Let X be uniform on [0,1/2] . Find the PDF fY(y) of Y=X/(1−X) .
For 0≤y≤1 :
fY(y)=
To find the PDF (probability density function) of Y, we need to go through a few steps:
Step 1: Find the cumulative distribution function (CDF) of Y.
Step 2: Differentiate the CDF of Y to find the PDF of Y.
Let's start with Step 1:
Step 1: Find the CDF of Y.
To find the CDF of Y, we need to express Y in terms of X and then determine the range of X for which Y ≤ y.
Given Y = X / (1 - X), we can substitute Y = y and solve for X:
Y = X / (1 - X)
y = X / (1 - X)
y(1 - X) = X
y - yX = X
y = X + yX
X(1 + y) = y
X = y / (1 + y)
Now, we need to determine the range of X for which Y ≤ y:
0 ≤ Y ≤ 1 implies 0 ≤ X ≤ 1 / (1 + y)
Step 2: Find the PDF of Y.
To find the PDF of Y, we differentiate the CDF of Y with respect to y:
CDF of Y, FY(y) = P(Y ≤ y)
= P(X / (1 - X) ≤ y)
= P(X ≤ y / (1 + y))
= ∫[0, y / (1 + y)] fX(x) dx
where fX(x) is the PDF of X, which is uniform on [0, 1/2]. Therefore, fX(x) = 2 for 0 ≤ x ≤ 1/2, and 0 otherwise.
Now, differentiating the CDF with respect to y:
d/dy [FY(y)] = d/dy [∫[0, y / (1 + y)] fX(x) dx]
= d/dy [2 * (y / (1 + y))]
= 2 / (1 + y)^2
So, the PDF of Y, fY(y), is:
fY(y) = 2 / (1 + y)^2 for 0 ≤ y ≤ 1
Therefore, for 0 ≤ y ≤ 1:
fY(y) = 2 / (1 + y)^2