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Update: The suggested Answer didn't solve the problem

Note: This question can be solved without knowing probability at all. (all you need is the bold text)

Today my lecturer wrote the following on board:

enter image description here

where the text in red specefies for which values the multiplication under integral sign isn't zero.

I understand that we need to split into conditions:

  1. where $\max\{0,z+a\}=0$ which means: $z<-a$

  2. where $\max\{0,z+a\}=z+a$ which means: $z>-a$

but in the first condition where did we get $-b<=z$ from? I understand that without it something will be wrong since for all values the integral will not be 0 (while we know from the text in red that for some it's zero for sure) but I don't understand where it specifically came from...

I have been thinking on this for hours.

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    $\begingroup$ That strange word in bottom right means "else" $\endgroup$ Jan 10, 2021 at 19:15
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    $\begingroup$ So, $Y$ and $X$ are exponential and uniform rv.s. And you are looking for the distribution of their difference? $\endgroup$
    – user140541
    Jan 10, 2021 at 19:23
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    $\begingroup$ Same recent questions: math.stackexchange.com/q/3979063/321264, math.stackexchange.com/q/3975018/321264 $\endgroup$ Jan 10, 2021 at 19:28
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    $\begingroup$ @StubbornAtom how it's the same? did you read my question? Here I am asking where the b came from $\endgroup$ Jan 10, 2021 at 19:29
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    $\begingroup$ what are the definitions of $f_\text{y}$ and $f_{-\text{x}}$? The $b$ has to come from the support of $f_{-\text{x}}$. $\endgroup$
    – robjohn
    Jan 10, 2021 at 22:02

1 Answer 1

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It looks like $Y \sim \text{Exp}(\lambda)$ and $X \sim \text{Unif}(a,b)$. In that case $Y \geq 0$ and $a \leq X \leq b$. So $Y-X \geq -b$. Thus $f_{Y-X}(z)=0$ when $z <-b$.

Another way to think about it is combining the inequality $y \geq 0$ with the inequality $z+b \geq y$ gives $z+b \geq 0$, so $z \geq -b$.

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    $\begingroup$ The red inequality implies that both $0 \leq y$ and $y \leq z+b$ and so $0 \leq z+b$. $\endgroup$
    – kccu
    Jan 10, 2021 at 19:32
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    $\begingroup$ But if you look at 1 and 2 which I wrote they don't include this requirement (the one you said) my main point is why you took specially: 𝑦β‰₯0 with the inequality 𝑧+𝑏β‰₯𝑦 while they are many many more combinations on the input. $\endgroup$ Jan 10, 2021 at 19:34
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    $\begingroup$ I'm not sure I follow. In the case $z \leq -a$, i.e., $\max(0,z+a)=0$, the inequality in red simplifies to $0 \leq y \leq z+b$. Thus the integral in $y$ goes from $y=0$ to $y=z+b$, and this formula is only valid when $z \geq -b$ and $z \leq -a$. Put another way, when $z<-b$ we have $f_{-X}(z-y)=0 \neq \frac{1}{b-a}$. $\endgroup$
    – kccu
    Jan 10, 2021 at 19:42
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    $\begingroup$ "and this formula is only valid when 𝑧β‰₯βˆ’π‘ and π‘§β‰€βˆ’π‘Ž." this is the problem for me how did you conclude this. what is "this formula" and why it's valid only for those 2 cases? The only requirement we had in this path was π‘§β‰€βˆ’π‘Ž $\endgroup$ Jan 10, 2021 at 20:00
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    $\begingroup$ "Put another way, when 𝑧<βˆ’π‘ we have π‘“βˆ’π‘‹(π‘§βˆ’π‘¦)=0" why this is correct too? why z-y isn't in [a,b] in this case? $\endgroup$ Jan 10, 2021 at 20:41

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