Suppose that X is a continuous random variable and that Y = X^4.
Then, for y >= 0, we have:
fY(y) = ay^b fX(-cy^d) + ay^b fX(cy^d)
For some a, b, d and some c > 0.
Find a, b, c and d.
a =
b =
c =
d =
To find a, b, c, and d for the given equation, we can use the probability density function (PDF) transformation method.
We know that Y = X^4, which implies that X = Y^(1/4).
Now, let's find the PDF of Y, denoted as fY(y).
To find fY(y), we need to determine the relationship between X and Y using the transformation method.
First, we find the cumulative distribution function (CDF) of Y, denoted as FY(y), which is the probability that Y takes on a value less than or equal to y.
FY(y) = P(Y ≤ y)
= P(X^4 ≤ y)
= P(X ≤ y^(1/4)) (since X is a continuous random variable)
To find the PDF of Y, we differentiate the CDF of Y with respect to y:
fY(y) = d/dy (FY(y))
= d/dy (P(X ≤ y^(1/4)))
= d/dy (F_X(y^(1/4))) (since X is a continuous random variable)
= f_X(y^(1/4)) * (1/4) * y^(-3/4) (chain rule of differentiation)
where f_X is the PDF of X.
Now, let's express fY(y) in terms of the PDF of X. We have:
fY(y) = f_X(y^(1/4)) * (1/4) * y^(-3/4)
Comparing this expression with the given equation:
fY(y) = ay^b fX(-cy^d) + ay^b fX(cy^d)
We see that a = 1/4, b = -3/4, c = 1, and d = 1/4.
Therefore, the values of a, b, c, and d are:
a = 1/4
b = -3/4
c = 1
d = 1/4.