Joint problem density function problem

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The discussion revolves around understanding the joint probability density function f(x,y) = e^{-x} for the range 0≤x≤y≤∞. Participants clarify that the density function is indeed a function of both variables x and y, and emphasize the importance of correctly determining the range for integration. There is a consensus that to find the value of A, the total probability must equal 1, leading to the equation A + e^{-A} = 2. Once A is determined, participants confirm that it will be possible to calculate P(x+y≤1). The conversation highlights the need for precise interpretation of the density function and its implications for probability calculations.
Lewis7879
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I need help guys I can't understand this
Can anyone explain thoroughly how do I form the range for this question?
f(x,y)= e-x for 0≤x≤y≤∞
0 Otherwise

Find P(x+y≤1)
I attempted this by integrating through the range of
0≤y≤(1-x) and 0≤x≤∞ but that doesn't seem right
 
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The statement is confusing. You appear to have a density function in x and y, which is a function of x only.
 
mathman said:
The statement is confusing. You appear to have a density function in x and y, which is a function of x only.
Hello mathman there's a slight error with range I made in the question which is 0≤y≤x≤∞
There was no other problems with the question as I was asked this way.
 
No, f(x,y)= e^{-x} for 0\le x\le y\le A is a function of both x and y. To determine "A", use the fact that the "total" probability must be 1:
\int_{y= 0}^A\int_{x= 0}^y e^{-x} dx dy= 1
 
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Getting an equation for A is easy enough. A+e^{-A}=2. I am confused as to what is the question.
 
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HallsofIvy said:
No, f(x,y)= e^{-x} for 0\le x\le y\le A is a function of both x and y. To determine "A", use the fact that the "total" probability must be 1:
\int_{y= 0}^A\int_{x= 0}^y e^{-x} dx dy= 1
Will I be able to find P(x+y≤1) after determine A?
 
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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