values is not requested here). 
(b) Compute the joint density P(y, z) (y, z) for z > 0 (Formulae for z < 0 are slightly more complicated and not asked). 
Conclude that the conditional density p(y|z) for z > 0 is given by: p(y|z) = y(z + 1)e^(-y(z+1)) for y > 0.

4 another paradox in continuous conditional probability let xy be two independent exponential random variables ie with cdf 1 e i i 2 0 fxx fyo otherwise let a 0 and 0 with x and y we cook up 50465

Added by Karl G.

Question

Please give Ace some feedback