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?

Did you already find that convolution? It surely depends on $\lambda$. — drhab, Sep 14 '18 at 13:49
No, I didn't find it myself, but from the answers here:https://math.stackexchange.com/questions/1439969/finding-convolution-of-exponential-and-uniform-distribution-how-to-set-integral it didn't seem that it depended on that, would you be able to explain a little more? — dlp, Sep 14 '18 at 13:51
Would I have to consider four possibilities, two where $0 \le a \le 1$ depending on $λ $ and two where $1 \le a \le 2$ depending on $λ $? — dlp, Sep 14 '18 at 13:53
In general if $X,Y$ are independent then $f_{X+Y}(a)=\int f_X(a-u)f_Y(u);du$. Applying that here leads to $f_{X+Y}(a)=\int1_{[0,1]}(a-u)1_{[0,\infty)}(u)\lambda e^{-\lambda u}du=\int_{\max(0,1-a)}^{a}\lambda e^{-\lambda u}du$ for $a>0$. For $a<0$ the integrand is $0$. Especially the expression $\max(0,1-a)$ asks for discerning the cases $a>1$ and $a<1$. — drhab, Sep 14 '18 at 14:16
I always find it annoying when someone refers to "an exponential distribution with parameter $\alpha.$" Does that mean $\alpha$ is the rate or that $\alpha$ is the expectation? I.e. $$ \begin{align} & e^{-\alpha x} (\alpha,dx) \quad \text{for } x\ge0, \ \ \text{or } \quad & e^{-x/\alpha} (dx/\alpha) \quad \text{for } x\ge 0 \quad \text{?} \end{align} $$ Both conventions are used. $\qquad$ — Michael Hardy, Sep 14 '18 at 16:45
Here I am pretty sure it is the former. $\lambda e^{-\lambda y}$ — dlp, Sep 14 '18 at 16:48
@drhab, I am not familiar that notation you used for the integral, would you be able to explain how it works? - $1$. Is there a reason that $u$ is being used in the convolution rather than $x$ or $y$? $2$. How do those expressions in parentheses under the $1$'s work? $3$. How do they disappear and turn into the bounds of that last integral? — dlp, Sep 14 '18 at 16:59

score 1 · Accepted Answer · answered Sep 16 '18 at 14:00

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.

score 0 · Answer 2 · answered Sep 16 '18 at 14:08

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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 ($X$,$Y$).

answered Sep 16 '18 at 14:08

rrv

561

Thanks @rrv, I edited it to say $1≤a$ – dlp Sep 16 '18 at 15:50
@agblt. Please kindly note that your new version "I know that where $X+Y=a$ the solution involves two possibilities, one where $0≤a≤1$ and the other where $1≤a$." is still wrong. For any positive $a$ there are infinitely many possible values of a pair $(X,Y)$. – rrv Sep 17 '18 at 06:47

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

2 Answers2