Finding PDF of Z: X_1, X_2 Exponential RVs

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Discussion Overview

The discussion revolves around finding the probability density function (PDF) of the random variable Z, defined as Z = X_1 + X_2 + X_1X_2, where X_1 and X_2 are independent and identically distributed exponential random variables. The participants explore various methods to derive the PDF, including convolution and joint distributions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the event {Z = z} can be expressed as {X1 = (z - X2)/(1 + X2)}, proposing the use of convolutions to find the PDF.
  • Another participant expresses a desire to find the PDF directly without differentiating the cumulative distribution function (CDF), indicating that the random variables are not actually exponential, which complicates the situation.
  • A different participant mentions the possibility of checking the F distribution for the PDF, noting its complexity.
  • One participant asserts that the PDF can be derived directly through convolution and discusses the need to find the joint PDF of W and Z, where W = X_1X_2 and Z = X_1 + X_2, using a Jacobian transformation.
  • Another participant suggests writing X1 in terms of X2 and comparing the problem to a wiki example involving normal distribution.

Areas of Agreement / Disagreement

Participants express differing views on the methods to derive the PDF, with some advocating for convolution while others seek alternative approaches. No consensus is reached on the best method to find the PDF of Z.

Contextual Notes

Participants note that the complexity arises from the nature of the random variables involved, and there are unresolved mathematical steps related to the joint PDF and Jacobian transformation.

EngWiPy
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Hello,

Suppose that:

[tex]Z=X_1+X_2+X_1X_2[/tex]

where [tex]X_i[/tex] for i=1 and 2 are independent and identically distribuited exponential RVs.

can we find the PDF of Z?

Regards
 
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You should note that the event {Z = z} is equivalent to {X1 + X2 + X1 X2 = z} or {X1 = (z - X2)/(1 + X2)}. You can use this and use convolutions (example).
 
EnumaElish said:
You should note that the event {Z = z} is equivalent to {X1 + X2 + X1 X2 = z} or {X1 = (z - X2)/(1 + X2)}. You can use this and use convolutions (example).

Right, But I want to find the PDF directly, not from differentiating the CDF, if possible. Because these RVs are, actually, not exponentials, but I said so to simplify the problem statement. So I want to avoid the derivative operation, which complicates the whole stituation.

I say the following:

let [tex]W=X_1+X_2[/tex] and [tex]Y=X_1X_2[/tex], then [tex]Z=W+Y[/tex]. But we need to evaluate joint PDF of W and Y. Is this approach in the right way?
 
Last edited:
saeddawoud said:
Hello,

Suppose that:

[tex]Z=X_1+X_2+X_1X_2[/tex]

where [tex]X_i[/tex] for i=1 and 2 are independent and identically distribuited exponential RVs.

can we find the PDF of Z?

Regards

You may want to check the f distribution. The PDF is a bit complicated and I don't have Latex, but you can look it up.
 
Last edited:
You can derive the pdf directly through convolution.
 
EnumaElish said:
You can derive the pdf directly through convolution.

If we assume that [tex]Y=W+Z[/tex] where [tex]W=X_1X_2[/tex] and [tex]Z=X_1+X_2[/tex], then we need to find the joint PDF [tex]f_{W,Z}(w,z)[/tex], which can be found using Jacobian transformation.

If we proceed using this, we have:

[tex]X_1=T_1^{-1}=\frac{W+Z-X_2}{1+X_2}[/tex] and [tex]X_2=T_2^{-1}=\frac{W+Z-X_1}{1+X_1}[/tex]

Then

[tex]F_{W,Z}(w,z)=f_{X_1,X_2}(x_1=T_1^{-1},x_2=T_2^{-1})|J|[/tex]

where [tex]|J|[/tex] is the magnitude of the Jacobian which will be zero in this case!

Is here anything wrong I did?

Regards
 
You can write X1 as (z - X2)/(1 + X2). Then study the wiki example with normal distribution. How is that example similar to your problem?
 

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