Finding PDF of Z: X_1, X_2 Exponential RVs

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Hello,

Suppose that:

Z=X_1+X_2+X_1X_2

where X_i 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 W=X_1+X_2 and Y=X_1X_2, then Z=W+Y. 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:

Z=X_1+X_2+X_1X_2

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

If we proceed using this, we have:

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

Then

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

where |J| 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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