Uniform Distributions Part 2: Beware of Transformations

Ok, now getting to the post that I actually meant to do to start the series on uniformity. The previous post was more about the thrilling adventure of bringing together a bunch of disparate math tools to resolve a seemingly simple question about uniform random variables. It was, ironically, not very visual, but it still in the spirit of the blog, to encourage the attitude that experiment, exploration, and trying different approaches to things is as much a part of math as it is for other scientific endeavors. 

Today we bring something much more visual and play around a lot with random number generation. It is a modernization and generalization of this post (it was the featured post on NestedTori for a looong time). In some sense, this might have more fun submission SoME, but given the time constraints (and the admitted surprise of it being earlier this year!), I opted for the adventure story that I had a much more solid narrative for, and something much more self-contained (and in fact, a requirement for the competition). We'll see how I do, but it speaks to the kind of choices and gambles one takes when choosing something to present to a larger event.

But anyway, in this post, we work toward obtaining uniform distributions of spheres and balls (this post will still be primarily setup on the disk in the plane; I'm splitting the already written post up in an effort to cap the length of my posts). We're working up to exploring a somewhat surprising way that uniformity can propagate itself. There has been some online buzz about balls and spheres and their very interesting volume formulas involving one of our frequent friends, the Gamma function, our factorial generalizer. Actually, I'd wanted to do a post about that formula for years, ever since the Pi vs. Tau debate (more specifically: volumes of spheres and balls are where it seems $\pi$ is better), I'm glad it got some visual stuff eventually, but of course, if I wanted to do my own post, maybe it would have more impact if I could rework it in some way in line with the theme of this blog. And that way, as we have been really enjoying lately, is through the lens of probability. However, we will not yet get to the full generalization today, as there is considerable material to get through (considerable FUN material, of course, but still, yours truly does admittedly have a bit of a problem keeping things short).

Summing the Squares: Not So Triangular?

First off, if we ask ourselves the question, what is the distribution of the sum of the squares of two uniform random variables, which is different from just adding them together to form the triangular distribution. This tells us what is the squared distance from the origin in the plane. Two uniform variables in $[-1, 1]$ mean that this radial squared function actually has a very complicated dependence because it has a different value in the corners than on the sides. We'd expect it to be uniform up to the unit disk, touching the boundary, and then it behaves differently when it goes into the corners. To see the real deal, we need to compute the distribution of squares of a uniform $[-1, 1]$ variables, and adding two independent instances of these. This, just as in the un-squared case, uses convolution.

So first let's examine the square of a uniform variable (first on $[0,1]$; the case $[-1,1]$ is a rescaling but causes some complications with having two signs in a square root): the pdf is $dx$ (keeping pdfs multiplied by $dx$ helps me remember how it transforms. One read/understand them more formally as infinitesimals—a "little bit of x"—or differential forms, my favorite mathematical tool in the known universe), but then taking $y = x^2$ means $x = \sqrt{y}$ and $$dx = \frac{1}{2\sqrt{y}} dy,$$ on the interval $[0,1]$ (and zero elsewhere). Write the resulting random variable $Z$ and $\rho_Z(z) = \frac{1}{2\sqrt{z}}$. To get the sum of two independent versions (call it $S$) of this requires us to convolve this function with itself. This turns out to be one tricky problem. First, sums of independent random variables are given by convolution (a fact we have used before):

\[\rho_{S}(\xi) = \int_{\mathbb{R}} \rho_{Z}(\xi - z) \rho_{Z}(z)\; dz\]

Looks straightforward enough to plug in the formula, but the trickiness comes in determining the right bounds of integration. Central to it is to reflect one of them and determine three different stages of overlap. There's 4 stages: no overlap with the reflected-translated function way to the left (so no overlap and so the function is $0$), overlap of the two unbounded sections, overlap of the two bounded sections, and no overlap due to the reflected-translated being far to the right. It's not the most inspiring calculation, so let me just show the result:

\[\rho_Y(y) = \frac{1}{2} \sin^{-1}\left(\frac{2}{y} -1\right)\]

pdf of the sum of two squared uniform random variables

Testing it with random number generators (see accompanying notebook), with a million values of $X$ and $Y$, we similarly bin the squared distance values $S$:

Very cool! You never would have thought of that shape arising from such a simple process. However, if you think about it a bit, this directly reflects that the disk is uniformly covered, but going into the corners is more difficult. The sharp break occurs because the area of the circle hits the boundary square and with distance > 1 cannot be achieved in the middle of the sides, but can be achieved at the corners. 

Now what we do is get rid of the samples $> 1$ which is basically tossing out the fraction $1 - \frac{\pi}{4}$ of the ones that exceed 1. The expected number of trials we need to run to get that is $4/\pi$ (as it's a geometric random variable), so we expect to have $\frac{\pi}{4} \approx 0.7853$ proportion of the generated samples to satisfy that condition. Then now we get to the good part: we can restrict to where $S \leq 1$. Let's just rename it to be $S$ even though technically it is different variable, with a restriction. In our random sample of a million, it did indeed leave us 784560 points (in the notebook version of this, you may get something slightly different; all that is important is it is close to $1-\pi/4$ of a million).

Now if we go back to our random variables $X$ and $Y$ (going from $[-1, 1]$ but whose squares, summed to $S$, gives a distribution exactly like the above by symmetry), also now both restricted so as to never be greater distance than $1$, we scatterplot the random samples corresponding to what produced the above:

Does it look uniform?

A slight lie: we actually showed only 1/100 of the sampled points (about 7845 in this image) because a million is wayy dense for an image like this. Looks very uniform! Again, we should emphasize the fact that although the original random variables for the $x$ and $y$ coordinates were independent uniform random variables their restrictions to the disk, $X$ and $Y$, are NOT independent. It is intuitive to understand at a big-picture, global level: $X$ and $Y$ cannot be independent, because the value of $Y$ is constrained to a very specific range once a value of $X$ is picked. The reason why this can be confusing is because when you look at an integral, the integrand $dx\; dy$ looks like a product. If we explicitly write an indicator function $\chi_D(x, y)$ then it becomes more clear, but often a double integral is written with the domain as a subscript, which hides the fact that there is a nontrivial dependence between the variables. We'll look at this in detail shortly.

Beware of Transformations

Of course, when we routinely dealing with something with circular symmetry, like a disk, often the concept of polar coordinates comes to mind. It would seem intuitive that every radius from $0$ to $1$ is equally possible, and every angle in $[0, 2\pi]$ is, also. But let's take a look at what happens:

Oops! All that uniformity in $\theta$ is scrunched down near the origin, making the sampling very concentrated here. The intuition is that the polar coordinate transformation takes what is evenly spread out in the rectangle $[0,1] \times [0,2\pi]$ (imagine it as a grid of rectangles of equal area), and then warps a bunch of stuff, so that the transformed rectangles have different sizes. This demonstrates exactly why we need Jacobians when we deal with transforming the domains of integration.  To remedy this situation, recall that the area element of the disk is not $\frac{1}{2\pi} dr \; d\theta$ which is effectively what we have here, but rather $\frac{1}{\pi}r \; dr \; d\theta$ instead. The solution to understand things in these terms is to factor it $\left(2r\; dr \right)\left(\frac{1}{2\pi} d\theta\right)$. The CDF for $r$ is simply $r^2$, so we use its inverse n the generation of $r$ from a uniform random variable. A variant of this fix is essentially what we did in our previous post on uniform distribution on the sphere. The moral of the story is, just because we start out with uniform (independed) variables, it doesn't mean we can apply an arbitrary transformation to them and still get something uniform. We may have to tweak one or more of the variables.

Breaking Down Multidimensional Uniformity via Conditioning

We explore another way of generating multidimensional (jointly distributed) random variables, which will be important in the coming posts, and really, the most general way we can understand such variables one dimension at a time. We again begin with an approach that seems natural, but doesn't work. For this we just consider two uniform $[0, 1]$ variables and reject if the sum of their squares $> 1$. This should uniformly fill the upper right quarter disk (it's for convenience so that I don't have to deal with a bunch of plus/minus signs with square roots and state all sorts of caveats with inverse trig functions and calculating the integrals. You already know how much I don't like inverse trig functions).

The reasoning for this sampling method is as follows. If we are already given a uniform sample $y$, then the draw that produced it needs to make it fall within the disk, but with no other restrictions. Namely $x$ would be uniformly be within $[0, \sqrt{1-y^2}]$. Thus a seemingly good strategy would be to get the sample $v$, and synthesize $x$ as the appropriate uniform random variable. The density function for $x$ in this case can be computed by some manipulations of indicator functions:

$$\int_0^1 \frac{1}{\sqrt{1-y^2}} \chi_{[0, \sqrt{1-y^2}]}(x) \; dy$$

The reason for this function is that we are simply using integration to take some weighted averages of sample values continuously indexed by $y$. Namely, given a fixed $y$, we have that $x$ is a some possible value for the random variable, with pdf being the conditional distribution of $x$ given $v$. Then the probability density evaluated at $v$ is the weighting in this weighted average. This is called a mixture distribution. By manipulating the inequality $0 \leq x \leq \sqrt{1-y^2}$ to get it in a form that gives its explicit dependence on the variable we actually want to integrate, $y$, we see it is equivalent to $0 \leq y \leq \sqrt{1-x^2}$, not surprising due to the symmetry of a circle:

$$\chi_{[0, \sqrt{1-y^2}]}(x) = \chi_{[0, \sqrt{1-x^2}]}(y).$$

We need this, because a indicator function on $y$ instead of $x$ enables us to just change the limits of integration. Therefore 

$$p_X(x) = \int_0^{\sqrt{1-x^2}} \frac{1}{\sqrt{1-y^2}} \; dy$$

the integrand of which we recognize as the derivative of $\sin^{-1}(y)$. Thus the density is

$$p_X(x) = \sin^{-1}(\sqrt{1-x^2}) - x \sin^{-1}(0) = \sin^{-1}(\sqrt{1-x^2})$$

on $[0, 1]$. What does this actually look like? (To recap: we select the $y$-coordinate with the usual uniform randomness, and then we select the $x$-coordinate uniformly among all the remaining possible values that are possible once $y$ is fixed):

It totally gets scrunched up top there in the upper left. This is because the uniformity in 2 dimensions means that the frequency of sampling of the horizontal coordinate decreases as the vertical coordinate increases, but this is not accounted for if you just sample the vertical coordinate independently first. This is somewhat subtle and not intuitively obvious; it comes from conflicting definitions of uniformity and equality, and that:

One has to realize uniformity does not persist under transformations and projections.

In order to "fix" the problem of the nonuniformity happening here, we consider the uniform distribution in the disk as having density form $\frac{4}{\pi} dx \; dy$. If the conditioning on a specific value of $y$ gives a uniform distribution in $[0, \sqrt{1-y^2}]$, then we have to split it up between the numerator and denominator: essentially all we're doing is "undoing" the extra scrunching. In some sense, it compromises by thinning out what happens vertically, so that things sample less frequently in the small scrunched up horizontal intervals near the upper left corner.

$$\begin{multline}\frac{4}{\pi} dx\; dy = \frac{4}{\pi} \frac{1}{\sqrt{1-y^2}} \chi_{[0, \sqrt{1-y^2}]}(x) dx\; \sqrt{1-y^2}\; dy \\ = \left( \frac{1}{\sqrt{1-y^2}} \chi_{[0, \sqrt{1-y^2}]}(x) dx\right)\left(\frac{4}{\pi} \sqrt{1-y^2}\;dy\right).\end{multline}$$

The first factor is the conditional distribution $p_{X|Y}(x|y)\; dx$, not $p_X(x)\; dx$. Conditional distributions are how you can factor random variable generation into a sequence of generations each depending on the previous result. To evaluate $p_X(x)\; dx$, one integrates the expression over all $y$. This gives $p_X(x)\; dx = \frac{4}{\pi}\sqrt{1-x^2}\; dx$ which is exactly the same a the distribution of $y$! This should not actually be surprising, given that the circle is symmetric between $x$ and $y$ But it should be noted that using this distribution for $x$ after using it for $y$ will not give correct results, simply because once we've simulated $y$, and we act based on that value, the distribution is by definition conditional. How do we simulate $y$ with the distribution $\frac{4}{\pi}\sqrt{1-y^2}$? The (not so fun) cumulative distribution function corresponding to this is:

$$\frac{2}{\pi} \left(y \sqrt{1-y^2} +\sin^{-1}(y)\right)$$ for $y \in [0, 1]$.

We would need to apply the inverse of this function on a uniform random variable to get what we want. This will take a bit of finessing with Newton's method if you really want to solve it. (At the end of this series, I'll have a Jupyter notebook that will allow you to follow along). We've already described how to get the $x$ coordinate from this, but to recap: we just sample uniformly from $0$ to $\sqrt{1-y^2}$ which is simply generating the uniform from $0$ to $1$ and then rescaling.

But the real takeaway here is basically a granular understanding of a formula of conditional probability that is often committed to memory long before building intuition:

$$\rho_{X,Y}(x,y) = \rho_X(x) \rho_{Y|X}(y|x).$$

The first factor is the marginal pdf, namely, what happens when we forget about $Y$ entirely, and the second is the conditional pdf. This formula always holds, and is a replacement of the more classic formula $\rho_{X,Y}(x,y) = \rho_X(x) \rho_Y(y)$ when $X$ and $Y$ are independent (usually learned much earlier). Anyway, in this formula lies the key to bootstrapping ourselves up to higher dimensions! See you in the next post. (Here's a SNEAK PREVIEW!!)

Can Two Independent Random Variables Sum to Uniform?

As you well know, I'm a big fan of probability and thinking of it in geometric terms. This is because the sample spaces in probability are almost always some sort of high dimensional state space, that describe the possible outcomes (or configurations) of a system. Probability often boils down to measuring what fraction of the overall total space of outcomes, that your particular experiment resulted in. In all of I've read about and studied over the years, understanding the intuition behind probability has been the theme that recurs the most. Today I want to start on an exploration of the notion of uniformity. Now we actually have visited this theme before as well, and the next couple of posts will actually expand upon those, exploring interesting facts about what happens in higher dimensions. There will be plenty of geometry to visualize!

But today I want to start very simple. What can be simpler than a uniform random variable between 0 and 1? It's basically everyone's random number generator: many distributions are simply calculated by combining multiple uniform random variables in different ways, for example, by arithmetic operations (adding, multiplying, raising to powers), applying functions (exponentials, sin, cos, etc), applying an inverse cumulative distribution function, etc. And we will attempt to do something that may look like it breaks all sorts of rules! Break rules? Consummate rule-follower mathematicians BREAKING RULES? What has the world come to? Now I don't encourage literally breaking rules and deliberately getting into trouble. At least, I'm not supposed to. All I am saying to do is to play around a bit and develop intuition, and try to apply things in ways you're not entirely sure of. It's the sense of adventure and some risk-taking, and what the nature of math exploration actually is. At the end of the day, we will need to have firmer justifications for things and we do need to verify. But, thinking of math as living, breathing, experimental science is not what students are used to thinking. Sometimes the intuitions behind things are lost, as well as a sense of adventure. Let me tell you, this was definitely quite the adventure... You will see me get very excited, and I hope it'll rub off on you, too.

Triangular Distribution

Now, adding independent random variables is a very well-known operation. Given $X$ and $Y$ independent uniform random variables (say on $[0, 1]$), if $Z = X + Y$, this is another random variable. What may be surprising at first is that such a sum is actually NOT uniformly distributed. It is easy enough to see that its maximum range is from 0 to 2. Next, we do something like, ask ourselves, when can the two variables sum to something in the range of 1.5 to 2? Once one of the variables is pinned down, this puts a constraint on the possibilities of the other variable. If we get, say, 0.9 for the first, then the second variable can only be in the range of 0.6 and onward, and actually, you would not actually be able to achieve the value 2, since it requires a value of 1.1, which is not an acceptable value. 

In fact, we get something called the triangular distribution. Numbers close to 1 have the highest probability (density) for a sum, while those closer to the extreme ends have lower probabilities, because they rely on each uniform variable to have values clustered very close to 0 or very close to 1. 

So having been sort of surprised about summing two uniform random variables to be non-uniform, one natural question may be to ask, is it actually possible to sum two independent random variables to be uniform? At least, that's the question that occurred to me. If you like, think about it for a bit and pause... Don't spend more than 10 minutes, though.

....

....

Ok. So it turns out that it actually is impossible. There is a very nice, visual proof at Stats StackExchange, which I actually like a lot. It would have made a nice Nested Tori post by itself. Perhaps I'll try to take it apart and explain it someday, but I also think they give a pretty good exposition as it is.

Is That All There Is to it? Should we just give up?

I mean, yes. The theorem says so. At least, we should give up on actually finding those two random variables... And that is well and good. However, should it be the end of the story? What happens if we try anyway? There's many interesting further explorations one can do here to shore up the probabilistic intuition of what is going on. First, knowing that there ARE random variables that can be subdivided in the manner we just described. The exponential (actually, the negative log of a uniform variable), in particular, is the sum of two identically distributed chi-square random variables. The exponential distribution is also implicated in another somewhat related and very interesting theorem about the product of identically distributed uniform random variables, both raised to a power of another such, discussed here by our Summer of Math host.

A second interesting direction is to realize that there are actually ways to sum random variables to uniform. They just cannot be independent. This has come up in some attempts I made to gain intuition on Bose-Einstein statistics, a quantum phenomenon in which indistinguishable particles or states actually behave as if all that binomial stuff collapses, as if, flipping two coins produced three outcomes of equal probability, {heads, heads}, {heads, tails}, and {tails, tails} and somehow that mixed outcome is weighted less. I could do a whole different post about this, as it is also fun, but fundamentally, it boils down to asking this exact same question about summing to uniform. It's just that, in order to solve it, I had to give up independence, in which I would flip different (unfair) coins depending on the outcome. 

But the third adventure is a real fun one, and that's what the rest of this post is about. Essentially, it had partially been started before learning of that Stats StackExchange proof and I worked it out in different phases. Had I known the StackExchange proof, I might well have never gotten started.

Probability distributions are characterized by their density functions, and we actually very systematically understand what effect does taking a sum of two independent random variables has on their density functions. It's a very well-known operation called convolution.

Convolutions

Given the density functions $\rho_X$ and $\rho_Y$, the density function $\rho_{X+Y}$ is given by the convolution integral

$$\rho_{X+Y}(z) = (\rho_X * \rho_Y)(z) = \int_{-\infty}^\infty \rho_X (x)\rho_Y(z-x) \; dx.$$

The intuition here is using one function to smear out another function by weighted averages. It's somewhat related to the convolutions encountered in machine learning and AI, especially if multidimensional, but convolution is very interesting and important operation in math, useful in all sorts of real analysis contexts. It's also an engineering favorite. This has various nice properties reminiscent of multiplication, like being associative, commutative, and distributive over addition of functions. As for an "identity" ... well... we'll get to that. The reason why convolution is very popular is because of its behavior under the Laplace transform (and Fourier transform as well, but the Laplace transform is what turns out to be the key to really solving this problem). This theorem is the

Convolution Theorem.

Given $f, g$ "nice" functions, the Laplace transform of the convolution, $\mathscr{L}[f*g](s)$, is equal to the product of the Laplace transforms $\mathscr{L}[f](s) \mathscr{L}[g](s)$.

In other words, on the transform side, it goes from being "like" multiplication to just, well, being multiplication. It is for this reason that Laplace transforms are useful in differential equations, and is often quoted as "transforming calculus problems into algebra problems". It should also be noted that convolutions are well-defined for functions that may not necessarily be probability density functions, and the theorem still holds, when the Laplace transforms of such functions exist. Or even, when they don't ... that's a different part of the adventure....

So what is happening in terms of probability density functions (pdfs) is that if one takes Laplace transforms of them, then the Laplace transform (LT) of the pdf of the random variable representing the sum, is the product of the LTs of both pdfs. The LT of uniform with pdf $\chi_{[0,1]}$, the indicator (or characteristic) function that is $1$ on $[0, 1]$ and $0$ elsewhere,

$$\int_{0}^\infty e^{-st} \chi_{[0, 1]} \; dt = \int_{0}^1 e^{-st} \; dt = \frac{1-e^{-s}}{s}.$$

So, now, if we want two independent random variables $\rho_X$ and $\rho_Y$ to represent something that sums to uniform, this means their convolution is the uniform pdf.

But that means ... their Laplace transforms multiply to $ \frac{1-e^{-s}}{s}$! (that's me exclaiming, not taking the factorial). But if they are identically distributed... This means the our purported density function for this is $$\mathscr{L}[\rho_X] = \mathscr{L}[\rho_Y] = \sqrt{\frac{1-e^{-s}}{s}}.$$

Great!!! All we have to do, then, is find the inverse Laplace Transform! No problem at all!!

Except... well... ::opens up a reference book:: looks at the Bromwich Integral... ok where am I supposed to put the contours? What are the singularities? What about that branch cut? Didn't we spend a lot of time last year and years before that hating on branch cuts? ARRRRGH WHAT?? :: slinks away in defeat... ::

Highly Suspect Distributional Calculus

So back to the drawing board. Let's pick apart what that Laplace transform of a uniform pdf really means. It is $\chi_{[0,1]}$, with transform $\frac{1 - e^{-s}}{s}$. As Laplace transform fans know, multiplication by $s$ is differentiation, and division by $s$ is integration starting at some point. Now let's forget that denominator for a moment, and look at pieces separately.

Laplace (and Fourier) transform fans are may be familiar with some (Schwartzian) distributions, or generalized functions (we'll use the latter term since the word "distribution" is overused) namely the Dirac delta (impulse "function") and its derivatives. I actually first learned about that function in an engineering class, not in a math class. And the dang thing is sure useful. The important things to know about this "function" (As a mathematician, I'm obligated to put function in quotes):

1. $\int_{-\infty}^{\infty} \delta(t) \; dt = 1$.
2. $\delta * f = f*\delta = f$ (it is the identity for convolution).
3. The Laplace transform of $\delta$ is $1$ (actually this follows from the above by the Convolution theorem).
4. $\delta$ is the (suitably generalized) derivative of the unit step function (Heaviside function) $\chi_{[0, \infty)}$, the function that is $1$ on the positive real line, and $0$ elsewhere.

It can be visualized as having a graph with infinitely skinny, infinitely high peak at $0$, a value of $0$ elsewhere, yet has a total area under it equal to $1$ (no actual function on the real line can do this: being nonzero at only one point always means it integrates to $0$). When the derivative of something is a $\delta$, that represents a jump discontinuity of size $1$. Yes, I know some of those listed properties do need some more justification, since $\delta$ is not in fact actually a function, but rest assured, there are firm foundations for this.

So if we just look at $1 - e^{-s}$, this corresponds to the Dirac delta "function", minus the Dirac delta "function" shifted to the right by $1$. This is also intuitive: the uniform pdf $\chi_{[0,1]}$  has two jumps, the value going from $0$ to $1$ at the point $0$, and from value $1$ back to $0$, at the point $1$. In general, $e^{-as}\mathscr{L}[f](s)$ the transform of the function $f(x-a)$, another useful property (the translation property)... we'll be using this a lot, soon.

Now if we take the square root:

$$\sqrt{1-e^{-s}}$$

then corresponds to the convolutional square root of that difference of deltas. What can we do with that? There is another thing we know: we know how to expand $\sqrt{1 - e^{-s}}$ in a power series (binomial series) (but crucially, and this is what took some time for me to see, in powers of $e^{-s}$, not $s$). For $\operatorname{Re}(s) > 0$, the series converges uniformly, so there are a number of good analytic operations we can do. Namely, the series is

$$\sqrt{1 - e^{-s}} = (1-e^{-s})^{1/2} = \sum_{n=0}^\infty {1/2 \choose n}(-1)^n e^{-ns}$$

The suspicious looking binomial coefficent ${ 1/2 \choose n}$ is defined using the Gamma function, the usual function that generalizes factorial to non-integers. The notation really is the generalization of binomial coefficient to the non-integer case and is written that way precisely because it generalizes a binomial expresion like $(1+y)^n$ to a non-integer power. It's just that the coefficients, once nonzero, never become zero again, so expanding it out makes it an infinite series. We find that we actually can massage the coefficient a bit to express it in terms of the more normal (integer) binomial coefficients:

$${1/2 \choose n} = \frac{1}{2^{2n-1} n} (-1)^{n+1} { 2n-2 \choose n-1}$$ (and conventionally taking it to be $1$ at $0$). Then $e^{-ns}$ is the transform of $\delta$ translated by $n$ (it's using the property that its LT is 1, and the translation property). Taking this inverse transform we have:

$$\mathscr{L}^{-1}\left[\sqrt{1-e^{-s}}\right] = \sum_{n=0}^\infty { 1/2 \choose n}(-1)^n\delta_n = \delta_0 - \sum_{n=1}\frac{1}{2^{2n-1} n}{ 2n-2 \choose n-1} \delta_n =: \rho(t).$$

The latter sum rather nicely gets rid of the signs, so that only the first term has a plus sign and the rest of the terms are subtracted from it.

A more "illegal" way of seeing this is to consider $\delta_0$ the identity operator as a convolution, and $\delta_0 - \delta_1$ as being a "perturbation" of the identity, and we formally take the square root of that difference using Binomial series:

$$\sqrt{\delta_0 - \delta_1} = \delta_0 - \frac{1}{2} \delta_1 - \frac{1}{8} \delta_1^2 - \dots = \sum_{n=0}^\infty { 1/2 \choose n}(-1)^n\delta_1^n$$

and using the fact that $\delta_1^n = \delta_n$ (because convolution by $\delta_1$ multiple times will translate by 1 over and over again). This is the best of Euler and he would do it with abandon, without regard to convergence. The advantage of the Laplace transform method is that we have all the series converging and we know that stuff like term-by-term integration is guaranteed to work in such circumstances.

In either case, this is indeed a generalized function in the same sense as $\delta$ is, and in fact, a tempered one: given a function $\phi$ the very nice Schwartz class of smooth functions vanishing at infinity after multiplication by polynomial of any degree, when $\rho$ operates on it, the resultant infinite series of translations of $\phi$ must converge, since each discrete value will be dominated by any negative $n$th power. The situation is even nicer with smooth functions of compact support (vanishing outside a finite interval): $\rho$ will always be a finite sum at any evaluation point. Suffice to say, the convergence is good enough.

Next We Will Need: The Square Root of .... Integral?

Ah. But we're missing something. We said to ignore that denominator $s$, so we're still not done yet. It's time to bring it back. Now, just like multiplying by $s$ corresponds to differentiation, dividing by $s$ corresponds to integration (from 0). Since the convergence is nice, dividing by $s$ we can just integrate that crazy sum above term-by-term, and be done..........

Oh wait. We divide by the square root of $s$, not $s$. Oops. What does dividing by the square root of $s$ mean? Dividing by $s^2$ is integrating twice. So dividing by the square root of $s$ is, of course, just integrating half a time. WHAT??? Yes the square root of integral!! (a double exclamation. Not the double factorial of integral, although I wonder if it's possible to..........)

One additional way to motivate things is that differentiation plays very well with convolution. Namely $\partial^{i+j} ( f * g) = (\partial^{i+j} f) * g = f*(\partial^{i+j} g) = \partial^i f * \partial^j g$ in any combination that makes sense. You can split the differentiations up to happen on each component that can take that many derivatives, you can put them all on one factor, or the other factor, you can re-split it up in different ways, etc. Similarly, if we think of integration from $0$ to $t$ as a negative order differential, it all continues to work out, too.

So here if we define $\rho$ to be the binomial-series sum of deltas above, we have $\rho * \rho = \delta_0 - \delta_1$ by construction. If we have to do it on a test function, fine. The second convolution will stagger the additional terms, so they all cancel exactly outside of $[0, 1]$. Then the function we seek, $\tilde \rho$ satisfies $\tilde \rho * \tilde \rho = \chi_{[0,1]} = \int_0^x \delta_0(t) - \delta_1(t) \; dt$.

Thus $\tilde \rho * \tilde \rho = \int_0^x \rho * \rho \; dt = \partial^{-1}(\rho * \rho)$, that is, it is the integral of the whole thing. For our next illegal trick, we're going to split that integration in exactly half, and put it on each individual $\rho$:

$$\tilde \rho * \tilde \rho = \partial^{-1}(\rho * \rho) = \partial^{-1/2} \rho * \partial^{-1/2} \rho.$$

Total nonsense?? Maybe. I have not actually justified for generalized functions in general, but we'll see in a short bit how Laplace transforms take care of all of it, at least with the exponent of ½. For functions vanishing at $0$, it sure is convenient (vanishing at $0$ makes it work well with Laplace transforms, since we often need to subtract derivatives of initial conditions). It turns out the iterated integral operation has, what is called, a kernel representation: you can represent iterated integrals as one single integral. The "kernel" in kernel representation is analogous to continuous version of a matrix, and the integral against a function is like a continuous sum version of matrix multiplication.

The kernel representation of iterated integration is:

$$\int_0^t \int_0^{t_{n-1}} \dots \int_0^{t_0} f(t_0) \; dt_0 \; dt_1 \; \dots \; dt_{n-1} = \frac{1}{(n-1)!}\int_{0}^t f(\tau)(t-\tau)^{n-1}\; d\tau.$$

The proof is actually very pretty straightforward: interchanging the order of integration and proof by induction. When the innermost variables are switched, the inside is constant with respect to the switched variable, so integrates to a difference. Stacking them up will give you the powers, so now it's only one integral.

But the amazing thing is... that expression is defined for all real numbers $\nu$ (we can rely on our trusty old friend, the Gamma function, for the factorial). We DEFINE the fractional integral

$$I^{\nu}(f)(t) = \partial^{-\nu}(f)(t) = \frac{1}{\Gamma(\nu)}\int_0^t (t - \tau)^{\nu - 1}f(\tau) \; d\tau.$$

The important thing to know here is $I^{\nu + \mu} = I^\nu I^\mu$. This can be shown by composing two integrals and using beta functions (of Bayesian statistics fame) to rewrite the product $\Gamma(\alpha) \Gamma(\beta)$. It will work for non-integer $\nu$ and $\mu$. It's a bit of a mess, so we won't reproduce all of it here. Anyway, this makes integration operations of all orders a form of convolution—the telltale sign being, the $\tau$ and the $t-\tau$, though there is some funny business with the limits of integration. Also, there's trouble if $\nu$ is a nonpositive integer (because the gamma function is infinite in such places), with the function $\frac{1}{\Gamma(\nu)}t^{\nu-1} \chi_{[0, \infty)}(t)$. Thus if we want to write it for $\nu = -\frac{1}{2}$, we have (recalling $\Gamma(1/2) = \sqrt{\pi}$, another amazing fact, and another place where $\pi$ has a habit of just inviting itself to random parties):

$$I^{1/2}(f)(t) = \frac{1}{\sqrt{\pi t}} \chi_{[0, \infty)}(t) * f(t).$$

We can also justify this a bit more by showing that the Laplace transform of $\frac{1}{\sqrt{\pi t}}\chi_{[0, \infty)}(t)$ is in fact $\frac{1}{\sqrt{s}}$ so that it just becomes $\frac{1}{\sqrt{s}}$ times the original series with $\sqrt{1-e^{-s}}$. Then this means the original transform is just a series of translated $\frac{1}{\sqrt{\pi t}}\chi_{[0, \infty)}(t)$'s. But STILL, ISN'T IT TOTALLY AWESOME THAT WE TOOK THE SQUARE ROOT OF INTEGRAL???

AND FINALLY...

So what's left to do? We need to take the ½th integral of that sum of delta's above. So this is just convolving the function $$\frac{1}{\sqrt{\pi t}} \chi_{[0, \infty)}(t)$$ with the majestic

$$\delta_0 - \sum_{n=1}\frac{1}{2^{2n-1} n}{ 2n-2 \choose n-1} \delta_n$$

Convolutions with deltas are supposed to be trivial, because they're the identity op, remember? We have some shifted deltas, however. Convolving with it gives a shifted function, namely, we shift that $\frac{1}{\sqrt{\pi t}} \chi_{[0,\infty)}(t)$ over by $n$ when we apply $\delta_n$. The result is therefore

$$\tilde \rho(t) = \frac{1}{\sqrt{\pi t} } \chi_{[0, \infty)}(t)* \rho(t) = \frac{1}{\sqrt{\pi t} }\chi_{[0, \infty)}(t)- \sum_{n=1}^\infty\frac{1}{2^{2n-1} n}{ 2n-2 \choose n-1} \frac{1}{\sqrt{\pi (t-n)} }\chi_{[0, \infty)}(t-n)$$

Those indicator functions are bloody annoying looking. I kept them in specifically because it allows you to keep track of the bounds of integration better. But there's a nice reformulation, since here, the action of these indicator functions is merely to cut off the sums depending on whether we have encountered the relevant delta term. In fact, its effect is to finitize the sum (which turns out to allow us to dodge some convergence questions): at any given $t$, only the terms up to $\lfloor t \rfloor$ have been added! We thus rewrite:

$$\tilde \rho(t) = \frac{1}{\sqrt{\pi t} } - \sum_{n=1}^{\lfloor t \rfloor}\frac{1}{2^{2n-1} n}{ 2n-2 \choose n-1} \frac{1}{\sqrt{\pi (t-n)} }$$

for $t > 0$ and $0$ otherwise. This includes a value of $\pm \infty$ at every integer. To avoid that, we can use $\lceil t \rceil -1$ instead of $\lfloor t \rfloor$, but infinite values on a discrete set of measure zero (set with no length) don't affect the result, especially when it comes to convolution, so we'll not bother.

Of course, we should actually probably take a look at what we just constructed. This is the graph of $\tilde{\rho}$:

You can very plainly see it has negative values, which prevents it from being a pdf. We've shown that a convolutional square root does in fact exist, but it is not a pdf. Because the uniform pdf has a downward jump, no fix can be made to eliminate the negative values (since a downward jump requires convolution by something with vertical asymptotes, whose values approach $-\infty$). And that, my friends, is my very satisfying negative answer to the this post's title! The the number of different tools and areas of math that this touches upon, not to mention, having just a random factor of $\pi$ showing up where there are no circles, is a real delight!