g(y)=1/(3y)
a=e^(-2)
b=e^1
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???
a=e^(-2)
b=e^1
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.
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.