Let X be uniform on [0,1/2] . Find the PDF fY(y) of Y=X/(1−X) .
For 0≤y≤1
To find the PDF (probability density function) of Y, we need to determine the distribution of Y and derive its density function.
First, let's determine the cumulative distribution function (CDF) of Y.
The CDF of Y, denoted as FY(y), is defined as the probability that Y is less than or equal to y. Mathematically, it can be represented as:
FY(y) = P(Y ≤ y)
Since Y = X/(1-X), we can substitute this expression into the CDF:
FY(y) = P(X/(1-X) ≤ y)
To solve this inequality, we first need to make sure that 0 ≤ y ≤ 1, as given in the problem.
Rearranging the inequality, we have:
X/(1-X) ≤ y
Simplifying, we get:
X ≤ y(1-X)
Expanding, we have:
X ≤ y - yX
Rearranging terms, we get:
X + yX ≤ y
Factoring out X, we have:
X(1+y) ≤ y
Now, we have two cases to consider:
Case 1: If 1 + y > 0, which means y > -1, then we can divide both sides of the inequality by (1+y) without changing the sign:
X ≤ y / (1+y)
Case 2: If 1 + y < 0, which means y < -1, then dividing by (1+y) would change the sign of the inequality:
X ≥ y / (1+y)
Now, let's consider these cases separately:
Case 1:
If y > -1, then the inequality X ≤ y / (1+y) represents the valid range for X. However, in our given problem, X is uniformly distributed in the range [0, 1/2]. Therefore, the valid range for X in this case is 0 ≤ X ≤ y / (1+y).
Case 2:
If y < -1, then the inequality X ≥ y / (1+y) represents the valid range for X. In our problem, X is uniformly distributed in the range [0, 1/2]. However, the inequality X ≥ y / (1+y) cannot be satisfied in this range for any y < -1. Therefore, this case does not contribute to the PDF of Y for 0 ≤ y ≤ 1.
Now, let's proceed with finding the PDF for Case 1.
PDF is defined as the derivative of the CDF. Therefore, we need to differentiate the CDF with respect to y and find fY(y).
Differentiating the CDF, FY(y), with respect to y, we get:
fY(y) = d/dy (FY(y))
To find fY(y), we'll differentiate the CDF:
fY(y) = d/dy (P(Y ≤ y))
Now, we can plug in the valid range for X derived from Case 1:
fY(y) = d/dy (P(0 ≤ X ≤ y / (1+y)))
Since X is uniformly distributed on [0, 1/2], the probability P(0 ≤ X ≤ y / (1+y)) is equal to the length of the interval (0, y / (1+y)) divided by the total length of the interval (0, 1/2). Therefore, we have:
fY(y) = d/dy (length of (0, y / (1+y)) / length of (0, 1/2))
The length of the interval (0, y / (1+y)) is y / (1+y), and the length of the interval (0, 1/2) is 1/2. Therefore, the PDF of Y can be written as:
fY(y) = d/dy (y / (1+y)) / (1/2)
Differentiating d/dy (y / (1+y)) with respect to y yields:
fY(y) = 1 / (1+y)^2 / (1/2)
Simplifying further:
fY(y) = 2 / (1+y)^2
Finally, we have obtained the PDF of Y, denoted as fY(y):
fY(y) = 2 / (1+y)^2 for 0 ≤ y ≤ 1.
This is the probability density function of Y.