The random variable X is exponential with parameter λ=1 . The random variable Y is defined by Y=g(X)=1/(1+X) .
a) The inverse function h , for which h(g(x))=x , is of the form ay^b+c . Find a , b , and c .
b) For y∈(0,1] , the PDF of Y is of the form fY(y)=y^a*e^((b/y)+c) . Find a , b , and c .
a)
a = 1
b= -1
c = -1
b)
a = -2
b = -1
c = 1
All are correct!
a) To find the inverse function h such that h(g(x)) = x, we need to substitute g(x) into h and solve for x.
Given that g(x) = 1/(1 + x), we can rewrite this equation as:
y = 1/(1 + x)
Now, solve for x:
1 + x = 1/y
x = 1/y - 1
Comparing this result to the form ay^b + c, we can see that:
a = 1
b = -1
c = 0
So, the inverse function h is h(y) = 1/y - 1.
b) The probability density function (PDF) of Y can be derived by finding the cumulative distribution function (CDF) of Y and then taking its derivative.
First, let's find the CDF of Y. For y ∈ (0, 1], the CDF of Y is defined as:
F_Y(y) = P(Y ≤ y) = P(g(X) ≤ y) = P(X ≥ h(y))
Since X is an exponential random variable with parameter λ = 1, the CDF of X is given by:
F_X(x) = 1 - e^(-λx) = 1 - e^(-x)
Using the inverse function h(y) = 1/y - 1, we can rewrite the inequality as:
X ≥ h(y) ⟹ X ≥ 1/y - 1 ⟹ X + 1 ≥ 1/y ⟹ 1/(X + 1) ≤ y
Now, substitute the CDF of X into the inequality:
F_Y(y) = P(Y ≤ y) = P(X ≥ h(y)) = 1 - P(X < h(y)) = 1 - F_X(h(y)) = 1 - (1 - e^(-h(y))) = e^(-h(y))
Substituting h(y) = 1/y - 1, we have:
F_Y(y) = e^(-(1/y - 1))
To find the PDF, we take the derivative of the CDF with respect to y:
f_Y(y) = d/dy [e^(-(1/y - 1))] = e^(-(1/y - 1)) * (1/y^2)
Comparing this result to the form f_Y(y) = y^a * e^((b/y) + c), we have:
a = -2
b = -1
c = 1
Therefore, the PDF of Y is f_Y(y) = y^(-2) * e^(-1/y + 1).