Let $Y$ be an exponential random variable with parameter $λ$ and $X$ be a uniform random variable on $[0,1]$ independent of $Y$. Find the probability density function of $X+Y$:
I know that where $X+Y=a$ the solution involves two possibilities, one where $0\le a \le1$ and the other where $1\le a $.
Here the purple line is $X+Y=a$ and the red curve is the exponential random variable. As you can see, $a$ starts after the red curve.
Another possibility is this:
Here $a$ starts before the red curve.
Why would not have to consider four possibilities, two where $0≤a≤1$ depending on the positioning of $a$ to to the exponential random variable, and two where $1≤a$ depending on the positioning of $a$ to the exponential random variable?