Trong một thời gian dài, tôi không hiểu tại sao "tổng" của hai biến ngẫu nhiên là tích chập của chúng , trong khi đó hàm tổng mật độ hỗn hợp của và là$f(x)$$g(x)$$p\,f(x)+(1-p)g(x)$; tổng số học và không tích chập của chúng. Cụm từ chính xác "tổng của hai biến ngẫu nhiên" xuất hiện trong google 146.000 lần và có hình elip như sau. Nếu người ta coi RV mang lại một giá trị duy nhất, thì giá trị đơn đó có thể được thêm vào một giá trị RV khác, không liên quan gì đến tích chập, ít nhất là không trực tiếp, tất cả chỉ là tổng của hai số. Tuy nhiên, một kết quả RV trong thống kê là một tập hợp các giá trị và do đó, một cụm từ chính xác hơn sẽ giống như "tập hợp các cặp giá trị riêng lẻ được liên kết từ hai RV là tích chập riêng của chúng" ... và có thể được xấp xỉ bằng tích chập các hàm mật độ tương ứng với các RV. Ngôn ngữ đơn giản hơn: 2 RV's of$n$-samples có hiệu lực hai vectơ n chiều thêm vào làm tổng vectơ của chúng.

Vui lòng cho biết chi tiết về cách tổng của hai biến ngẫu nhiên là một tích chập và tổng.

Tính toán kết hợp liên quan đến phân phối các biến ngẫu nhiên là tất cả các biểu hiện toán học của Định luật xác suất tổng .

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Trong ngôn ngữ của bài viết của tôi tại Điều gì có nghĩa là một biến ngẫu nhiên của người Viking? ,

Một cặp biến ngẫu nhiên $(X,Y)$ bao gồm một hộp vé trên mỗi số được viết hai số, một số $X$ được chỉ định và $Y$ . Tổng của các biến ngẫu nhiên này có được bằng cách thêm hai số được tìm thấy trên mỗi vé.

Tôi đã đăng một bức ảnh của một hộp như vậy và vé của nó tại Làm rõ khái niệm tổng các biến ngẫu nhiên .

Tính toán này theo nghĩa đen là một nhiệm vụ bạn có thể giao cho lớp học lớp ba. (Tôi đưa ra quan điểm này để nhấn mạnh cả tính đơn giản cơ bản của hoạt động cũng như cho thấy mức độ kết nối mạnh mẽ với những gì mọi người hiểu "tổng" có nghĩa là gì.)

Làm thế nào tổng của các biến ngẫu nhiên được biểu thị bằng toán học phụ thuộc vào cách bạn thể hiện nội dung của hộp:

• Về mặt hàm khối lượng xác suất (pmf) hoặc hàm mật độ xác suất (pdf), đó là hoạt động của tích chập .

• Xét về các hàm tạo mô men (mgf), nó là sản phẩm (theo nguyên tố).

• Về mặt chức năng phân phối (tích lũy) (cdf), đây là một hoạt động liên quan chặt chẽ đến tích chập. (Xem tài liệu tham khảo.)

• Về mặt chức năng đặc trưng (cf), nó là sản phẩm (theo nguyên tố).

• Xét về các hàm tạo tích lũy (cgf), nó là tổng.

Hai trong số này là đặc biệt trong trường hợp hộp có thể không có pmf, pdf hoặc mgf, nhưng nó luôn có cdf, cf và cgf.

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Để xem tại sao chập là phương pháp thích hợp để tính toán PMF hoặc pdf của một tổng của các biến ngẫu nhiên, xem xét các trường hợp cả ba biến $X,$ $Y,$ và $X+Y$ có một PMF: theo định nghĩa, các PMF cho $X+Y$ tại bất kỳ số $z$ cho tỷ lệ vé trong hộp trong đó tổng $X+Y$ bằng $z,$ được viết $\Pr(X+Y=z).$

Pmf của tổng được tìm thấy bằng cách chia nhỏ bộ vé theo giá trị của $X$ được ghi trên đó, theo Luật Xác suất Tổng, xác nhận tỷ lệ (của các tập hợp rời rạc) thêm. Kỹ thuật hơn,

Tỷ lệ vé được tìm thấy trong một tập hợp các tập hợp rời rạc của hộp là tổng tỷ lệ của các tập hợp con riêng lẻ.

Nó được áp dụng như vậy:

Tỷ lệ vé trong đó $X+Y=z$ , viết $\Pr(X+Y=z),$ phải bằng tổng trên tất cả các giá trị có thể $x$ của tỷ lệ vé trong đó $X=x$ và $X+Y=z,$ viết $\Pr(X=x, X+Y=z).$

Vì $X=x$ và $X+Y=z$ ngụ ý $Y=z-x,$ biểu thức này có thể được viết lại trực tiếp theo các biến ban đầu $X$ và $Y$ là

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

Đó là sự chập chững.

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Chỉnh sửa

Xin lưu ý rằng mặc dù các kết quả có liên quan đến tổng các biến ngẫu nhiên, nhưng các kết quả không phải là các kết quả của chính các biến ngẫu nhiên!

Thật vậy, trong hầu hết các trường hợp, không thể kết hợp hai biến ngẫu nhiên. Để làm việc này, các miền của họ phải có cấu trúc toán học bổ sung. Cấu trúc này là một nhóm tôpô liên tục.

Không cần đi sâu vào chi tiết, đủ để nói rằng tích chập của bất kỳ hai hàm $X, Y:G \to H$ phải trừu tượng trông giống như

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

(Tổng có thể là một tích phân và, nếu điều này sẽ tạo ra các biến ngẫu nhiên mới từ các biến hiện có, $X\star Y$ phải có thể đo được bất cứ khi nào $X$ và $Y$ ;

Công thức này gọi hai thao tác. Một là nhân trên $H:$ nó phải có ý nghĩa với các giá trị nhân $X(h)\in H$ và $Y(k)\in H.$ Người kia là sự bổ sung trên $G:$ nó phải làm cho tinh thần để thêm các yếu tố của $G.$

Trong hầu hết các ứng dụng xác suất, $H$ là một tập hợp số (thực hoặc phức) và phép nhân là số thường. Nhưng $G,$ không gian mẫu, thường không có cấu trúc toán học nào cả. Đó là lý do tại sao tích chập của các biến ngẫu nhiên thường không được xác định. Các đối tượng liên quan đến kết cấu trong luồng này là các biểu diễn toán học của các phân phối của các biến ngẫu nhiên. Chúng được sử dụng để tính toán phân phối của một tổng các biến ngẫu nhiên, dựa vào phân phối chung của các biến ngẫu nhiên đó.

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Tài liệu tham khảo

Stuart và Ord, Lý thuyết thống kê nâng cao của Kendall, Tập 1. Phiên bản thứ năm, 1987, Chương 1, 3 và 4 ( Phân phối tần số, Khoảnh khắc và tích lũy và Chức năng đặc trưng ).

Ký hiệu, chữ hoa và chữ thường

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

• 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 $z_i=x_i+y_i \qquad \forall x_i,y_i$

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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

$$f_{mixed roll}(z) = 0.5 \, f_{6-sided}(z) + 0.5 \, f_{12-sided}(z)$$

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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 $x+y=z$.

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Written by StackExchangeStrike

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 $\{1,2,3,4,5,6\}$.

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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.

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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.

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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):

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

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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.

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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).

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^{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. ;-)}

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.

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.

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[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!]

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.

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.

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):

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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

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

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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

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

https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0216422

"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.