Let X be uniform on [0,1/2]. Find the PDF fY(y) of Y=X/(1−X).
For 0≤y≤1:
fY(y)= ?
2/(1+y)^2
To find the probability density function (PDF) of Y, we first need to find the cumulative distribution function (CDF) of Y. Then, we can differentiate the CDF to obtain the PDF.
Let's begin by finding the CDF of Y, denoted as FY(y):
FY(y) = P(Y ≤ y)
Since Y = X / (1 - X), we can rewrite the inequality as:
X / (1 - X) ≤ y
Multiplying both sides of the inequality by (1 - X) (we can do this because (1 - X) > 0 for 0 ≤ X ≤ 1/2), we get:
X ≤ y(1 - X)
Now, we can solve this inequality for X:
X ≤ y - yX
X + yX ≤ y
Factor out X:
X(1 + y) ≤ y
Divide through by (1 + y) (we can do this because (1 + y) > 0 for 0 ≤ y ≤ 1):
X ≤ y / (1 + y)
Since X is uniformly distributed on [0, 1/2], we know that the probability that X is less than or equal to any value c in [0, 1/2] is c / (1/2) = 2c. Therefore, we have:
FY(y) = 2 * (y / (1 + y))
Now, to find the PDF of Y, we differentiate the CDF with respect to y:
fY(y) = d/dy [FY(y)]
Using the quotient rule of differentiation, we have:
fY(y) = (2 * (1 + y) - 2 * y) / (1 + y)^2
Simplifying this expression, we get:
fY(y) = 2 / (1 + y)^2
Therefore, the PDF of Y is:
fY(y) = 2 / (1 + y)^2, for 0 ≤ y ≤ 1.
To find the probability density function (PDF) of Y, we'll first find the cumulative distribution function (CDF) of Y and then differentiate it with respect to y to get the PDF.
Let's begin:
Given that X is uniformly distributed on [0,1/2], we can find the cumulative distribution function (CDF) of X as follows:
F_X(x) = P(X ≤ x)
Since X is uniformly distributed on [0,1/2], the CDF of X is given by:
F_X(x) = x / (1/2) = 2x, for 0 ≤ x ≤ 1/2
Next, we need to find the cumulative distribution function (CDF) of Y. This can be done by considering the following:
Y = X / (1 - X)
1 - Y = 1 - (X / (1 - X)) = (1 - X - X) / (1 - X) = (1 - 2X) / (1 - X)
To find the CDF of Y, we need to calculate P(Y ≤ y), which can be written as:
P(Y ≤ y) = P(X / (1 - X) ≤ y)
= P(1 - 2X ≤ y(1 - X))
= P(-X ≤ y - yX)
= P(X - yX ≤ y)
= P(X(1 - y) ≤ y)
= 1 - P(X(1 - y) > y)
= 1 - P(X > y / (1 - y))
Since X is uniformly distributed on [0, 1/2], we can write this as:
P(Y ≤ y) = 1 - P(X > y / (1 - y))
= 1 - (1 - 2(y / (1 - y)))
= 2y / (1 - y)
Now, to find the PDF f_Y(y) of Y, we differentiate the CDF of Y with respect to y:
f_Y(y) = d/dy [2y / (1 - y)]
= 2 / (1 - y)^2
Therefore, the PDF of Y, for 0 ≤ y ≤ 1, is:
f_Y(y) = 2 / (1 - y)^2