Let x1,x2,x3 be random sample with pdf
f(x)=exp(-x),x>0
find the joint pdf of y1=x1/x2,
y2=x3/(x1+x2),y3=x1+x2
find pdf of y1
Are y1,y2,y3 independent?
To find the joint PDF of y1=x1/x2, y2=x3/(x1+x2), and y3=x1+x2, we need to determine the relationship between the random variables x1, x2, and x3.
First, we need to determine the joint probability distribution function (PDF) of x1, x2, and x3. Since x1, x2, and x3 are random samples, they are independent and identically distributed with the PDF f(x) = exp(-x), for x > 0.
To find the joint PDF of y1, y2, and y3, we need to consider the transformation between the variables (x1, x2, x3) and (y1, y2, y3).
Let's write the transformation equations:
y1 = x1/x2
y2 = x3/(x1 + x2)
y3 = x1 + x2
To find the joint PDF of y1, y2, and y3, we need to calculate the Jacobian determinant of this transformation. The Jacobian determinant is given by:
J = | ∂(x1, x2, x3)/∂(y1, y2, y3) |
Let's calculate the Jacobian determinant in this case. We have:
∂x1/∂y1 = 1/x2 , ∂x1/∂y2 = -x3/(x1 + x2)^2 , ∂x1/∂y3 = 1 ,
∂x2/∂y1 = -x1/x2^2 , ∂x2/∂y2 = x3/(x1 + x2)^2 , ∂x2/∂y3 = 1 ,
∂x3/∂y1 = 0 , ∂x3/∂y2 = 1/(x1 + x2) , ∂x3/∂y3 = 1.
So, the Jacobian determinant is:
J = |1/x2 -x3/(x1 + x2)^2 1 |
|-x1/x2^2 x3/(x1 + x2)^2 1 |
|0 1/(x1 + x2) 1 |
Taking the determinant of this matrix gives:
J = 1/(x1 + x2)^2
Now, we can write the joint PDF of y1, y2, and y3 as a function of x1, x2, and x3:
f(y1, y2, y3) = f(x1, x2, x3) * |J|
Since the joint PDF of x1, x2, and x3 is the product of their individual PDFs, we have:
f(x1, x2, x3) = f(x1) * f(x2) * f(x3) = exp(-x1) * exp(-x2) * exp(-x3) = exp(-(x1 + x2 + x3))
Therefore, the joint PDF of y1, y2, and y3 is given by:
f(y1, y2, y3) = exp(-(x1 + x2 + x3)) / (x1 + x2)^2
Now we need to find the PDF of y1. In order to do that, we need to marginalize the joint PDF of y1, y2, and y3 with respect to y2 and y3:
f(y1) = ∫∫ f(y1, y2, y3) dy2 dy3
Integrating f(y1, y2, y3) with respect to y2 and y3 gives:
f(y1) = ∫∫ exp(-(x1 + x2 + x3)) / (x1 + x2)^2 dy2 dy3
Since x1, x2, and x3 are independent random samples, we can rewrite this as:
f(y1) = ∫∫∫ exp(-(x1 + x2 + x3)) / (x1 + x2)^2 dx1 dx2 dx3
Simplifying this expression is not straightforward, and it may require numerical methods to obtain the PDF of y1.