The PDF of exp(X)
Let X be a random variable with PDF f_X. Find the PDF of the random variable Y=e^X for each of the following cases:
For general f_X, when y>0, f_Y(y)=
f_X(ln y)
---------
y
When f_X(x) = {1/3,0,if −2<x≤1,otherwise,
we have f_Y(y) = {g(y),0,if a<y≤b,otherwise.
Give a formula for g(y) and the values of a and b using standard notation . (In your answers, you may use the symbol 'e' to denote the base of the natural logarithm.)
g(y)= unanswered
a= unanswered
b= unanswered
When f_X(x) = {2e−2x,0,if x>0,otherwise,
we have f_Y(y) = {g(y),0,if a<y,otherwise.
Give a formula for g(y) and the value of a using the standard notation .
g(y)= unanswered
a= 1
When X is a standard normal random variable, we have, for y>0, f_Y(y)=
(2*π)^-1/2 * e^-((ln y)^2)/2)
----------------
y
PLEASE, COULD YOU PROVIDE THE OTHER ANSWERS TO THIS PROBLEM???
g(y)=1/(3y)
a=e^(-2)
b=e^1
Sure! Here are the answers to the remaining parts of the problem:
For the case when f_X(x) = {1/3, 0, if -2 < x ≤ 1, otherwise},
g(y) = 1/3 * (ln y)
a = e^(-2)
b = e
For the case when f_X(x) = {2e^(-2x), 0, if x > 0, otherwise},
g(y) = 2 * e^(-ln y / 2)
a = 0
I hope this helps! Let me know if you have any further questions.
Sure! Let's go through each case step by step and find the PDF of the random variable Y.
1. For general f_X, when y>0, the formula for f_Y(y) is:
f_Y(y) = f_X(ln y) / y
The formula is obtained by substituting y = e^X into the PDF of X, f_X, and then dividing by y.
2. When f_X(x) = {1/3, 0, if −2 < x ≤ 1, otherwise}, the formula for g(y) and the values of a and b are:
g(y) = f_X(ln y) / |y|
Here, we need to consider two cases: when ln y ≤ 1 and when ln y > 1. In the first case, a = e^(-2) and b = e. In the second case, a = 0 and b = e. This is because ln y ≤ 1 corresponds to -2 < x ≤ 1, and ln y > 1 corresponds to x > 1.
3. When f_X(x) = {2e^(-2x), 0, if x > 0, otherwise}, the formula for g(y) and the value of a are:
g(y) = f_X(ln y) / |y|
Here, a = 0 because x > 0 corresponds to ln y > 0. Since there is no upper limit for y, we do not have a specific value for b.
4. When X is a standard normal random variable, for y > 0, the formula for f_Y(y) is:
f_Y(y) = (2π)^(-1/2) * e^(-(ln y)^2/2) / y
This formula is derived by substituting y = e^X into the standard normal distribution formula and then dividing by y.
I hope this helps! Let me know if you have any further questions.