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: enter image description here

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:

enter image description here

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?

  • 1
    Did you already find that convolution? It surely depends on $\lambda$. – drhab Sep 14 at 13:49
  • No, I didn't find it myself, but from the answers here:math.stackexchange.com/questions/1439969/… it didn't seem that it depended on that, would you be able to explain a little more? – agblt Sep 14 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 $? – agblt Sep 14 at 13:53
  • 1
    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 at 14:16
  • 3
    1) $u$ is commonly used as the dummy variable in integration. It does not matter if you use any other variable, as long as it is different from the variable $a$, which is the argument of the resulting pdf. 2) $1$ is the indication function, $1_A(x) = 1$ when $x \in A$ and $0$ otherwise. 3) After putting in the definition, it help you to get the support of the integrand, by noting that $1_A1_B = 1_{A\cap B}$ – BGM Sep 15 at 3:53
up vote 0 down vote accepted
+150

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.

  • I realized my question makes no sense... – agblt Sep 16 at 23:36
  • Thanks for the guidance! – agblt Sep 17 at 14:02

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 ($X$,$Y$).

  • Thanks @rrv, I edited it to say $1≤a$ – agblt Sep 16 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 at 6:47

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