Let X have a uniform distribution in the unit interval [0; 1], and let Y have an expo-
nential distribution with parameter lambda = 2. Assume that X and Y are independent.
Let Z = X + Y .
(a) Find P(Y => X).
(b) Find the conditional PDF of Z given that Y = y.
(c) Find the conditional PDF of Y given that Z = 3.
what I did for the first one, was finding the point of intersection of y=1 and y=2e^(2x) which is x=(ln2)/2 and then, as X>Y only for (ln2)/2<X<1, find its probability and subtract it from 1 to find P(Y=>X). However, using the uniform(1-(1-(ln2)/2)) and using the exponential(1-(e^(-ln2)-e^(-2))) gave me two different values for this probability, and I do not know why.
Can someone help me with this question, please?
Thank you!
nential distribution with parameter lambda = 2. Assume that X and Y are independent.
Let Z = X + Y .
(a) Find P(Y => X).
(b) Find the conditional PDF of Z given that Y = y.
(c) Find the conditional PDF of Y given that Z = 3.
what I did for the first one, was finding the point of intersection of y=1 and y=2e^(2x) which is x=(ln2)/2 and then, as X>Y only for (ln2)/2<X<1, find its probability and subtract it from 1 to find P(Y=>X). However, using the uniform(1-(1-(ln2)/2)) and using the exponential(1-(e^(-ln2)-e^(-2))) gave me two different values for this probability, and I do not know why.
Can someone help me with this question, please?
Thank you!