Suppose $X,Y$ are independent $Uniform(-1,1)$ random variables. Determine the distribution of $Z=X-Y$.
I do not really think I should add my work here because whatever I have tried until now, has turned out to be extremely complicated in terms of inequalities. The main difficulty of this problem I would say, lies in carving out the appropriate regions in which certain probabilities are non-zero. But anyway, I am sketching out my method which is apparently a failure.
$P(X-Y<t)=\int_{y}P(X<t+y)f_Y(y)dy$
It is to be observed that if $t+y<-1$ then $P(X<t+y)=0$. So we need $t+y>-1$ i.e. $y>-1-t$. But $y>-1$ (otherwise $f_Y(y)=0$ ) so $y>\max\{-1,-1-t\}$. So we have got the lower bound for $y$.
If $t+y>1$ i.e. if $y>1-t$ then $P(X<t+y)=1$ so for this case, $y>\max\{-1,-1-t,1-t\}$. But, $-1-t<1-t$ hence we need just $y>\max\{-1,1-t\}$.
After this point, I gave up. There are too many inequalities and things happening. It is just too hard to keep track of which region I am working with.
Is there a better method? If yes, I would appreciate it. If not, please comment and I have to proceed with this tiresome logic.
1) Draw the square $[-1,1]\times [-1,1]$. in the cartesian plane. The value of $X$ is on the $x$ axis, $Y$ on the $Y$ axis.
2) Asking the probability of $Z\leq s$ is equivalent to finding the area under the the line $x-y=s$, bounded by the square.
3) Find this area using simple geometry.