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) (y^-1)-1 So a=1,b=-1,c=-1
b)y^-2*e^(-1/y+1). so a=-2,b=-1,c=1
h(x) = 1/x - 1 = (1-x)/x
so, ...
so yes h(x) = 1/x - 1 = (1-x)/x
but what do you do then?
derivative of (1-x)/x = -1/x^2
a) To find the inverse function h(g(x)) = x, we need to solve for x in terms of y. Let's start by substituting g(x) = y into the equation:
1/(1+X) = y
Now we can solve for X:
X = 1/y - 1
Comparing this equation to the form ay^b + c, we can see that a = -1, b = -1, and c = -1.
Therefore, the inverse function h(g(x)) = -y^-1 - 1.
b) To find the PDF of Y, we need to find the derivative of the CDF (cumulative distribution function) with respect to y.
Since X is exponentially distributed with parameter λ = 1, the CDF of X is given by:
FX(x) = 1 - e^(-λx)
= 1 - e^(-x)
Now we can substitute Y = g(X) into the CDF equation:
FY(y) = P(Y ≤ y)
= P(g(X) ≤ y)
= P(X ≥ 1/(1+y))
= 1 - P(X < 1/(1+y))
= 1 - FX(1/(1+y))
= 1 - (1 - e^(-1/(1+y)))
= e^(-1/(1+y))
Taking the derivative of FY(y) with respect to y, we get:
fY(y) = d/dy(e^(-1/(1+y)))
= e^(-1/(1+y)) * (-1) * (-1/(1+y)^2)
= e^(-1/(1+y)) / (1+y)^2
Comparing this equation to the form fY(y) = y^a * e^((b/y) + c), we can see that a = 2, b = -1, and c = 0.
Therefore, the PDF of Y is fY(y) = y^2 * e^(-1/(1+y)).
Note: The valid range for y is (0, 1] since Y is defined as g(X) = 1/(1+X).
a) To find the inverse function h(g(x)) = x, we need to solve for x in terms of y. So, let's substitute g(x) = 1/(1+x) into the equation h(g(x)) = x:
h(g(x)) = x
h(1/(1+x)) = x
Now, let's solve for h(y):
h(y) = x
h(y) = x - 0
h(y) = x - 0y^0
h(y) = ay^b + c
Comparing this equation with h(y) = ay^b + c, we can see that a = 1, b = 0, and c = 0. Therefore, the inverse function h(y) = y^0 + 0 = 1.
b) To find the PDF of Y, we need to determine the probability density function fY(y) for y∈(0,1].
First, let's find the cumulative distribution function (CDF) of Y, denoted as FY(y):
FY(y) = P(Y ≤ y) = P(g(X) ≤ y)
FY(y) = P(1/(1+X) ≤ y)
FY(y) = P(X ≥ 1/y - 1)
Since X follows an exponential distribution with parameter λ=1, the cumulative distribution function (CDF) of X, denoted as FX(x), is given by:
FX(x) = P(X ≤ x) = 1 - e^(-λx)
So, we can express the event X ≥ 1/y - 1 as X ≤ -(1/y - 1) using the complement rule:
FY(y) = 1 - FX(-(1/y - 1))
Now, let's differentiate FY(y) with respect to y to obtain the PDF of Y:
fY(y) = d/dy [FY(y)]
fY(y) = d/dy [1 - FX(-(1/y - 1))]
To simplify the expression, we'll substitute λ=1 and use the chain rule:
fY(y) = d/dy [1 - e^(y - 1)]
fY(y) = -d/dy [e^(y - 1)]
fY(y) = -(e^(y - 1))(d/dy)(y - 1)
fY(y) = -(e^(y - 1))(1)
fY(y) = -e^(y - 1)
Comparing this equation with fY(y) = y^a * e^((b/y) + c), we can see that a = 1, b = -1, and c = 0.
Therefore, the PDF of Y is fY(y) = y^1 * e^((-1/y) + 0) = y * e^(-1/y).