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 find the mapping from (x1, x2, x3) to (y1, y2, y3) and determine the corresponding Jacobian.
Step 1: Define the mappings:
y1 = x1/x2
y2 = x3/(x1+x2)
y3 = x1+x2
Step 2: Solve the equations for x1, x2, and x3:
x1 = y1x2
x2 = y1x3/(1-y2)
x3 = y2(y1x3)/(1-y2)
Step 3: Determine the Jacobian:
J = ( ∂(x1, x2, x3) / ∂(y1, y2, y3) ) = | ∂x1/∂y1 ∂x1/∂y2 ∂x1/∂y3 |
| ∂x2/∂y1 ∂x2/∂y2 ∂x2/∂y3 |
| ∂x3/∂y1 ∂x3/∂y2 ∂x3/∂y3 |
Calculating the partial derivatives, we get:
∂x1/∂y1 = x2, ∂x1/∂y2 = -x1/(1-y2), ∂x1/∂y3 = 0
∂x2/∂y1 = 0, ∂x2/∂y2 = -x3/(1-y2), ∂x2/∂y3 = 0
∂x3/∂y1 = (y1x3)/(1-y2), ∂x3/∂y2 = (y2x3)/(1-y2), ∂x3/∂y3 = 1
Therefore, the Jacobian is:
J = | x2 -x1/(1-y2) 0 |
| 0 -x3/(1-y2) 0 |
| (y1x3)/(1-y2) (y2x3)/(1-y2) 1 |
Step 4: Find the joint PDF:
Joint PDF = PDF of x1 * PDF of x2 * PDF of x3 * |J|
= exp(-x1) * exp(-x2) * exp(-x3) * |J|
= exp(-y1x2) * exp(-y1x3/(1-y2)) * exp(-x3) * |J|
Step 5: Simplify the joint PDF:
Joint PDF = exp(-y1x2 - y1x3/(1-y2) - x3) * |J|
Now, to find the PDF of Y1, we need to integrate out the other variables Y2 and Y3 from the joint PDF of Y1, Y2, and Y3.
PDF of Y1 = ∫(∫(Joint PDF * dy2) * dy3)
After performing the integration, the PDF of Y1 will be obtained.
To determine if Y1, Y2, and Y3 are independent, we can check if the joint PDF can be expressed as the product of the individual PDFs of Y1, Y2, and Y3. If the joint PDF can be factorized in this way, then the variables are independent. If not, they are dependent.