How Do You Find the PDF of Z=X+Y When X and Y Are Not Independent?

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To find the PDF of Z=X+Y when X and Y are not independent, the joint PDF f(x,y) is given as f(x,y) = (1/x) for 0≤y≤x≤1. The cumulative distribution function F(z) is defined as F(z) = P(Z≤z) = ∫∫fXY(x,y) dx dy with specific limits. The differentiation to find the PDF f(z) involves using Leibnitz's rule, but the integral does not converge, indicating potential issues with the validity of f(x,y) as a PDF. The discussion highlights the complexities of working with dependent random variables and suggests that the problem may be more suited for a homework forum.
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f(x,y) = (1/x) for 0≤y≤x≤1

A new rv Z=X+Y where X,Y not independent find the pdf of z

My approach

F(z) = P(Z≤z) = ∫∫fXY(x,y) dx dy x= -∞ to ∞ y= 0 to z-y

f(z) = d/dz(F(z)) = ∫fXY(z-y,y) dy y= -∞ to ∞ (using Leibnitz)

where i am stuck is this doesn't converge
 
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(this thread belongs in the homework forums)

f(x,y) might not be a valid pdf
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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