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,
and find pdf of Y1.
Are Y1,Y2,Y3 independent?
To find the joint probability density function (pdf) of Y1, Y2, and Y3, we need to use the transformation method. Let's go step by step:
Step 1: Determine the transformation equations:
Y1 = X1/X2
Y2 = X3/(X1+X2)
Y3 = X1+X2
Step 2: Express the X1, X2, and X3 in terms of Y1, Y2, and Y3:
Solving for X1, we have:
X1 = Y1Y3/(Y1+1)
Similarly, solving for X2,
X2 = Y3 - X1 = Y3 - Y1Y3/(Y1+1) = Y3(1 - Y1/(Y1+1))
Finally, X3 = Y2(Y1+1)Y3 - Y2Y1Y3 = Y3(Y1+Y2(Y1+1))
Step 3: Find the Jacobian determinant:
The Jacobian determinant, J, is calculated as follows:
J = |∂(X1,X2,X3)/∂(Y1,Y2,Y3)| = |(∂X1/∂Y1)(∂X2/∂Y2)(∂X3/∂Y3)|
Calculating the partial derivatives, we get:
∂X1/∂Y1 = (Y1+1)Y3/(Y1+1)² = Y3/(Y1+1)
∂X2/∂Y2 = 0
∂X3/∂Y3 = Y1 + Y2(Y1+1)
So, J = |Y3/(Y1+1)*0*(Y1 + Y2(Y1+1))| = 0
Step 4: Compute the joint pdf of Y1, Y2, Y3:
The joint pdf of Y1, Y2, Y3 is given as the product of the pdfs of X1, X2, and X3 multiplied by the absolute value of the Jacobian determinant:
f(Y1, Y2, Y3) = f(X1, X2, X3) · |J|
Substituting the pdf of X1, X2, X3, and the Jacobian determinant:
f(Y1, Y2, Y3) = exp(-Y3(Y1+Y2(Y1+1))) * 0 = 0
Hence, the joint pdf of Y1, Y2, Y3 is zero, meaning Y1, Y2, and Y3 are not independent.
Now, to find the pdf of Y1, we integrate out Y2 and Y3 from the joint pdf:
f(Y1) = ∫∫f(Y1, Y2, Y3) dY2 dY3
Since the joint pdf is zero, the integral will also be zero, indicating that Y1 does not have a valid pdf and is not a random variable.
In summary:
- The joint pdf of Y1, Y2, Y3 is 0, implying they are not independent.
- There is no valid pdf for Y1.