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I am trying to solve the problems about finding $X+Y+Z$ where $X, Y, Z$ are uniformly distributed random variables, each in the interval of $[0,1]$. (please reference: Finding the distribution of the sum of three independent uniform random variables)

I understand most of the derivations, but the only part I do not understand is from the point where $0 \leq s \leq 1,$ and $w-1 \leq s \leq w$.

The solutions that I have seen jump directly to divide $w$ into three different situations:

  1. $w < 0$ and $w >3$
  2. $1<w < 2$
  3. $2 < w < 3$

Would anyone please show me how I can know the ranges of values to consider by looking at the two inequalities? $$(0 \leq s \leq 1,\text{ and }w-1 \leq s \leq w)$$

To clarify my confusion, I started with s = 0, and I will get 0 < = w <= 1.

Then, I let s = 1, and I get 1 < = w <= 2.

Comparing with the solution, the interval of [0,1] is not considered. Also, I have no way of getting the interval of (-infinity, 0) and (3, infinity) since I have no way of knowing 3 is a value that I should consider.

What have I missed in my analysis?

Amos Ku
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    you already knows that the density of $X+Y+Z$, assuming they are independent of each other, is generated by convolution of it individual densities? – Masacroso Jan 22 '20 at 08:21
  • What is $v$? And you seem to be missing $0 \lt w \lt 1$ – Henry Jan 22 '20 at 09:03
  • My apology. My textbook uses V while the forum uses S. I have already update my question above. – Amos Ku Jan 23 '20 at 02:20
  • They have used the change of variable method in integration. See the argument of function has changed from $w-u$ to $u$. Hence limits are updated accordingly. – AxyuS Jul 05 '20 at 07:36

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