Finding PDF of Z: X_1, X_2 Exponential RVs

In summary, the conversation discusses finding the PDF of a variable Z, which is a sum and product of two independent and identically distributed exponential random variables. The speaker wants to avoid finding the PDF through differentiation and suggests using convolutions. Another speaker suggests using the f distribution and deriving the PDF directly through convolution. The conversation ends with a discussion on finding the joint PDF of two variables, W and Z, using Jacobian transformation.
  • #1
EngWiPy
1,368
61
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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  • #2
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).
 
  • #3
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:
  • #4
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:
  • #5
You can derive the pdf directly through convolution.
 
  • #6
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
 
  • #7
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?
 

Related to Finding PDF of Z: X_1, X_2 Exponential RVs

1. What is the PDF of Z if X1 and X2 are independent exponential random variables?

The PDF of Z is given by fZ(z) = λ2e-λz, where λ is the rate parameter for the exponential distribution and z is the value of the random variable Z.

2. How do you find the mean of Z if X1 and X2 are independent exponential random variables?

The mean of Z can be found by taking the sum of the mean of X1 and X2, since the exponential distribution is memoryless and the mean is equal to the inverse of the rate parameter. Therefore, the mean of Z is 2/λ.

3. What is the relationship between the variance of Z and the variances of X1 and X2?

The variance of Z is equal to the sum of the variances of X1 and X2 since the random variables are independent. This can be represented as Var(Z) = Var(X1) + Var(X2) = 2/λ2.

4. Can the PDF of Z be used to find the probability of a specific value of Z?

Yes, the probability of a specific value of Z can be found by plugging in the desired value of z into the PDF function fZ(z). This will give the probability density at that specific value of Z.

5. How can you use the PDF of Z to find the cumulative distribution function (CDF) of Z?

The CDF of Z can be found by integrating the PDF of Z from 0 to the desired value of Z. This will give the probability that Z is less than or equal to that value. The CDF can also be expressed as FZ(z) = 1 - e-λz.

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