Warum ist die Summe zweier Zufallsvariablen eine Faltung?


33

Für lange Zeit habe ich nicht verstanden , warum die „Summe“ von zwei Zufallsvariablen ist ihre Faltung , während eine Mischung Dichtefunktion Summe von und istf(x)g(x)pf(x)+(1p)g(x); die arithmetische Summe und nicht ihre Faltung. Der genaue Ausdruck "die Summe von zwei Zufallsvariablen" erscheint in Google 146.000 mal und ist wie folgt elliptisch. Wenn man davon ausgeht, dass ein RV einen einzelnen Wert ergibt, dann kann dieser einzelne Wert zu einem anderen RV-Einzelwert addiert werden, der nichts mit Faltung zu tun hat, zumindest nicht direkt. Alles, was ist, ist eine Summe von zwei Zahlen. Ein RV-Ergebnis in der Statistik ist jedoch eine Sammlung von Werten, und daher wäre ein genauerer Ausdruck so etwas wie "die Menge der koordinierten Summen von Paaren von zugeordneten Einzelwerten aus zwei RVs ist ihre diskrete Faltung" ... und kann durch die angenähert werden Faltung der Dichtefunktionen, die diesen RVs entsprechen. Noch einfachere Sprache: 2 Wohnmobile vonn-samples sind zwei n-dimensionale Vektoren, die sich zu ihrer Vektorsumme addieren.

Bitte zeigen Sie die Details, wie die Summe zweier Zufallsvariablen eine Faltung und eine Summe ist.


6
Ich glaube nicht wirklich, dass es eine abstrakte algebraische Summe ist . Wenn wir eine 'Summe von Variablen' bilden, beziehen wir uns auf die typische Rechenoperation, wie wir sie kennen, wenn wir natürliche oder reelle Zahlen addieren. Das bedeutet, dass wir eine neue Variable erstellen, indem wir die anderen Variablen addieren. Der Begriff der "Summe von Variablen" existiert auch außerhalb des Bereichs der Statistik und ist unabhängig von den Ausdrücken über Windungen und Wahrscheinlichkeiten. Also, in der Tat ‚die Summe der Variablen ist eine Faltung‘, ist falsch. Aber niemand deutet dies an. Wir sollten das Wort "ist" in dieser Aussage ändern.
Sextus Empiricus

5
Dies ist so, als würde man argumentieren, dass nicht 'das Produkt zweier Funktionen f und g' genannt werden sollte (oder nur als ein abstrakter algebraischer Begriff von 'Produkt' interpretiert wird), weil es eine Faltung in Begriffen ist der Fourier-Transformationen dieser Funktionen. f(x)g(x)
Sextus Empiricus

16
Der "Hinweis" ist irreführend. Eine Summe der Zufallsvariablen und bedeutet genau das Gleiche. "Summe" wird von Schülern verstanden: Für jedes wird der Wert durch Addition der Zahlen undEs ist nichts Abstraktes daran. Diese Wohnmobile haben Verteilungen. Es gibt viele Möglichkeiten, die Verteilungen darzustellen. Die Verteilungsfunktion von ist die Faltung der DFs von und ; Die charakteristische Funktion von ist das ProduktXYω(X+Y)(ω)X(ω)Y(ω).X+YXYX+Yihrer CFs; die kumulativ erzeugende Funktion von ist die Summe ihrer CGFs; und so weiter. X+Y
whuber

3
Ich sehe weder Zufallsvariablen noch Verteilungen in Ihrer Berechnung.
whuber

8
In der Sprache meines Beitrags unter stats.stackexchange.com/a/54894/919 besteht ein Paar zufälliger Variablen aus einer Schachtel mit Tickets, auf die jeweils zwei Zahlen geschrieben sind, eine mit und die andere Die Summe dieser Zufallsvariablen ergibt sich durch Addition der beiden auf jedem Ticket gefundenen Zahlen. Die Berechnung ist buchstäblich eine Aufgabe, die Sie einem Klassenzimmer der dritten Klasse zuweisen könnten. (Ich betone dies, um sowohl die grundlegende Einfachheit der Operation zu betonen als auch zu zeigen, wie stark sie mit dem zusammenhängt, was jeder unter einer "Summe" versteht.)X Y .(X,Y)XY.
whuber

Antworten:


14

Faltungsberechnungen, die mit Verteilungen von Zufallsvariablen verbunden sind, sind alle mathematischen Manifestationen des Gesetzes der Gesamtwahrscheinlichkeit .


In der Sprache meines Beitrags unter Was ist mit einer „Zufallsvariablen“ gemeint? ,

Ein Paar von Zufallsvariablen (X,Y) besteht aus einer Schachtel von Tickets, auf die jeweils zwei Zahlen geschrieben sind, von denen eine mit X und die andere mit Y . Die Summe dieser Zufallsvariablen ergibt sich durch Addition der beiden auf jedem Ticket gefundenen Zahlen.

Ich habe ein Bild von einer solchen Schachtel und ihren Eintrittskarten bei Clarifying the concept of sum of random variables gepostet .

Bildbeschreibung hier eingeben

Diese Berechnung ist buchstäblich eine Aufgabe, die Sie einem Klassenzimmer der dritten Klasse zuweisen könnten. (Damit möchte ich sowohl die grundlegende Einfachheit der Operation betonen als auch zeigen, wie stark sie mit dem zusammenhängt, was jeder unter einer "Summe" versteht.)

Wie die Summe der Zufallsvariablen mathematisch ausgedrückt wird, hängt davon ab, wie Sie den Inhalt der Box darstellen:

Die ersten beiden sind insofern besonders, als die Box möglicherweise keine pmf, pdf oder mgf hat, aber immer eine cdf, cf und cgf.


X, Y,X+YX+YzX+Yz,Pr(X+Y=z).

Die PMF der Summe wird durch Zerlegen der Menge von Tickets gemäß dem Wert von geschrieben ist, nach dem Gesetz der Gesamtwahrscheinlichkeit, das Proportionen (von disjunkten Teilmengen) addiert, ermittelt. Technisch gesehenX

Der Anteil von Tickets, der in einer Sammlung nicht zusammenhängender Untergruppen der Box gefunden wird, ist die Summe der Anteile der einzelnen Untergruppen.

Es gilt also:

Der Anteil der Tickets mit , geschrieben muss gleich der Summe aller möglichen Werte des Anteils der Tickets mit und geschriebenX+Y=zPr(X+Y=z),xX=xX+Y=z,Pr(X=x,X+Y=z).

Da und bedeuten dieser Ausdruck direkt in Bezug auf die ursprünglichen Variablen und als umgeschrieben werdenX=xX+Y=zY=zx,XY

Pr(X+Y=z)=xPr(X=x,Y=zx).

Das ist die Faltung.


Bearbeiten

Bitte beachten Sie, dass Faltungen zwar mit Summen von Zufallsvariablen verknüpft sind, die Faltungen jedoch keine Faltungen der Zufallsvariablen selbst sind!

Tatsächlich ist es in den meisten Fällen nicht möglich, zwei Zufallsvariablen zu falten. Damit dies funktioniert, müssen ihre Domänen eine zusätzliche mathematische Struktur haben. Diese Struktur ist eine kontinuierliche topologische Gruppe.

Ohne auf Details einzugehen, genügt es zu sagen, dass die Faltung von zwei beliebigen Funktionen X,Y:GH abstrakt ungefähr so ​​aussehen muss

(XY)(g)=h,kGh+k=gX(h)Y(k).

(Die Summe könnte ein Integral sein, und wenn dies neue Zufallsvariablen aus vorhandenen erzeugen soll, muss XY messbar sein, wann immer X und Y sind. Hier muss eine gewisse Berücksichtigung der Topologie oder der Messbarkeit erfolgen.)

Diese Formel ruft zwei Operationen auf. Eine ist die Multiplikation mit H: Es muss sinnvoll sein, die Werte X(h)H und Y(k)H. zu multiplizieren . Der andere ist der Zusatz zu G: Es muss sinnvoll sein, Elemente von G hinzuzufügen .G.

In den meisten Wahrscheinlichkeitsanwendungen ist H eine Menge von Zahlen (reell oder komplex) und die Multiplikation ist die übliche. Aber G, der Probenraum, hat oft überhaupt keine mathematische Struktur. Deshalb ist die Faltung von Zufallsvariablen in der Regel nicht einmal definiert. Die Objekte, die an den Windungen in diesem Thread beteiligt sind, sind mathematische Darstellungen der Verteilungen von Zufallsvariablen. Sie werden verwendet, um die Verteilung einer Summe von Zufallsvariablen unter Berücksichtigung der gemeinsamen Verteilung dieser Zufallsvariablen zu berechnen.


Verweise

Stuart und Ord, Kendalls Advanced Theory of Statistics, Band 1. Fünfte Ausgabe, 1987, Kapitel 1, 3 und 4 ( Häufigkeitsverteilungen, Momente und Kumulanten sowie charakteristische Funktionen ).


Assoziativität mit Skalarmultiplikation aus algebraischen Eigenschaften bedeutet , dass für jede reelle (oder komplexe) Zahl a ist . Während eine schöne Eigenschaft darin besteht, dass die Faltung zweier Dichtefunktionen eine Dichtefunktion ist, eine nicht auf faltende Dichtefunktionen beschränkt ist und die Faltung im Allgemeinen keine Wahrscheinlichkeitsbehandlung ist, kann dies sicher sein, aber es kann eine Zeitreihenbehandlung sein.
a(fg)=(af)g
a
Carl

@Carl Wie passt dieser Kommentar zu Ihrer ursprünglichen Frage, bei der nach Summen zufälliger Variablen gefragt wird ? Bestenfalls ist es tangential.
Whuber

Ich bitte Sie, nicht zu stark zu verallgemeinern. Einen Satz mit "Convolution is" zu beginnen, ohne zu sagen, dass "Convolution of RV's is" elliptisch ist. Mein ganzes Problem war hier mit der elliptischen Notation. Die Vektoraddition von zwei Raum-Vektoren ist eine Faltung, unabhängig davon, ob diese Vektoren normalisiert sind oder nicht. Wenn sie normalisiert sind, müssen sie keine Wahrscheinlichkeiten sein. Das ist die ganze Wahrheit, nicht nur ein Teil davon. n
Carl

Danke: Ich werde den ersten Satz klarstellen, um zu betonen, dass ich Ihre Frage beantworte.
Whuber

Neu hinzugekommen ist die Faltung von Wohnmobilen, was ich technisch gefragt habe. Und vielleicht bin ich uneindeutig, aber Faltung ist nicht immer von RVs, sondern kann immer auf einige Skalierungsfaktoren von Dichtefunktionen mal jene Dichtefunktionen reduziert werden, bei denen die Skalare multiplikativ sind und bei denen die Dichtefunktionen manchmal RVs sind, in welchem ​​Fall die Skalierungsfaktoren sind die multiplikative Identität, dh 1.
Carl

41

Notation, Groß- und Kleinschreibung

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

  • Zufallsvariablen werden normalerweise in Großbuchstaben geschrieben: X , Y usw.
  • Bestimmte Realisierungen einer Zufallsvariablen werden in entsprechenden Kleinbuchstaben geschrieben. Zum Beispiel könnte x1 , x2 , ..., xn eine Stichprobe sein, die der Zufallsvariablen X und es wird eine kumulative Wahrscheinlichkeit P(X>x) formell geschrieben , um die Zufallsvariable von der Realisierung zu unterscheiden.

Z=X+Y bedeutetzi=xi+yixi,yi


Variablenmischung -> Summe der PDFs

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

Sie verwenden eine Summe der Wahrscheinlichkeitsdichtefunktionen fX1 und fX2 wenn die Wahrscheinlichkeit (von beispielsweise Z) durch eine einzige Summe verschiedener Wahrscheinlichkeiten definiert ist.

Wenn zum Beispiel Z ein Bruchteil s der Zeit ist, die durch X1 definiert ist, und ein Bruchteil 1s der Zeit, die durch X2 definiert ist , dann erhalten Sie

P(Z=z)=sP(X1=z)+(1s)P(X2=z)
und
fZ(z)=sfX1(z)+(1s)fX2(z)

. . . . Ein Beispiel ist eine Wahl zwischen Würfeln mit entweder 6-seitigen Würfeln oder 12-seitigen Würfeln. Angenommen, Sie machen in 50-50 Prozent der Fälle den einen oder anderen Würfel. Dann

fmixedroll(z)=0.5f6sided(z)+0.5f12sided(z)


Summe der Variablen -> Faltung von PDFs

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

Sie verwenden eine Faltung der Wahrscheinlichkeitsdichtefunktionen fX1 und fX2 wenn die Wahrscheinlichkeit (von z. B. Z) durch mehrere Summen verschiedener (unabhängiger) Wahrscheinlichkeiten definiert ist.

Wenn zum Beispiel Z=X1+X2 (dh eine Summe!) Und mehrere verschiedene Paare x1,x2 summieren sich x 2 zu z mit jeweils der Wahrscheinlichkeit fX1(x1)fX2(x2) . Dann erhält man die Faltung

P(Z=z)=all pairs x1+x2=zP(X1=x1)P(X2=x2)

und

fZ(z)=x1 domain of X1fX1(x1)fX2(zx1)

oder für stetige Variablen

fZ(z)=x1 domain of X1fX1(x1)fX2(zx1)dx1

. . . . Ein Beispiel ist eine Summe von zwei Würfeln fX2(x)=fX1(x)=1/6 für x{1,2,3,4,5,6} und

fZ(z)=x{1,2,3,4,5,6} and zx{1,2,3,4,5,6}fX1(x)fX2(zx)

note Ich entscheide mich für die Integration und Summierung von x1 domain of X1 , was ich intuitiver finde, aber es ist nicht erforderlich und Sie können von bis wenn Sie fX1(x1)=0 außerhalb der Domain definieren .

Bildbeispiel

example of 'sum of variables' resulting in 'convolution of pdfs'

Lassen Z seine X+Y . Um P(z12dz<Z<z+12dz)Sie müssen die Wahrscheinlichkeiten für alle Realisierungen vonx,y, die zuz-1führen,z12dz<Z=X+Y<z+12dz.

So that is the integral of f(x)g(y) in the region ±12dz along the line x+y=z.


Written by StackExchangeStrike


6
@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?
Sextus Empiricus

2
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'.
Sextus Empiricus

5
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.
Sextus Empiricus

8
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
Sextus Empiricus

2
@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.

28

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 12. 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=(12X1)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]=16 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 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=ca 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 ca on the second die is

Pr[X=a and Y=ca]=Pr[X=a]Pr[Y=ca].

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]+

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=12) 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+YPr[T=c]=aZPr[X=a]Pr[Y=ca].

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=XY, 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=BCPr[A=a]=b,cPr[B=b and C=c][a=bc],
where stands for an arbitrary binary operation and [a=bc] is an Iverson bracket, i.e.
[a=bc]={1if a=bc, and0otherwise.

(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 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. ;-)


4
This should be the accepted answer! Very intuitive and clear!
Vladislavs Dovgalecs

3
@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

3
@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

5
@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 x+yx+y, so why should you be confused by the fact that Pr[X+Y=c]Pr[X=c]+Pr[Y=c]?
Ilmari Karonen

3
@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

7

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)=pY(Sx)pX(x)dx
or, equivalently:
p(S)=pX(Sy)pY(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

6
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

4
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

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

1
("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 - Reinstate Monica

6

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 ω; 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 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(ω)+Y(ω)", to show their arguments explicitly) bears exactly the same meaning as the "+" in "sin(θ)+cos(θ)". 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

It is the elliptic notation that confused me: Si=Xi+Yi for all i=1,2,3,...,n1,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

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(θ)+cos(ϕ)' using vectors, or say that the '+' in that expression signifies vector addition?
Scortchi - Reinstate Monica

To do what? The context was discrete data, e.g., RV's, not continuous functions, e.g., PDF's or sin(θ), and sin(θ)+cos(ϕ) is an ordinary sum.
Carl

1
@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 - Reinstate Monica

3

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=ZX. You make these two choices, in this order, when you write

P(Z=z)=P(X=x)P(Y=zx)dx.
But that's a convolution.

Notice gone. (+1) to you for caring.
Carl

2

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)

(fg)(n)=kf(k)g(nk)

then you see that the sum of the parameters to the functions(nk 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

(a0+a1x1+a2x2++anxn)(b0+b1x1+b2x2++bmxm)=i=0m+nkakbikxi

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.


2

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 fX(x) and fY(y), respectively. Show that the density function of the sum X+Y is the convolution of the functions fX(x) and fY(y).

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

FZ(z)=P(X+Yz)=(x,y):x+yzfX(x)fY(y)dydx
=fX(x)[yzxfY(y)dy]dx=fX(x)[FY(zx)]dx.

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

fZ(z)=dFZ(z)dz=fX(x)fY(zx)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. enter image description here

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

3
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

@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

1
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

1
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

1

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 X=(X1,X2,...,Xm) have the joint p.d.f. fX(x1,x2,...,xm). Define a new random vector Y=(Y1,Y2,...,Ym) by

Yi=gi(X1,X2,...,Xm),i=1,2,...,m

where gi is continuously differntiable and (g1,g2,...,gm) is invertible with the inverse

Xi=hi(Y1,Y2,...,Ym),i=1,2,...,m

Then the joint p.d.f. of Y (in the domain of invertibility) is

fY(y1,y2,...,ym)=fX(h1(x1,x2,...,xm),h2(x1,x2,...,xm),...,hm(x1,x2,...,xm))|J|

where J is the Jacobian determinant

J=|x1y1x1y2...x1ymx2y1x2y2...x2ymxmy1xmy2...xmym|


Now, let's apply this formula to obtain the joint p.d.f. of a sum of i.r.vs X1+X2:

Define the random vector X=(X1,X2) with unknown joint p.d.f. fX(x1,x2). Next, define a random vector Y=(Y1,Y2) by

Y1=g1(X1,X2)=X1+X2Y2=g2(X1,X2)=X2.

The inverse map is then

X1=h1(Y1,Y2)=Y1Y2X2=h2(Y1,Y2)=Y2.

Thus, because of this and our assumption that X1 and X2 are independent, the joint p.d.f. of Y is

fY(y1,y2)=fX(h1(y1,y2),h2(y1,y2))|J|=fX(y1y2,y2)|J|=fX1(y1y2)fX2(y2)|J|

where the Jacobian J is