The random variable X has a PDF of the form:
fX(x) = {1/x^2, for x >= 1,
0, otherwise.}
Let Y = X^2. For Y >= 1, the PDF of Y takes the form fY(y) = a/y^b. Find the values of a and b.
a =
b =
To find the PDF of the new random variable Y = X^2, we need to determine the transformation by which X is related to Y.
Let's start by finding the cumulative distribution function (CDF) of Y.
The CDF of Y, denoted as FY(y), is given by:
FY(y) = P(Y ≤ y) = P(X^2 ≤ y) = P(|X| ≤ √y) = P(X ≤ √y) - P(X ≤ -√y)
Since X has a known PDF, we can integrate it to obtain the CDF:
FX(x) = ∫(1/x^2)dx = -1/x|∞ to x = -1/x - (-1/∞) = -1/x
Now, we can find the CDF of Y:
FY(y) = P(X ≤ √y) - P(X ≤ -√y)
= -1/√y - (-1/-∞)
= -1/√y + 0
Next, we differentiate the CDF of Y with respect to y to obtain the PDF:
fY(y) = d/dy(FY(y))
= d/dy(-1/√y + 0)
= -1/(2y^3/2)
Therefore, the PDF of Y, fY(y), takes the form fY(y) = a/y^b.
Comparing the derived expression with the given form, we can see that:
a = -1/2
b = 3/2
Hence, the values of a and b for the PDF of Y are:
a = -1/2
b = 3/2