Probability Question - Joint density function

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SUMMARY

The joint probability density function (JPDF) for the random variables Z and W, defined as Z = X + Y and W = X/(X + Y), can be derived using convolution for Z and transformation techniques for W. The density function for Z is calculated by convolving the individual densities f(x) = exp(-x)u(x) and f(y) = exp(-y)u(y). For W, the JPDF requires the use of Jacobians, which are derived from the transformations of Z and W, rather than simply summing the individual densities.

PREREQUISITES
  • Understanding of independent random variables and their probability density functions (PDFs).
  • Knowledge of convolution for finding the density of sums of random variables.
  • Familiarity with Jacobians in the context of transformation of variables in probability.
  • Basic principles of integration in probability theory.
NEXT STEPS
  • Study the convolution theorem for independent random variables in probability theory.
  • Learn about Jacobians and their application in transforming random variables.
  • Explore examples of deriving joint probability density functions from transformations.
  • Review the integration of joint densities over specified regions in probability space.
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Students and professionals in statistics, data science, and applied mathematics who are working with joint probability distributions and transformations of random variables.

cpatel23
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Before I begin, here is the question:

If the PDF of two independent random variables X and Y are:
f(x) = exp(-x)u(x)
f(y) = exp(-y)u(y)
Determine the join probability density function (JPDF) of Z&W defined by:
Z = X+Y
W = X/(X+Y).

So, I know how to solve this except for one thing. How do I get the expression for Z and W.
For Z do I literally just add f(x) + f(y) meaning z = exp(-x) + exp(-y) for (x,y) >0? Same with W?
Once I get the expressions I just find fxy(x,y) and divide by the determinant of the Jacobian. The problem is that the Jacobian depends on the derivative of Z and W which I do not know how to get an expression for.

Please help.
 
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The density function for Z is the convolution of the density functions for X and Y. For W it is much more complicated.
 
cpatel23 said:
Before I begin, here is the question:

If the PDF of two independent random variables X and Y are:
f(x) = exp(-x)u(x)
f(y) = exp(-y)u(y)
Determine the join probability density function (JPDF) of Z&W defined by:
Z = X+Y
W = X/(X+Y).

So, I know how to solve this except for one thing. How do I get the expression for Z and W.
For Z do I literally just add f(x) + f(y) meaning z = exp(-x) + exp(-y) for (x,y) >0? Same with W?
Once I get the expressions I just find fxy(x,y) and divide by the determinant of the Jacobian. The problem is that the Jacobian depends on the derivative of Z and W which I do not know how to get an expression for.

Please help.

The formula for the density of a sum ##X+Y## of two independent random variables with densities ##f_X(x)## and ##f_Y(y)## is found in every probability textbook, as well as on-line. If you don't know it you can work it out from first principles; one way is to get density ##f_Z(z)## from the fact that
f_Z(z) \, \Delta z \doteq P(z < X+Y < z + \Delta z) \;\text{as} \: \Delta z \to 0
and to integrate the joint density ##f_X(x) f_Y(y)## over the region ##\{z < x+y < z + \Delta z \}## in ##(x,y)-## space.

To get the joint density of ##(Z,W)##, use the standard transformation formulas that involve Jacobians, etc. The joint density of ##(X,Y)## is ##f_X(x) f_Y(y)##, and certainly does not involve a sum ##f_X(x) + f_Y(y)##, or anything like it.
 

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