Help on the density of sum of two uniform variables.

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To calculate the density function of Z=X+Y, where X and Y are independent uniform variables on [0,1], the correct approach involves integrating over the appropriate domains. For 0<z<1, the density function f(z) equals z. For 1<z<2, the density function is f(z)=2-z, as the integration limits for x should be from z-1 to 1. The initial mistake was in misidentifying the integration limits for the second case. Understanding the relationship between z, x, and y is crucial for accurate calculations.
gimmytang
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Hi, I need to calculate the density function of Z=X+Y, where X and Y are independent uniform distributed on [0,1]. The calculation is in the following:
f_{Z}(z)=\int_{A}dx
a. If 0<z<1, A={x:0<x<z} then f(z) = z;
b. If 1<z<2, A={x:0<x<1} then f(z) = 1;
Step b is wrong, but I don't know where I am wrong. Any hint will be appreciated!
Thanks
gim :cry:
 
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For step b, the domain of x is z-1 to 1, so f(z)=2-z.

The reason for that is z-x=y, which is restricted to (0,1).
 
yep, you are right. thanks!
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

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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