The random variable X has a PDF of the form
fX(x)= 1/x^2,for x≥1,
0,otherwise.
Let Y=X2. For y≥1, the PDF of Y it takes the form fY(y)=a / y^b. Find the values of a and b.
a = 1/2 and b = 3
To find the PDF of Y, we need to first determine the cumulative distribution function (CDF) of Y and then differentiate it to obtain the PDF.
The CDF of Y is given by:
FY(y) = P(Y ≤ y) = P(X² ≤ y) = P(X ≤ √y)
Since X has a PDF of fX(x) = 1/x² for x ≥ 1, we can integrate fX(x) over the range [1, √y] to find P(X ≤ √y):
P(X ≤ √y) = ∫₁^√y (1/x²) dx
= ∫₁^√y (x^(-2)) dx
= -x^(-1) evaluated from 1 to √y
= -(√y)^(-1) + 1
= 1 - 1/√y
Now, to find the PDF of Y, we differentiate FY(y) with respect to y:
fY(y) = d/dy [FY(y)]
= d/dy [1 - 1/√y]
= 0 + 1/(2y^(3/2))
= 1/(2y^(3/2))
Therefore, the PDF of Y is fY(y) = a / y^b, with a = 1/2 and b = 3/2.
To find the values of a and b for the PDF of Y, we need to determine the distribution of Y based on the transformation Y = X^2.
First, let's find the cumulative distribution function (CDF) of X and use it to find the CDF of Y.
The CDF of X is given by:
FX(x) = ∫(1 to x) 1/t^2 dt
= [-1/t] from 1 to x
= 1 - 1/x
To find the CDF of Y, we substitute Y = X^2:
FY(y) = P(Y ≤ y) = P(X^2 ≤ y) = P(|X| ≤ √y) = P(-√y ≤ X ≤ √y)
We know that X ≥ 1, so we can write it as:
FY(y) = PX -^(√y) = PX ≤ √y - PX ≤ -√y
= FX(√y) - FX(-√y)
= (1 - 1/√y) - (1 - 1/(-√y))
= 2 - 2/√y
Now, to find the PDF of Y, we differentiate the CDF with respect to y:
fY(y) = d/dy(FY(y))
= d/dy (2 - 2/√y)
= 2/(2√y)
= 1/√y
We can rewrite this in the form fY(y) = a/y^b, where a = 1 and b = 1/2.
Therefore, the PDF of Y is given by fY(y) = 1/y^(1/2), where y ≥ 1, and the values of a and b are a = 1 and b = 1/2.