Problem 1: The PDF of exp(X)
(6/6 points)
Let X be a random variable with PDF fX. Find the PDF of the random variable Y=eX for each of the following cases:
For general fX, when y>0, fY(y)=
Solution: f_x(ln(y))/y
When fX(x) = {1/3,0,if −2<x≤1,otherwise,
we have fY(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.)
Solution: g(y) = 1/(3*y)
a = e^(-2)
b = e
When fX(x) = {2e−2x,0,if x>0,otherwise,
we have fY(y) = {g(y),0,if a<y,otherwise.
Give a formula for g(y) and the value of a using the standard notation .
Solution: g(y) = 2/(y^(3))
a= 1
When X is a standard normal random variable, we have, for y>0, fY(y)=
Solution: 1/sqrt(2pi) * (e^(-ln(x))^2/2)/y
Enjoj :-)
To find the PDF of the random variable Y=eX, we can use the change of variables method. We need to consider three different cases based on the given fX(x).
Case 1: General fX(x)
For this case, we assume that we have a general PDF fX(x). We want to find the PDF fY(y) when y > 0.
To find fY(y), we can use the change of variables formula. Let g(y) be the PDF of Y.
g(y) = fX(ln(y)) / |dy/dx|
Here, |dy/dx| represents the absolute value of the derivative of y with respect to x.
In this case, dy/dx = e^x since Y = eX.
Substituting dy/dx = e^x into the formula, we get
g(y) = fX(ln(y)) / e^x
Since Y = eX, we can rewrite the formula as
g(y) = fX(ln(y)) / y
So, for general fX, when y > 0, the PDF of Y is fY(y) = fX(ln(y)) / y.
Case 2: Specific fX(x) = {1/3, 0, if −2 < x ≤ 1, otherwise}
For this case, we have a specific piecewise function for fX(x). We want to find the PDF g(y) when a < y ≤ b.
To find g(y), we need to substitute the given values of fX(x) into the general formula g(y) = fX(ln(y)) / y.
First, we find g(y) when a < y ≤ b.
Since fX(x) = 1/3 for -2 < x ≤ 1, we replace fX(x) with 1/3 in the formula:
g(y) = (1/3) / y
So, for this specific fX, when a < y ≤ b, the PDF of Y is g(y) = 1/(3y).
Next, we find the values of a and b.
In the given fX(x), fX(x) = 0 for x ≤ -2 and x > 1. So, we need to find the values of a and b such that fX(x) = 0 for y ≤ a and y > b.
Using Y = eX, we have Y ≤ a if and only if X ≤ ln(a) and Y > b if and only if X > ln(b).
Since fX(x) = 0 for x ≤ -2 and x > 1, we have fX(x) = 0 for X ≤ ln(a) and X > ln(b). This implies ln(a) ≤ -2 and ln(b) > 1.
To find the values of a and b, we solve these inequalities for a and b:
ln(a) ≤ -2 => e^(-2) ≤ a
ln(b) > 1 => b > e
Therefore, for this specific fX, the PDF of Y is g(y) = 1/(3y) when a < y ≤ b, where a = e^(-2) and b = e.
Case 3: Specific fX(x) = {2e^(-2x), 0, if x > 0, otherwise}
For this case, we have another specific function for fX(x). We want to find the PDF g(y) when y > a.
Using the general formula g(y) = fX(ln(y)) / y, we substitute fX(x) = 2e^(-2x):
g(y) = (2e^(-2*ln(y))) / y
Since ln(e^x) = x, we simplify the formula:
g(y) = (2*y^(-2)) / y^2
Simplifying further, we have:
g(y) = 2 / (y^3)
So, for this specific fX, when y > a, the PDF of Y is g(y) = 2/(y^3).
Finally, in the special case where X is a standard normal random variable, we know that:
- The PDF of a standard normal distribution is given by fX(x) = (1/√(2π)) * e^(-x^2/2).
- We want to find the PDF of Y = eX for y > 0.
Substituting fX(x) into the general formula g(y) = fX(ln(y)) / y:
g(y) = [(1/√(2π)) * e^(-ln(y)^2/2)] / y
Simplifying, we have:
g(y) = (1/√(2π)) * (e^(-ln(y)^2/2)) / y
So, for X as a standard normal random variable, when y > 0, the PDF of Y is fY(y) = (1/√(2π)) * (e^(-ln(y)^2/2)) / y.