Question. Suppose $X \sim U([1,3])$ and $Y \sim U([1,2] \cup [4,5])$ are two independent random variables (but obviously not identically distributed). Find the pdf of $X + Y$.

So far. I'm familiar with the theoretical mechanics to set up a solution. So, if we let $\lambda$ be the Lebesgue measure and notice that $[1,2]$ and $[4,5]$ disjoint, then the pdfs are

$$f_X(x) = \begin{cases} \frac{1}{2}, &x \in [1,3] \\ 0, &\text{otherwise} \end{cases} \quad\text{and}\quad f_Y(y) = \begin{cases} \frac{1}{\lambda([1,2] \cup [4,5])} = \frac{1}{1 + 1} = \frac{1}{2}, &y \in [1,2] \cup [4,5] \\ 0, &\text{otherwise} \end{cases} $$

Now, let $Z = X + Y$. Then, the pdf of $Z$ is the following convolution $$f_Z(t) = \int_{-\infty}^{\infty}f_X(x)f_Y(t - x)dx = \int_{-\infty}^{\infty}f_X(t -y)f_Y(y)dy.$$

To me, the latter integral seems like the better choice to use. So, we have that $f_X(t -y)f_Y(y)$ is either $0$ or $\frac{1}{4}$. But I'm having some difficulty on choosing my bounds of integration?

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    $\begingroup$ If you draw a suitable picture, the pdf should be instantly obvious ... and you'll also get relevant information about what the bounds would be for the integration $\endgroup$ – Glen_b Sep 26 at 6:13
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    $\begingroup$ Does this answer your question? general solution sum of two uniform random variables aY+bX=Z? $\endgroup$ – Xi'an Sep 26 at 7:18
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    $\begingroup$ I find it convenient to conceive of $Y$ as being a mixture (with equal weights) of $Y_1,$ a Uniform$(1,2)$ distribution, and $Y_,$ a Uniform$(4,5)$ distribution. Thus $X+Y$ is an equally weighted mixture of $X+Y_1$ and $X+Y_2.$ $\endgroup$ – whuber Sep 26 at 16:50
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    $\begingroup$ I was hoping for perhaps a cleaner method than strictly plotting. Consider if the problem was $X \sim U([1,5])$ and $Y \sim U([1,2] \cup [4,5] \cup [7,8] \cup [10, 11])$. It becomes a bit cumbersome to draw now. Using the comment by @whuber, I believe I arrived at a more efficient method to reach the solution. $\endgroup$ – EzioBosso Sep 26 at 21:29

Here is a plot as suggested by comments

enter image description here

What I was getting at is it is a bit cumbersome to draw a picture for problems where we have disjoint intervals (see my comment above). It's not bad here, but perhaps we had $X \sim U([1,5])$ and $Y \sim U([1,2] \cup [4,5] \cup [7,8] \cup [10, 11])$.

Using @whuber idea: We notice that the parallelogram from $[4,5]$ is just a translation of the one from $[1,2]$. So, if we let $Y_1 \sim U([1,2])$, then we find that

$$f_{X+Y_1}(z) = \begin{cases} \frac{1}{4}z - \frac{1}{2}, &z \in (2,3) \tag{$\dagger$}\\ \frac{1}{2}z - \frac{3}{2}, &z \in (3,4)\\ \frac{5}{4} - \frac{1}{4}z, &z \in (4,5)\\ 0, &\text{otherwise} \end{cases} $$

Since, $Y_2 \sim U([4,5])$ is a translation of $Y_1$, take each case in $(\dagger)$ and add 3 to any constant term. Then you arrive at ($\star$) below.

Brute force way:

  • $\mathbf{2 < z < 3}$: $y=1$ to $y = z-1$, which gives $\frac{1}{4}z - \frac{1}{2}$.
  • $\mathbf{3 < z < 4}$: $y=1$ to $y = z-1$, such that $2\int_1^{z-1}\frac{1}{4}dy = \frac{1}{2}z - \frac{3}{2}$.
  • $\mathbf{4 < z < 5}$: $y=z-3$ to $y=2$, which gives $\frac{5}{4} - \frac{1}{4}z$.
  • $\mathbf{5 < z < 6}$: $y=4$ to $y = z-1$, which gives $\frac{1}{4}z - \frac{5}{4}$.
  • $\mathbf{6 < z < 7}$: $y = 4$ to $y = z-2$, such that $2\int_4^{z-2}\frac{1}{4}dy = \frac{1}{2}z - 3$.
  • $\mathbf{7 < z < 8}$: $y = z-3$ to $y=5$, which gives $2 - \frac{1}{4}z$.


$$f_Z(z) = \begin{cases} \frac{1}{4}z - \frac{1}{2}, &z \in (2,3) \tag{$\star$}\\ \frac{1}{2}z - \frac{3}{2}, &z \in (3,4)\\ \frac{5}{4} - \frac{1}{4}z, &z \in (4,5)\\ \frac{1}{4}z - \frac{5}{4}, &z \in (5,6)\\ \frac{1}{2}z - 3, &z \in (6,7)\\ 2 - \frac{1}{4}z, &z \in (7,8)\\ 0, &\text{otherwise} \end{cases}$$

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    $\begingroup$ +1 For more methods of solving this problem, see stats.stackexchange.com/a/43075/919. $\endgroup$ – whuber Sep 26 at 21:31
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    $\begingroup$ Thank you for the link! It's too bad there isn't a sticky section, which contains questions that contain answers that go above and beyond what's required (like yours in the link). $\endgroup$ – EzioBosso Sep 26 at 21:42

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