How my professor made the following conclusion?

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:

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.

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

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 $$z \geq -b$$.

• The red inequality implies that both $0 \leq y$ and $y \leq z+b$ and so $0 \leq z+b$.
– kccu
Jan 10, 2021 at 19:32
• 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: 𝑦≥0 with the inequality 𝑧+𝑏≥𝑦 while they are many many more combinations on the input. Jan 10, 2021 at 19:34
• 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 $f_{-X}(z-y)=0 \neq \frac{1}{b-a}$.
– kccu
Jan 10, 2021 at 19:42
• "and this formula is only valid when 𝑧≥−𝑏 and 𝑧≤−𝑎." 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 𝑧≤−𝑎 Jan 10, 2021 at 20:00
• "Put another way, when 𝑧<−𝑏 we have 𝑓−𝑋(𝑧−𝑦)=0" why this is correct too? why z-y isn't in [a,b] in this case? Jan 10, 2021 at 20:41