Convolution of exponential and uniform distribution-why is there not four possibilities to consider

Question

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 $.

My question is, why is it not also dependent upon where the exponential random variable intersects $X+Y=a$? I will illustrate with a diagram:

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?

Answer 1

You have two piecewise functions, one nonzero on $[0,\infty)$ and the other nonzero on $[0,1]$. This is all that matters. When sliding the uniform $\text{pdf}$, you have three regimes: $[0,1]$ before $0$, $[0,1]$ over $0$, and $[0,1]$ after $0$. The first regime can be trivially ignored, and in fact there are just two cases to be considered.

The valuations of the $\text{pdf}$ are not relevant in the case analysis.

Answer 2

I know that where $X+Y=a$ the solution involves two possibilities, one where $0≤a≤1$ and the other where $1≤a≤2$.

This is wrong. For any positive value of $a$ there is a corresponding pair of values ($X$,$Y$).