# Why is the sum of two random variables a convolution?

For long time I did not understand why the "sum" of two random variables is their convolution, whereas a mixture density function sum of $$f(x)$$ and $$g(x)$$ is $$p\,f(x)+(1-p)g(x)$$; the arithmetic sum and not their convolution. The exact phrase "the sum of two random variables" appears in google 146,000 times, and is elliptical as follows. If one considers an RV to yield a single value, then that single value can be added to another RV single value, which has nothing to do with convolution, at least not directly, all that is is a sum of two numbers. An RV outcome in statistics is however a collection of values and thus a more exact phrase would be something like "the set of coordinated sums of pairs of associated individual values from two RV's is their discrete convolution"...and can be approximated by the convolution of the density functions corresponding to those RV's. Even simpler language: 2 RV's of $$n$$-samples are in effect two n-dimensional vectors that add as their vector sum.

Please show the details of how the sum of two random variables are a convolution and a sum.

• I don't really believe that it is 'sum' in an abstract algebraic sense. When we make a 'sum of variables' then we refer to the typical arithmetic operation as we know when adding natural numbers or real numbers. That means that we make a new variable by 'adding' the other variables together. The notion of 'a sum of variables' also exist outside the realm of statistics and is independent from the expressions about convolutions and probabilities. So, indeed 'the sum of variables is a convolution', is wrong. But nobody is implying this. We should change the word 'is' in that statement. – Martijn Weterings Mar 6 '18 at 20:37
• This is like arguing that $f(x) \cdot g(x)$ should not be called 'the product of two functions f and g' (or only interpreted as some abstract algebraic notion of 'product') because it is a convolution in terms of the Fourier transforms of those functions. – Martijn Weterings Mar 6 '18 at 20:51
• The "notice" is misleading. A sum of random variables $X$ and $Y$ is meant in precisely the same sense "sum" is understood by schoolchildren: for each $\omega$, the value $(X+Y)(\omega)$ is found by adding the numbers $X(\omega)$ and $Y(\omega).$ There's nothing abstract about it. These RVs have distributions. There exist many ways to represent the distributions. The distribution function of $X+Y$ is the convolution of the DFs of $X$ and $Y$; the characteristic function of $X+Y$ is the product of their CFs; the cumulant generating function of $X+Y$ is the sum of their CGFs; and so on. – whuber Mar 6 '18 at 22:01
• I don't see either random variables or distributions in your calculation. – whuber Mar 7 '18 at 0:21
• In the language of my post at stats.stackexchange.com/a/54894/919, a pair of random variables $(X,Y)$ consists of a box of tickets on each of which are written two numbers, one designated $X$ and the other $Y.$ The sum of these random variables is obtained by adding the two numbers found on each ticket. The computation literally is a task you could assign to a third-grade classroom. (I make this point to emphasize both the fundamental simplicity of the operation as well as showing how strongly it is connected with what everybody understands a "sum" to mean.) – whuber Mar 7 '18 at 14:35

Convolution calculations associated with distributions of random variables are all mathematical manifestations of the Law of Total Probability.

In the language of my post at What is meant by a “random variable”?,

A pair of random variables $$(X,Y)$$ consists of a box of tickets on each of which are written two numbers, one designated $$X$$ and the other $$Y$$. The sum of these random variables is obtained by adding the two numbers found on each ticket.

I posted a picture of such a box and its tickets at Clarifying the concept of sum of random variables.

This computation literally is a task you could assign to a third-grade classroom. (I make this point to emphasize both the fundamental simplicity of the operation as well as showing how strongly it is connected with what everybody understands a "sum" to mean.)

How the sum of random variables is expressed mathematically depends on how you represent the contents of the box:

The first two of these are special insofar as the box might not have a pmf, pdf, or mgf, but it always has a cdf, cf, and cgf.

To see why convolution is the appropriate method to compute the pmf or pdf of a sum of random variables, consider the case where all three variables $$X,$$ $$Y,$$ and $$X+Y$$ have a pmf: by definition, the pmf for $$X+Y$$ at any number $$z$$ gives the proportion of tickets in the box where the sum $$X+Y$$ equals $$z,$$ written $$\Pr(X+Y=z).$$

The pmf of the sum is found by breaking down the set of tickets according to the value of $$X$$ written on them, following the Law of Total Probability, which asserts proportions (of disjoint subsets) add. More technically,

The proportion of tickets found within a collection of disjoint subsets of the box is the sum of the proportions of the individual subsets.

It is applied thus:

The proportion of tickets where $$X+Y=z$$, written $$\Pr(X+Y=z),$$ must equal the sum over all possible values $$x$$ of the proportion of tickets where $$X=x$$ and $$X+Y=z,$$ written $$\Pr(X=x, X+Y=z).$$

Because $$X=x$$ and $$X+Y=z$$ imply $$Y=z-x,$$ this expression can be rewritten directly in terms of the original variables $$X$$ and $$Y$$ as

$$\Pr(X+Y=z) = \sum_x \Pr(X=x, Y=z-x).$$

That's the convolution.

### Edit

Please note that although convolutions are associated with sums of random variables, the convolutions are not convolutions of the random variables themselves!

Indeed, in most cases it is not possible to convolve two random variables. For this to work, their domains have to have additional mathematical structure. This structure is a continuous topological group.

Without getting into details, suffice it to say that convolution of any two functions $$X, Y:G \to H$$ must abstractly look something like

$$(X\star Y)(g) = \sum_{h,k\in G\mid h+k=g} X(h)Y(k).$$

(The sum could be an integral and, if this is going to produce new random variables from existing ones, $$X\star Y$$ must be measurable whenever $$X$$ and $$Y$$ are; that's where some consideration of topology or measurability must come in.)

This formula invokes two operations. One is the multiplication on $$H:$$ it must make sense to multiply values $$X(h)\in H$$ and $$Y(k)\in H.$$ The other is the addition on $$G:$$ it must make sense to add elements of $$G.$$

In most probability applications, $$H$$ is a set of numbers (real or complex) and multiplication is the usual one. But $$G,$$ the sample space, often has no mathematical structure at all. That's why the convolution of random variables is usually not even defined. The objects involved in convolutions in this thread are mathematical representations of the distributions of random variables. They are used to compute the distribution of a sum of random variables, given the joint distribution of those random variables.

### References

Stuart and Ord, Kendall's Advanced Theory of Statistics, Volume 1. Fifth Edition, 1987, Chapters 1, 3, and 4 (Frequency Distributions, Moments and Cumulants, and Characteristic Functions).

• Associativity with scalar multiplication from algebraic properties relates that $$a ( f ∗ g ) = ( a f ) ∗ g$$ for any real (or complex) number $a$. Whereas one nice property is that the convolution of two density functions is a density function, one is not restricted to convolving density functions, and convolution is not in general a probability treatment, sure it can be, but it can be a time series treatment, e.g., a treatment of water runoff in lakes after a rainfall, a drug concentration model following dosing, etc. – Carl Feb 14 at 21:08
• @Carl How does that comment comport with your original question, which asks about sums of random variables? At best it is tangential. – whuber Feb 14 at 21:45
• I am asking you to not overgeneralize. To begin a sentence with "convolution is" without saying "convolution of RV's is" is elliptic. My whole problem here was with the elliptic notation. Vector addition of two $n$-space vectors is convolution, whether or not those vectors are normalized. If they are normalized, they need not be probabilities, That is the whole truth, not just part of it. – Carl Feb 14 at 21:59
• Thank you: I will clarify the first sentence to emphasize that I am answering your question. – whuber Feb 14 at 22:03
• New addition is true for convolution of RV's, which is technically what I asked. And perhaps I am equivocating but convolution is not always of RV's but can always be reduced to some scale factors of density functions times those density functions, where the scalars are multiplicative and where the density functions are sometimes RV's, in which case the scale factors are the multiplicative identity, i.e., 1. – Carl Apr 6 at 19:37

# Notation, upper and lower case

https://en.wikipedia.org/wiki/Notation_in_probability_and_statistics

• Random variables are usually written in upper case roman letters: $X$, $Y$, etc.
• Particular realizations of a random variable are written in corresponding lower case letters. For example $x_1$, $x_2$, …, $x_n$ could be a sample corresponding to the random variable $X$ and a cumulative probability is formally written $P ( X > x )$ to differentiate random variable from realization.

$Z=X+Y$ means $z_i=x_i+y_i \qquad \forall x_i,y_i$

# Mixture of variables -> sum of pdf's

https://en.wikipedia.org/wiki/Mixture_distribution

You use a sum of the probability density functions $f_{X_1}$ and $f_{X_2}$ when the probability (of say Z) is a defined by a single sum of different probabilities.

For example when $Z$ is a fraction $s$ of the time defined by $X_1$ and a fraction $1-s$ of the time defined by $X_2$, then you get $$\mathbb{P}(Z=z) = s \mathbb{P}(X_1=z) + (1-s) \mathbb{P}(X_2=z)$$ and $$f_Z(z) = s f_{X_1}(z) + (1-s) f_{X_2}(z)$$

. . . . an example is a choice between dice rolls with either a 6 sided dice or a 12 sided dice. Say you do 50-50 percent of the time the one dice or the other. Then $$f_{mixed roll}(z) = 0.5 \, f_{6-sided}(z) + 0.5 \, f_{12-sided}(z)$$

# Sum of variables -> convolution of pdf's

https://en.wikipedia.org/wiki/Convolution_of_probability_distributions

You use a convolution of the probability density functions $f_{X_1}$ and $f_{X_2}$ when the probability (of say Z) is a defined by multiple sums of different (independent) probabilities.

For example when $Z = X_1 + X_2$ (ie. a sum!) and multiple different pairs $x_1,x_2$ sum up to $z$, with each the probability $f_{X_1}(x_1)f_{X_2}(x_2)$. Then you get the convolution $$\mathbb{P}(Z=z) = \sum_{\text{all pairs }x_1+x_2=z} \mathbb{P}(X_1=x_1) \cdot \mathbb{P}(X_2=x_2)$$

and $$f_Z(z) = \sum_{x_1 \in \text{ domain of }X_1} f_{X_1}(x_1) f_{X_2}(z-x_1)$$

or for continuous variables

$$f_Z(z) = \int_{x_1 \in \text{ domain of }X_1} f_{X_1}(x_1) f_{X_2}(z-x_1) d x_1$$

. . . . an example is a sum of two dice rolls $f_{X_2}(x) = f_{X_1}(x) = 1/6$ for $x \in \lbrace 1,2,3,4,5,6 \rbrace$ and $$f_Z(z) = \sum_{x \in \lbrace 1,2,3,4,5,6 \rbrace \\ \text{ and } z-x \in \lbrace 1,2,3,4,5,6 \rbrace} f_{X_1}(x) f_{X_2}(z-x)$$

note I choose to integrate and sum $x_1 \in \text{ domain of } X_1$, which I find more intuitive, but it is not necessary and you can integrate from $-\infty$ to $\infty$ if you define $f_{X_1}(x_1)=0$ outside the domain.

# Image example

Let $Z$ be $X+Y$. To know $\mathbb{P}(z-\frac{1}{2}dz<Z<z+\frac{1}{2}dz)$ you will have to integrate over the probabilities for all the realizations of $x,y$ that lead to $z-\frac{1}{2}dz<Z=X+Y<z+\frac{1}{2}dz$.

So that is the integral of $f(x)g(y)$ in the region $\pm \frac{1}{2}dz$ along the line $x+y=z$.

• @Carl it is not jargonesque. The convolution can indeed be seen as a sum of many sums. But, this is not what 'the sum of variables' refers to. It refers to such things as when we speak of a 'a sum of two dice rolls', which has a very normal meaning and interpretation in every day life (especially when we play a board game). Would you rather like to say that we take a combination of two dice rolls when we do use the algebraic sum of two dice rolls? – Martijn Weterings Mar 6 '18 at 12:01
• The probability of rolling 7 with the (single) sum of two dice is the sum of (many) probabilities for rolling 1-6, 2-5, 3-4, 4-3, 5-2, 6-1. The term sum occurs two times and in the first case, when it refers to a single summation expression, it is what the statement 'sum of two variables' refers to, as in 'sum of two dice rolls'. – Martijn Weterings Mar 6 '18 at 12:12
• Indeed, the integral replaces the sum of probabilities. But, that relates to the second use of the term sum, not the first use of the term sum. So we can still refer to the sum of two variables (which is the first use of the term). That is because the term 'sum' is not used to refer to the convolution operation or summation operation of the probabilities, but to the summation of the variables. – Martijn Weterings Mar 6 '18 at 12:29
• at least it is not jargonesque to state 'the probability density for a sum of dice rolls is defined by the convolution of the probability densities for the individual dice rolls'. The term 'a sum of dice rolls' has a very normal interpretation in every day life when there are no statisticians around with their jargon. It is in this sense (sum of dice rolls) that you need to interpret (sum of variables). This step is neither not jargonesque. People use 'sums of variables' all the time. It is only the statistician who thinks about the probabilities for these sums and starts applying convolutions – Martijn Weterings Mar 6 '18 at 13:21
• @Carl: I think you misunderstood my statement. You were saying that it is not good to call a convolution integral a sum, implying that somebody calls the convolution integral a sum. But nobody here is saying this. What was said is that a convolution integral is the pdf of the sum of certain variables. You were changing the statement to something false, and then complained that it is false. – Pakk Feb 17 at 12:29

Your confusion seems to arise from conflating random variables with their distributions.

To "unlearn" this confusion, it might help to take a couple of steps back, empty your mind for a moment, forget about any fancy formalisms like probability spaces and sigma-algebras (if it helps, pretend you're back in elementary school and have never heard of any of those things!) and just think about what a random variable fundamentally represents: a number whose value we're not sure about.

For example, let's say I have a six-sided die in my hand. (I really do. In fact, I have a whole bag of them.) I haven't rolled it yet, but I'm about to, and I decide to call the number that I haven't rolled yet on that die by the name "$X$".

What can I say about this $X$, without actually rolling the die and determining its value? Well, I can tell that its value won't be $7$, or $-1$, or $\frac12$. In fact, I can tell for sure that it's going to be a whole number between $1$ and $6$, inclusive, because those are the only numbers marked on the die. And because I bought this bag of dice from a reputable manufacturer, I can be pretty sure that when I do roll the die and determine what number $X$ actually is, it's equally likely to be any of those six possible values, or as close to that as I can determine.

In other words, my $X$ is an integer-valued random variable uniformly distributed over the set $\{1,2,3,4,5,6\}$.

OK, but surely all that is obvious, so why do I keep belaboring such trivial things that you surely know already? It's because I want to make another point, which is also trivial yet, at the same time, crucially important: I can do math with this $X$, even if I don't know its value yet!

For example, I can decide to add one to the number $X$ that I'll roll on the die, and call that number by the name "$Q$". I won't know what number this $Q$ will be, since I don't know what $X$ will be until I've rolled the die, but I can still say that $Q$ will be one greater than $X$, or in mathematical terms, $Q = X+1$.

And this $Q$ will also be a random variable, because I don't know its value yet; I just know it will be one greater than $X$. And because I know what values $X$ can take, and how likely it is to take each of those values, I can also determine those things for $Q$. And so can you, easily enough. You won't really need any fancy formalisms or computations to figure out that $Q$ will be a whole number between $2$ and $7$, and that it's equally likely (assuming that my die is as fair and well balanced as I think it is) to take any of those values.

But there's more! I could just as well decide to, say, multiply the number $X$ that I'll roll on the die by three, and call the result $R = 3X$. And that's another random variable, and I'm sure you can figure out its distribution, too, without having to resort to any integrals or convolutions or abstract algebra.

And if I really wanted, I could even decide to take the still-to-be-determined number $X$ and to fold, spindle and mutilate it divide it by two, subtract one from it and square the result. And the resulting number $S = (\frac12 X - 1)^2$ is yet another random variable; this time, it will be neither integer-valued nor uniformly distributed, but you can still figure out its distribution easily enough using just elementary logic and arithmetic.

OK, so I can define new random variables by plugging my unknown die roll $X$ into various equations. So what? Well, remember when I said that I had a whole bag of dice? Let me grab another one, and call the number that I'm going to roll on that die by the name "$Y$".

Those two dice I grabbed from the bag are pretty much identical — if you swapped them when I wasn't looking, I wouldn't be able to tell — so I can pretty safely assume that this $Y$ will also have the same distribution as $X$. But what I really want to do is roll both dice and count the total number of pips on each of them. And that total number of pips, which is also a random variable since I don't know it yet, I will call "$T$".

How big will this number $T$ be? Well, if $X$ is the number of pips I will roll on the first die, and $Y$ is the number of pips I will roll on the second die, then $T$ will clearly be their sum, i.e. $T = X+Y$. And I can tell that, since $X$ and $Y$ are both between one and six, $T$ must be at least two and at most twelve. And since $X$ and $Y$ are both whole numbers, $T$ clearly must be a whole number as well.

But how likely is $T$ to take each of its possible values between two and twelve? It's definitely not equally likely to take each of them — a bit of experimentation will reveal that it's a lot harder to roll a twelve on a pair of dice than it is to roll, say, a seven.

To figure that out, let me denote the probability that I'll roll the number $a$ on the first die (the one whose result I decided to call $X$) by the expression $\Pr[X = a]$. Similarly, I'll denote the probability that I'll roll the number $b$ on the second die by $\Pr[Y = b]$. Of course, if my dice are perfectly fair and balanced, then $\Pr[X = a] = \Pr[Y = b] = \frac16$ for any $a$ and $b$ between one and six, but we might as well consider the more general case where the dice could actually be biased, and more likely to roll some numbers than others.

Now, since the two die rolls will be independent (I'm certainly not planning on cheating and adjusting one of them based on the other!), the probability that I'll roll $a$ on the first die and $b$ on the second will simply be the product of those probabilities: $$\Pr[X = a \text{ and } Y = b] = \Pr[X = a] \Pr[Y = b].$$

(Note that the formula above only holds for independent pairs of random variables; it certainly wouldn't hold if we replaced $Y$ above with, say, $Q$!)

Now, there are several possible values of $X$ and $Y$ that could yield the same total $T$; for example, $T = 4$ could arise just as well from $X = 1$ and $Y = 3$ as from $X = 2$ and $Y = 2$, or even from $X = 3$ and $Y = 1$. But if I had already rolled the first die, and knew the value of $X$, then I could say exactly what value I'd have to roll on the second die to reach any given total number of pips.

Specifically, let's say we're interested in the probability that $T = c$, for some number $c$. Now, if I know after rolling the first die that $X = a$, then I could only get the total $T = c$ by rolling $Y = c - a$ on the second die. And of course, we already know, without rolling any dice at all, that the a priori probability of rolling $a$ on the first die and $c - a$ on the second die is $$\Pr[X = a \text{ and } Y = c-a] = \Pr[X = a] \Pr[Y = c-a].$$

But of course, there are several possible ways for me to reach the same total $c$, depending on what I end up rolling on the first die. To get the total probability $\Pr[T = c]$ of rolling $c$ pips on the two dice, I need to add up the probabilities of all the different ways I could roll that total. For example, the total probability that I'll roll a total of 4 pips on the two dice will be: $$\Pr[T = 4] = \Pr[X = 1]\Pr[Y = 3] + \Pr[X = 2]\Pr[Y = 2] + \Pr[X = 3]\Pr[Y = 1] + \Pr[X = 4]\Pr[Y = 0] + \dots$$

Note that I went a bit too far with that sum above: certainly $Y$ cannot possibly be $0$! But mathematically that's no problem; we just need to define the probability of impossible events like $Y = 0$ (or $Y = 7$ or $Y = -1$ or $Y = \frac12$) as zero. And that way, we get a generic formula for the distribution of the sum of two die rolls (or, more generally, any two independent integer-valued random variables):

$$T = X + Y \implies \Pr[T = c] = \sum_{a \in \mathbb Z} \Pr[X = a]\Pr[Y = c - a].$$

And I could perfectly well stop my exposition here, without ever mentioning the word "convolution"! But of course, if you happen to know what a discrete convolution looks like, you may recognize one in the formula above. And that's one fairly advanced way of stating the elementary result derived above: the probability mass function of the sum of two integer-valued random variable is the discrete convolution of the probability mass functions of the summands.

And of course, by replacing the sum with an integral and probability mass with probability density, we get an analogous result for continuously distributed random variables, too. And by sufficiently stretching the definition of a convolution, we can even make it apply to all random variables, regardless of their distribution — although at that point the formula becomes almost a tautology, since we'll have pretty much just defined the convolution of two arbitrary probability distributions to be the distribution of the sum of two independent random variables with those distributions.

But even so, all this stuff with convolutions and distributions and PMFs and PDFs is really just a set of tools for calculating things about random variables. The fundamental objects that we're calculating things about are the random variables themselves, which really are just numbers whose values we're not sure about.

And besides, that convolution trick only works for sums of random variables, anyway. If you wanted to know, say, the distribution of $U = XY$ or $V = X^Y$, you'd have to figure it out using elementary methods, and the result would not be a convolution.

Addendum: If you'd like a generic formula for computing the distribution of the sum / product / exponential / whatever combination of two random variables, here's one way to write one: $$A = B \odot C \implies \Pr[A = a] = \sum_{b,c} \Pr[B = b \text{ and } C = c] [a = b \odot c],$$ where $\odot$ stands for an arbitrary binary operation and $[a = b \odot c]$ is an Iverson bracket, i.e. $$[a = b \odot c] = \begin{cases}1 & \text{if } a = b \odot c, \text{ and} \\ 0 & \text{otherwise}. \end{cases}$$

(Generalizing this formula for non-discrete random variables is left as an exercise in mostly pointless formalism. The discrete case is quite sufficient to illustrate the essential idea, with the non-discrete case just adding a bunch of irrelevant complications.)

You can check yourself that this formula indeed works e.g. for addition and that, for the special case of adding two independent random variables, it is equivalent to the "convolution" formula given earlier.

Of course, in practice, this general formula is much less useful for computation, since it involves a sum over two unbounded variables instead of just one. But unlike the single-sum formula, it works for arbitrary functions of two random variables, even non-invertible ones, and it also explicitly shows the operation $\odot$ instead of disguising it as its inverse (like the "convolution" formula disguises addition as subtraction).

Ps. I just rolled the dice. It turns out that $X = 5$ and $Y = 6$, which implies that $Q = 6$, $R = 15$, $S = 2.25$, $T = 11$, $U = 30$ and $V = 15625$. Now you know. ;-)

• This should be the accepted answer! Very intuitive and clear! – Vladislavs Dovgalecs Mar 7 '18 at 1:12
• @Carl: The point I'm trying to make is that the sum of the random variables is indeed a simple sum: $T = X + Y$. If we wish to calculate the distribution of $T$, then we'll need to do something more complicated, but that's a secondary issue. The random variable is not its distribution. (Indeed, a random variable is not even fully characterized by its distribution, since the (marginal) distribution alone doesn't encode information about its possible dependencies with other variables.) – Ilmari Karonen Mar 7 '18 at 2:10
• @Carl: ... In any case, if you wanted to introduce a special symbol for "addition of random variables", then for consistency you should also have special symbols for "multiplication of random variables" and "division of random variables" and "exponentiation of random variables" and "logarithm of random variables" and so on. All of those operations are perfectly well defined on random variables, viewed as numbers with an uncertain value, but in all cases calculating the distribution of the result is far more involved than just doing the corresponding calculation for constants. – Ilmari Karonen Mar 7 '18 at 2:13
• @Carl: The confusion goes away when you stop confusing a random variable with its distribution. Taking the distribution of a random variable is not a linear operation in any meaningful sense, so the distribution of the sum of two random variables is (usually) not the sum of their distributions. But the same is true for any nonlinear operation. Surely you're not confused by the fact that $\sqrt{x + y} \ne \sqrt x + \sqrt y$, so why should you be confused by the fact that $\Pr[X + Y = c] \ne \Pr[X = c] + \Pr[Y = c]$? – Ilmari Karonen Mar 7 '18 at 3:24
• @Carl: Wait, what? I roll two dice, write down the results $X$ and $Y$, and then calculate $Z = X/Y$. How is that not ordinary division? (And yes, it's still ordinary division even if I do it before I roll the dice. In that case, the values of $X$ and $Y$ just aren't fixed yet, and therefore neither is the value of $Z$.) – Ilmari Karonen Mar 7 '18 at 3:58

Actually I don't think this is quite right, unless I'm misunderstanding you.

If $X$ and $Y$ are independent random variables, then the sum/convolution relationship you're referring to is as follows: $$p(X+Y) = p(X)*p(Y)$$ That is, the probability density function (pdf) of the sum is equal to the convolution (denoted by the $*$ operator) of the individual pdf's of $X$ and $Y$.

To see why this is, consider that for a fixed value of $X=x$, the sum $S=X+Y$ follows the pdf of $Y$, shifted by an amount $x$. So if you consider all possible values of $X$, the distribution of $S$ is given by replacing each point in $p(X)$ by a copy of $p(Y)$ centered on that point (or vice versa), and then summing over all these copies, which is exactly what a convolution is.

Formally, we can write this as: $$p(S) = \int p_Y(S-x)p_X(x)dx$$ or, equivalently: $$p(S) = \int p_X(S-y)p_Y(y)dy$$

Edit: To hopefully clear up some confusion, let me summarize some of the things I said in comments. The sum of two random variables $X$ and $Y$ does not refer to the sum of their distributions. It refers to the result of summing their realizations. To repeat the example I gave in the comments, suppose $X$ and $Y$ are the numbers thrown with a roll of two dice ($X$ being the number thrown with one die, and $Y$ the number thrown with the other). Then let's define $S=X+Y$ as the total number thrown with the two dice together. For example, for a given dice roll, we might throw a 3 and a 5, and so the sum would be 8. The question now is: what does the distribution of this sum look like, and how does it relate to the individual distributions of $X$ and $Y$? In this specific example, the number thrown with each die follows a (discrete) uniform distribution between [1, 6]. The sum follows a triangular distribution between [1, 12], with a peak at 7. As it turns out, this triangular distribution can be obtained by convolving the uniform distributions of $X$ and $Y$, and this property actually holds for all sums of (independent) random variables.

• Summing many sums is more combining than a single sum worth notating with a '+' sign. My preference would be to say that random variables combine by convolution. – Carl Mar 6 '18 at 11:44
• A convolution could be called a sum of many sums, sure. But what you have to understand is that the convolution applies strictly to the PDFs of the variables that are summed. The variables themselves are not convolved. They are just added one to the other, and there is no way to construe that addition as a convolution operation (so the basic premise of your question, as it is now stated, is incorrect). – Ruben van Bergen Mar 6 '18 at 12:37
• You are misunderstanding that reference. It states: The probability distribution of the sum of two or more independent random variables is the convolution of their individual distributions. It does not say that a sum of two random variables is the same as convolving those variables. It says that the distribution of the sum is the convolution of the distribution of the individual variables. A random variable and its distribution are two different things. – Ruben van Bergen Mar 6 '18 at 13:03
• Sure, you can convolve random variables. But the sum/convolution property that is widely known and discussed in that article (and in my answer above) does not deal with convolutions of random variables. It is specifically concerned with sums of random variables, and the properties of the distribution of that sum. – Ruben van Bergen Mar 6 '18 at 13:22
• ("Sure, you can convolve random variables". Can you? My understanding was that because to get the distribution function of the sum of random variables you convolve the mass/density functions of each, many people talk (loosely) of convolving distributions, & some talk (wrongly) of convolving random variables. Sorry to digress, but I'm curious.) – Scortchi Mar 6 '18 at 15:06

Start by considering the set of all possible distinct outcomes of a process or experiment. Let $$X$$ be a rule (as yet unspecified) for assigning a number to any given outcome $$\omega$$; let $$Y$$ be too. Then $$S=X+Y$$ states a new rule $$S$$ for assigning a number to any given outcome: add the number you get from following rule $$X$$ to the number you get from following rule $$Y$$.

We can stop there. Why shouldn't $$S=X+Y$$ be called a sum?

If we go on to define a probability space, the mass (or density) function of the random variable (for that's what our rules are now) $$S=X + Y$$ can be got by convolving the mass (or density) function of $$X$$ with that of $$Y$$ (when they're independent). Here "convolving" has its usual mathematical sense. But people often talk of convolving distributions, which is harmless; or sometimes even of convolving random variables, which apparently isn't—if it suggests reading "$$X + Y$$" as "$$X \ \mathrm{convoluted\ with} \ Y$$", & therefore that the "$$+$$" in the former represents a complex operation somehow analogous to, or extending the idea of, addition rather than addition plain & simple. I hope it's clear from the exposition above, stopping where I said we could, that $$X+Y$$ already makes perfect sense before probability is even brought into the picture.

In mathematical terms, random variables are functions whose co-domain is the set of real numbers & whose domain is the set of all outcomes. So the "$$+$$" in "$$X + Y$$" (or "$$X(\omega) + Y(\omega)$$", to show their arguments explicitly) bears exactly the same meaning as the "$$+$$" in "$$\sin(\theta)+\cos(\theta)$$". It's fine to think about how you'd sum vectors of realized values, if it aids intuition; but that oughtn't to engender confusion about the notation used for sums of random variables themselves.

[This answer merely tries to draw together succintly points made by @MartijnWeterings, @IlmariKaronen, @RubenvanBergen, & @whuber in their answers & comments. I thought it might help to come from the direction of explaining what a random variable is rather than what a convolution is. Thank you all!]

• (+1) For effort. Answer too deep for me fathom. However, it did lead me to one. Please read that and let me know your thoughts. – Carl Mar 11 '18 at 21:52
• It is the elliptic notation that confused me: $S_i=X_i+Y_i$ for all $i=1,2,3,...,n-1,n$, in other words, vector addition. If someone had said, "vector addition" rather than "addition", I would not have been scratching my head wondering what was meant, but not said. – Carl Feb 14 at 21:22
• Well, if you put realizations of $X$ & $Y$ into vectors, & wanted to calculate the vector of realizations of $S$, then you'd use vector addition. But that seems rather tangential. After all, would you feel the need to explain '$\sin(\theta) + \cos(\phi)$' using vectors, or say that the '$+$' in that expression signifies vector addition? – Scortchi Feb 16 at 11:37
• To do what? The context was discrete data, e.g., RV's, not continuous functions, e.g., PDF's or $\sin(\theta)$, and $\sin(\theta) + \cos(\phi)$ is an ordinary sum. – Carl Feb 16 at 16:29
• @Carl: (1) If a biologist models the no. eggs laid in a duck's nest as a Poisson r.v., they're not really countenancing the possibility of an infinity of eggs. If you've got a question about the role of infinite sets in Mathematics, ask it on Mathematics or Philosophy SE. (2) Though quite standard, the nomenclature can indeed mislead; hence my answer. – Scortchi Feb 23 at 18:33

In response to your "Notice", um, ... no.

Let $X$, $Y$, and $Z$ be random variables and let $Z = X+Y$. Then, once you choose $Z$ and $X$, you force $Y = Z - X$. You make these two choices, in this order, when you write $$P(Z = z) = \int P(X = x) P(Y = z - x) \mathrm{d}x \text{.}$$ But that's a convolution.

• Notice gone. (+1) to you for caring. – Carl Mar 7 '18 at 6:40

The reason is the same that products of power functions are related to convolutions. The convolution always appears naturally, if you combine to objects which have a range (e.g. the powers of two power functions or the range of the PDFs) and where the new range appears as the sum of the original ranges.

It is easiest to see for medium values. For $x + y$ to have medium value, either both have to have medium values, or if one has a high value, the other has to have a low value and vice versa. This matches with the form of the convolution, which has one index going from high values to low values while the other increases.

If you look at the formula for the convolution (for discrete values, just because I find it easier to see there)

$(f * g)(n) = \sum_k f(k)g(n-k)$

then you see that the sum of the parameters to the functions($n-k$ and $k$) always sums exactly to $n$. Thus what the convolution is actually doing, it is summing all possible combinations, which have the same value.

For power functions we get

$(a_0+a_1x^1+a_2x^2+\ldots+a_nx^n)\cdot(b_0+b_1x^1+b_2x^2+\ldots+b_mx^m)=\sum_{i=0}^{m+n}\sum_k a_k*b_{i-k}x^i$

which has the same pattern of combining either high exponents from the left with low exponents from the right or vice versa, to always get the same sum.

Once you see, what the convolution is actually doing here, i.e. which terms are being combined and why it must, therefore, appear in many places, the reason for convolving random variables should become quite obvious.

Let us prove the supposition for the continuous case, and then explain and illustrate it using histograms built up from random numbers, and the sums formed by adding ordered pairs of numbers such that the discrete convolution, and both random variables are all of length $$n$$.

From Grinstead CM, Snell JL. Introduction to probability: American Mathematical Soc.; 2012. Ch. 7, Exercise 1:

Let $$X$$ and $$Y$$ be independent real-valued random variables with density functions $$f_X (x)$$ and $$f_Y (y)$$, respectively. Show that the density function of the sum $$X + Y$$ is the convolution of the functions $$f_X (x)$$ and $$f_Y (y)$$.

Let $$Z$$ be the joint random variable $$(X, Y )$$. Then the joint density function of $$Z$$ is $$f_X (x)f_Y (y)$$, since $$X$$ and $$Y$$ are independent. Now compute the probability that $$X + Y ≤ z$$, by integrating the joint density function over the appropriate region in the plane. This gives the cumulative distribution function of $$Z$$.

$$F_Z(z)=\mathrm{P}(X+Y\leq z)= \int_{(x,y):x+y\leq z} f_X(x)\,f_Y (y)\,dy\,dx$$ $$= \int_{-\infty}^\infty f_X(x)\left[\int_{y\,\leq \,z-x} f_Y(y)\,dy \right] dx= \int_{-\infty}^\infty f_X(x)\left[F_Y(z−x)\right]\,dx.$$

Now differentiate this function with respect to $$z$$ to obtain the density function of $$z$$.

$$f_Z(z) = \frac{dF_Z(z)}{dz} = \int_{-\infty}^\infty f_X(x)\,f_Y ( z-x)\,dx.$$

To appreciate what this means in practice, this was next illustrated with an example. The realization of a random number element (statistics: outcome, computer science: instance) from a distribution can be viewed as taking the inverse cumulative density function of a probability density function of a random probability. (A random probability is, computationally, a single element from a uniform distribution on the [0,1] interval.) This gives us a single value on the $$x$$-axis. Next, we generate another $$x$$-axis second random element from the inverse CDF of another, possibly different, PDF of a second, different random probability. We then have two random elements. When added, the two $$x$$-values so generated become a third element, and, notice what has happened. The two elements now become a single element of magnitude $$x1+x2$$, i.e., information has been lost. This is the context in which the "addition" is taking place; it is the addition of $$x$$-values. When multiple repetitions of this type of addition take place the resulting density of realizations (outcome density) of the sums tends toward the PDF of the convolution of the individual densities. The overall information loss results in smoothing (or density dispersion) of the convolution (or sums) compared to the constituting PDF's (or summands). Another effect is location shifting of the convolution (or sums). Note that realizations (outcomes, instances) of multiple elements afford only sparse elements populating (exemplifying) a continuous sample space.

For example, 1000 random values were created using a gamma distribution with a shape of $$10/9$$, and a scale of $$2$$. These were added pairwise to 1000 random values from a normal distribution with a mean of 4 and a standard deviation of $$1/4$$. Density scaled histograms of each of the three groups of values were co-plotted (left panel below) and contrasted (right panel below) with the density functions used to generate the random data, as well as the convolution of those density functions.

As seen in the figure, the addition of summands explanation appears to be plausible as the kernel smoothed distributions of data (red) in the left hand panel are similar to the continuous density functions and their convolution in the right hand panel.

• @whuber Finally, I think I understand. The sum is of random events. Take a look at my explanation and tell me if it is clear now, please. – Carl Mar 11 '18 at 21:27
• It helps to be careful with the language. Events are sets. Rarely are they even sets of numbers (that's why their elements are termed "outcomes"). Events don't add--the values of random variables do. The issue about "impressively complicated" is just a distraction. Indeed, if you want to get to the heart of the matter, make sure one of the summands in your example is a zero-mean random variable, because the mean effects an overall shift in the location. You want to understand intuitively what convolution does otherwise than shift the location. – whuber Mar 11 '18 at 22:47
• @whuber Thanks-useful. Only in statistics is an outcome a single element of a sample space. For the rest of us an outcome is the result of an event. Smoothing AND shifting. What I show is the least confusing example of many as it reduces collision of the superimposed plots. – Carl Mar 12 '18 at 7:24
• I see now how you are thinking of mixture models. You are constructing what are sometimes known as "multisets." (Usually a constructor other than brackets $\{,\}$ is used in order to clarify the notation.) The idea appears to be that of an empirical distribution function: the empirical distribution of a multiset $A$ and the empirical distribution of a multiset $B$ give rise to the empirical distribution of their multiset union, which is the mixture of the two distributions with relative weights $|A|$ and $|B|.$ – whuber Apr 6 at 18:49
• I think I detect a potential source of confusion in these ongoing edits. Because it would take too long to explain in a comment, I have appended an edit to my answer in the hope it might help a little. Indeed, the original first line of my answer was misleading on that account, so I have fixed it, too, with apologies. – whuber Apr 6 at 19:21

General expressions for the sums of n continuous random variables are found here:

"Multi-stage models for the failure of complex systems, cascading disasters, and the onset of disease"

For positive random variables, the sum can be simply written in terms of a product of Laplace transforms and the inverse of their product. The method is adapted from a calculation that appeared in E.T. Jaynes "Probability Theory" textbook.

This question may be old, but I'd like to provide yet another perspective. It builds on a formula for a change in variable in a joint probability density. It can be found in Lecture Notes: Probability and Random Processes at KTH, 2017 Ed. (Koski, T., 2017, pp 67), which itself refers to a detailed proof in Analysens Grunder, del 2 (Neymark, M., 1970, pp 148-168):

Let a random vector $$\mathbf{X} = (X_1, X_2,...,X_m)$$ have the joint p.d.f. $$f_\mathbf{X}(x_1,x_2,...,x_m)$$. Define a new random vector $$\mathbf{Y}=(Y_1, Y_2,...,Y_m)$$ by

$$Y_i = g_i(X_1,X_2,...,X_m), \hspace{2em}i=1,2,...,m$$

where $$g_i$$ is continuously differntiable and $$(g_1,g_2,...,g_m)$$ is invertible with the inverse

$$X_i = h_i(Y_1,Y_2,...,Y_m),\hspace{2em}i=1,2,...,m$$

Then the joint p.d.f. of $$\mathbf{Y}$$ (in the domain of invertibility) is

$$f_\mathbf{Y}(y_1,y_2,...,y_m) = f_\mathbf{X}(h_1(x_1,x_2,...,x_m),h_2(x_1,x_2,...,x_m),...,h_m(x_1,x_2,...,x_m))|J|$$

where $$J$$ is the Jacobian determinant

$$J = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} & ... & \frac{\partial x_1}{\partial y_m}\\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} & ... & \frac{\partial x_2}{\partial y_m}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial x_m}{\partial y_1} & \frac{\partial x_m}{\partial y_2} & ... & \frac{\partial x_m}{\partial y_m}\\ \end{vmatrix}$$

Now, let's apply this formula to obtain the joint p.d.f. of a sum of i.r.vs $$X_1 + X_2$$:

Define the random vector $$\mathbf{X} = (X_1,X_2)$$ with unknown joint p.d.f. $$f_\mathbf{X}(x_1,x_2)$$. Next, define a random vector $$\mathbf{Y}=(Y_1,Y_2)$$ by

$$Y_1 = g_1(X_1,X_2) = X_1 + X_2\\ Y_2 = g_2(X_1,X_2) = X_2.$$

The inverse map is then

$$X_1 = h_1(Y_1,Y_2) = Y_1 - Y_2\\ X_2 = h_2(Y_1,Y_2) = Y_2.$$

Thus, because of this and our assumption that $$X_1$$ and $$X_2$$ are independent, the joint p.d.f. of $$\mathbf{Y}$$ is

$$\begin{split} f_\mathbf{Y}(y_1,y_2) &= f_\mathbf{X}(h_1(y_1,y_2),h_2(y_1,y_2))|J|\\ & = f_\mathbf{X}(y_1 - y_2, y_2)|J|\\ & = f_{X_1}(y_1 - y_2) \cdot f_{X_2}(y_2) \cdot |J| \end{split}$$

where the Jacobian $$J$$ is

$$J = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2}\\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{vmatrix} = \begin{vmatrix} 1 & -1\\ 0 & 1 \end{vmatrix} = 1$$

To find the p.d.f. of $$Y_1 = X_1 + X_2$$, we marginalize

$$\begin{split} f_{Y_1} &= \int_{-\infty}^\infty f_\mathbf{Y}(y_1,y_2) dy_2\\ &= \int_{-\infty}^\infty f_\mathbf{X}(h_1(y_1,y_2),h_2(y_1,y_2))|J| dy_2\\ &= \int_{-\infty}^\infty f_{X_1}(y_1 - y_2) \cdot f_{X_2}(y_2) dy_2 \end{split}$$

which is where we find your convolution :D