- #1
whitejac
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Homework Statement
X ~ Uniform (0,1)
Y = e-X
Find FY (y) - or the CDF
Find fY(y) - or the PDF
Find E[Y]
2. Homework Equations
E[Y] = E[e-X] = ∫0 , 1 e-xfx(x)dx
FY(y) = P(Y < y)
fY(y) = F'Y(y)
The Attempt at a Solution
FX(x) =
{
0 for x<0
x for 0<x<1
1 for 1<x
}
fX(x) =
{
1 for 0<x<1
0 otherwise
}
Then use this information to solve:
FY(y) = P(Y < y) = P(e-X < y) = P(X < -lny)
^This is a negative function which violates one of the axioms of our continuous random variables (fX(x) ≥ 0)
fY(y) is reliant on FY(y)
E[Y] = E[e-X]
= ∫0 , 1 e^-xfx(x)dx
= -e-x | 0 , 1
= 1 - 1/e
I know this answer is corrrect, which means that my CDF and PDF of Y are incorrect because using those will give me the negativ of this answer. So I'm a little bit stumped, or misinterpretting how to continue the CDF equation.