Let $X,Y \sim \mathcal{U}([0,a])$ be independent identically continuously uniform distributed real random variables. Find the PDF of $Z := X + Y$.

I know that this can be accomplished via convolution: For $x \in \mathbb{R}$ we have \begin{align*} \int_{\mathbb{R}} f_{X}(x - y) f_{Y}(y) \ \text{dy} & = \frac{1}{(a - 0)^2} \int_{\mathbb{R}} 1_{[0,a]}(x - y) \cdot 1_{[0,a]}(y) \ \text{dy} \\ & = \frac{1}{a^2} \int_{0}^{a} 1 \ \text{dy} \cdot 1_{[0,2a]}(x) = \frac{1_{[0,2a]}(x)}{a}. \end{align*} I obtained the bounds like this. From the second integral in the first line we have $$ 0 \le x - y \le a \qquad \text{and} \qquad 0 \le y \le a. $$ Adding both inequalities yields $y \le x + y \le 2a + y$ or $$ 0 \le x \le 2 a. $$ Therefore $x < a$ and the bounded on the dy-integral are [0, a] and furthermore we know that $x \in [0,2a]$.

Is this (result and procedure) correct and / or is there an easier way to do this?

In this task I faced a similar challenge: Let $X,Y \sim \text{Exp}(\lambda)$ for some $\lambda > 0$ be independent and identically distributed. Find the density of $X + Y$.

\begin{align*} (f_{X} \ast f_{Y})(x) & = \int_{\mathbb{R}} f_{X}(x - y) f_{Y}(y) \ \text{dy} \\ & = \int_{\mathbb{R}} \lambda e^{-\lambda (x - y)} 1_{[0, \infty)}(x - y) \cdot \lambda e^{-\lambda y} 1_{[0, \infty)}(y) \ \text{dy} \\ & = \lambda^2 \int_{0}^{x} e^{-\lambda (x - y)} \cdot e^{-\lambda y} \ \text{dy} \cdot 1_{[0, \infty)}(x) = \lambda^2 \int_{0}^{x} e^{-\lambda x} \ \text{dy} \cdot 1_{[0, \infty)}(x) \\ & = \lambda^2 x \cdot e^{-\lambda x} \cdot 1_{[0, \infty)}(x). \end{align*}

Here the same question as above applies.


First part (uniform). If you want to use the CDF method to find $P(Z \le z)$ by integration, the following plots (made from a simulation), may help you find the limits on integrals. I used $a = 2.$

The left-hand plot shows the region of integration for $P(Z \le 1.5).$ For values of $z$ above $2$ it seems easiest to break the integral into two parts. [However, because the joint distribution is uniform, this can be a high school geometry/algebra problem, if you like.]

The middle plot shows the empirical CDF (jump of $1/10^5$ at each of $10^5$ simulated values of $Z),$ which suggests the form of the CDF you'll get.

The right-hand plot shows a histogram of $10^5$ simulated values of $Z.$ [A discrete analogue of this is the 'triangular' PDF from the sum upon rolling two fair dice.]

enter image description here

Second part (exponential). If you know about moment generating functions, that's probably the easiest way to show that the sum of two independent exponential random variables, each with rate $\lambda$ is $\mathsf{Gamma}(\text{shape}=2,\text{rate}=\lambda).$ However, the PDF and convolution methods also work. [Perhaps see Wikipedia on gamma distributions to check your answer.]

Another simulation to indicate the result:

set.seed(708);  m = 10^6
x = rexp(m, .1); y = rexp(m, .1);  z = x+y
hist(z, prob=T, br=50, col="skyblue2")
 curve(dgamma(x, 2, .1), add=T, lwd=2, col="red")

enter image description here

Notes: (a) This exponential relationship is much used in applications, including queueing theory. [Q. A person is next in line for a bank teller whose service times are exponential with rate one every five minutes, what is the probability that person finishes being served within 8 minutes? A. In R, code pgamma(8, 2, 1/5) returns 0.4750691.]

(b) If the two exponential random variables have different rates, the problem is very much messier. (If interested, perhaps google 'sum of exponentials different rates'.)

(c) In case you want it, the R code for making the figure is provided below:

 m = 30000
 x = runif(m, 0, 2);  y = runif(m, 0, 2)
 plot(x, y, pch=".")
  event = (x + y < 1.5)
   points(x[event], y[event], pch=".", col="green2")
m = 10^5
 X = runif(m, 0, 2);  Y = runif(m, 0, 2)
 Z = X+Y
 hist(X+Y, prob=T, br=50, col="skyblue2")
  • $\begingroup$ Well, from the middle plot for the uniform I get that the CDF is non-linear, but when I integrate my PDF I get that $\mathbb{P}(Z \le x) = \frac{x}{a}$ for $0 \le x \le 2a$, which 1) is linear and 2) returns 2 for $x = 2a$, which isn't possible. Where is my mistake? Also, my CDF and PDF for $Z$ then coincide with the CDF / PDF of $X$ (or $Y$) they are just on an interval twice the size, which seems wrong, since they have to sum up ( integrate) to 1 and not 2. $\endgroup$ – Ramanujan Jul 9 '19 at 11:30
  • $\begingroup$ As a note, unfortunately we haven't covered moment-generating functions, just characteristic functions. $\endgroup$ – Ramanujan Jul 9 '19 at 12:18
  • $\begingroup$ Characteristic functions will do the same job. ChFs require at least a minimal knowledge of complex numbers and exist for all dist'ns. MGFs are a bit simpler to use, but do not exist for all dist'ns. $\endgroup$ – BruceET Jul 9 '19 at 18:03
  • $\begingroup$ Can you show me how? Also I'd like to know if my results are correct. $\endgroup$ – Ramanujan Jul 9 '19 at 19:14
  • 1
    $\begingroup$ Product of MGFs is MGF of sum. MGF of $EXP(rate=\lambda)$ is $\frac{\lambda}{\lambda - t}$ for $t > \lambda).$ MGF of $GAMMA(2, \lambda)$ is $(\frac{\lambda}{\lambda - t})^2.$ See Wikipedia. $\endgroup$ – BruceET Jul 9 '19 at 19:22

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