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)=
- unanswered fX(eyy) fX(ln yy) fX(ln y)y none of the above
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.)
g(y)= - unanswered
a= - unanswered
b= - unanswered
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 .
g(y)= - unanswered
a= - unanswered
Sorry, but I'm not able to help with that question.
To find the PDF of the random variable Y=eX, we need to use the change of variable method.
For the general case when y > 0, the formula for the PDF of Y is fY(y) = |(d/dy)(ln(y)) * fX(ln(y))|.
For the first case, where fX(x) = {1/3, 0, if -2 < x ≤ 1, otherwise}:
To find the formula for g(y), we substitute y = e^x into the range -2 < x ≤ 1:
-2 < ln(y) ≤ 1.
Next, we need to find the range of y for which fY(y) is non-zero. The range of y can be calculated by exponentiating the limits of x:
e^-2 < y ≤ e^1.
Therefore, the formula for g(y) is g(y) = fX(ln(y)), and the values of a and b are a = e^-2 and b = e^1.
For the second case, where fX(x) = {2e^-2x, 0, if x > 0, otherwise}:
To find the formula for g(y), we substitute y = e^x into the range x > 0:
0 < ln(y).
Next, we need to find the range of y for which fY(y) is non-zero. The range of y can be calculated by exponentiating the limits of x:
1 < y.
Therefore, the formula for g(y) is g(y) = fX(ln(y)), and the value of a is a = 1.
Note: The values of a and b or the range of y may vary depending on the specific input provided. Please make sure to check the conditions and values mentioned in the problem statement.
To find the PDF of the random variable Y = eX, we need to use the transformation method. Let's go through each case one by one.
1) For the general fX:
The first step is to find the cumulative distribution function (CDF) of Y. The CDF of Y is given by:
FY(y) = P(Y ≤ y) = P(eX ≤ y).
To solve this, we need to find the range of X values that satisfy eX ≤ y. This can be written as X ≤ ln(y) since eX = y when X = ln(y).
Using the CDF, we can find the PDF by differentiating FY(y) with respect to y:
fY(y) = d/dy (FY(y)).
Now, substitute the expression X ≤ ln(y) into fX(x):
fX(ln(y)) = fX(X ≤ ln(y)).
Therefore, the correct answer is fX(ln(y)).
2) For fX(x) = {1/3, 0, if -2 < x ≤ 1, otherwise}:
Since the function fX(x) is given, we need to determine the range of Y such that g(y) = fY(y) ≠ 0.
To find the range, we need to solve the inequality eX > 0, which implies X > 0.
Therefore, the values of a and b are 0 and infinity, respectively. Thus, a = 0 and b = ∞.
So, the formula for g(y) is g(y) = fX(ln(y)) for a ≤ y ≤ b, and 0 otherwise.
3) For fX(x) = {2e^(-2x), 0, if x > 0, otherwise}:
Similar to the previous case, we need to determine the range of Y such that g(y) = fY(y) ≠ 0.
Let's solve the inequality eX > y:
e^(-2x) > y.
Taking the natural logarithm on both sides, we get:
-2x > ln(y).
Solving for x, we have:
x < -ln(y)/2.
Hence, the range of X is x < -ln(y)/2.
Thus, the formula for g(y) is g(y) = fX(X < -ln(y)/2) for y > 0, and 0 otherwise.
In summary:
1) For general fX: g(y) = fX(ln(y)), a = 0, and b = ∞.
2) For fX(x) = {1/3, 0, if -2 < x ≤ 1, otherwise}: g(y) = fX(ln(y)), a = 0, and b = ∞.
3) For fX(x) = {2e^(-2x), 0, if x > 0, otherwise}: g(y) = fX(X < -ln(y)/2), and a = 0.
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