What is half a derivative, and why do actuaries care?

A student asked me this once and I did not have a good answer ready. We had just done the third derivative of something and she said, mostly to be funny, "what about the half derivative". I laughed and said it doesn't mean anything. That was wrong. It means something, people have been working on it since 1695, and this year I have been reading actuarial papers built on top of it while studying for exams.

So here is the answer I wish I'd given her. It takes a while, but you'll be able to compute one yourself by the end.

Where the idea comes from

Leibniz and l'Hopital were writing letters about the new calculus, and l'Hopital asked what \( \tfrac{d^n y}{dx^n} \) would mean if \( n \) were \( \tfrac{1}{2} \). Leibniz wrote back that it was an apparent paradox from which, one day, useful consequences would be drawn. He was right, just slowly. It took about three hundred years.

The instinct is the same one behind a lot of mathematics you already know. You learn \( 2^3 \) as three twos multiplied together, which makes \( 2^{1/2} \) meaningless, and then you find a rule that still works and suddenly \( \sqrt{2} \) is sitting there. You learn factorials on whole numbers and then the Gamma function slides in and gives you \( (1/2)! \). Same move every time: find the pattern that the whole number version obeys, then ask what else obeys it.

Integrating is the easier door

Differentiating half a time is awkward to attack directly. Integration is friendlier, because there's a formula for repeated integration that already has an \( n \) sitting in it.

If you integrate a function \( n \) times starting from \( a \), you don't need \( n \) nested integrals. Cauchy's formula collapses it to one:

\[ {}_aI_x^{\,n} f(x) \;=\; \frac{1}{(n-1)!}\int_a^x (x-t)^{\,n-1} f(t)\,dt \]

Look at what that's doing. To integrate repeatedly, you sweep over every earlier value of the function from \( a \) up to \( x \), and you weight each one by how long ago it happened, through the \( (x-t)^{n-1} \) factor. Hold onto that. It comes back later and it's the whole reason actuaries showed up in this article.

Now, the only thing stopping us putting \( n = \tfrac{1}{2} \) into that formula is the factorial. And the Gamma function already solved that, with \( \Gamma(n) = (n-1)! \) for whole numbers and sensible values everywhere else. Swap it in and you get the Riemann-Liouville fractional integral:

\[ {}_aI_x^{\,\alpha} f(x) \;=\; \frac{1}{\Gamma(\alpha)}\int_a^x (x-t)^{\,\alpha-1} f(t)\,dt \]

That's it. That's a fractional integral. Nothing exotic happened, we just let \( \alpha \) be any positive number instead of a whole one.

And now the derivative, by a slightly sneaky route

There's no equally direct formula for repeated differentiation, so the standard trick is to go the long way round. To take half a derivative: integrate half a time, then differentiate once, and you're left owing half a derivative. In symbols, for \( 0 < \alpha < 1 \):

\[ {}_aD_x^{\,\alpha} f(x) \;=\; \frac{d}{dx}\left[ {}_aI_x^{\,1-\alpha} f(x) \right] \;=\; \frac{1}{\Gamma(1-\alpha)}\,\frac{d}{dx}\int_a^x \frac{f(t)}{(x-t)^{\alpha}}\,dt \]

Two warnings before we compute anything, because both of them bite.

First, the lower limit \( a \) is part of the answer. An ordinary derivative doesn't care where you started measuring. A fractional one does, because it integrates over the whole stretch from \( a \) to \( x \). Change \( a \) and you change the result. Most of the time people use \( a = 0 \), and sometimes \( a = -\infty \), and these give genuinely different answers for the same function. I'll show you that happening.

Second, there is more than one definition. Swapping the order, integrating after differentiating, gives the Caputo derivative, which applied people usually prefer because it lets you state initial conditions in the ordinary way. For everything below the two agree, so I'll stay with Riemann-Liouville.

Four worked examples

The test for all of these: do it twice and you should get the ordinary first derivative back. If half a derivative means anything, two of them have to make one.

1. The polynomial, where everything behaves

Run \( f(t) = t^{p} \) through the definition with \( a = 0 \) and the integral is one you can do by hand with the substitution \( t = xu \). It lands on a rule worth memorising:

\[ {}_0D_x^{\,\alpha} x^{p} \;=\; \frac{\Gamma(p+1)}{\Gamma(p-\alpha+1)}\;x^{\,p-\alpha} \]

Check it against something you know. Whole \( \alpha = 1 \), \( p = 2 \): you get \( \Gamma(3)/\Gamma(2) \cdot x = 2x \). Correct.

Now half. With \( p = 2, \alpha = \tfrac12 \), using \( \Gamma(3) = 2 \) and \( \Gamma(5/2) = \tfrac{3\sqrt{\pi}}{4} \):

\[ {}_0D_x^{1/2} x^{2} \;=\; \frac{\Gamma(3)}{\Gamma(5/2)}\,x^{3/2} \;=\; \frac{2}{\tfrac{3\sqrt\pi}{4}}\,x^{3/2} \;=\; \frac{8}{3\sqrt{\pi}}\;x^{3/2} \approx 1.5045\,x^{3/2} \]

A strange object. Half of \( x^2 \) and half of \( 2x \), landing on \( x^{3/2} \) exactly between them. Do it again:

\[ {}_0D_x^{1/2}\left[\frac{8}{3\sqrt\pi} x^{3/2}\right] = \frac{8}{3\sqrt\pi}\cdot\frac{\Gamma(5/2)}{\Gamma(2)}\,x = \frac{8}{3\sqrt\pi}\cdot\frac{3\sqrt\pi}{4}\,x = 2x \]

Two halves make a whole. The \( \sqrt\pi \) that looked so out of place cancels itself.

2. The logarithm, which hides a nice surprise

For \( f(x) = \ln x \) with \( a = 0 \), the result is

\[ {}_0D_x^{1/2} \ln x \;=\; \frac{\ln(4x)}{\sqrt{\pi x}} \]

which is already odd, because a 4 has appeared from nowhere. Doing it twice should give \( 1/x \), and it does, but the reason is the best part of this whole section.

Split it up: \( \frac{\ln 4}{\sqrt\pi}x^{-1/2} + \frac{1}{\sqrt\pi}x^{-1/2}\ln x \). Now feed that first piece into the power rule with \( p = -\tfrac12, \alpha = \tfrac12 \). The denominator becomes \( \Gamma(0) \), which is infinite, so the whole term is zero.

So \( x^{-1/2} \) is invisible to the half derivative. It gets annihilated, the way a constant is annihilated by an ordinary derivative. Every fractional derivative has its own private set of things it can't see, which is a fact worth sitting with, and it is why adding "plus a constant" to a fractional integral is not the end of the story.

The surviving log term works out to \( \sqrt{\pi}/x \), the \( \sqrt\pi \) cancels again, and you land on \( 1/x \). I checked all of this numerically to twenty five decimal places before publishing it, because the first time I tried it my quadrature was wrong and I nearly wrote the wrong thing here.

3. The exponential, which refuses to cooperate

Here's the one that should bother you. The defining property of \( e^x \) is that differentiating leaves it alone. So half a derivative should too, surely.

It doesn't. With \( a = 0 \):

\[ {}_0D_x^{1/2} e^{x} \;=\; \frac{1}{\sqrt{\pi x}} \;+\; e^{x}\,\mathrm{erf}(\sqrt{x}) \]

An error function turns up, and a term that blows up at the origin. Do it twice and you do get \( e^x \) back, so the machinery is consistent, but the halfway house is nothing like \( e^x \).

The culprit is that lower limit. We started the clock at \( 0 \), which chopped off everything \( e^x \) was doing before then, and \( e^x \) very much does not vanish there. Start at \( a = -\infty \) instead, the Liouville form, and the mess evaporates:

\[ {}_{-\infty}D_x^{\,\alpha} e^{x} = e^{x} \quad\text{for any } \alpha \]

Same function, two different answers, and neither is wrong. Which one you want depends on whether your problem has a beginning. A process that started when the policy was written has a beginning. A wave that has always been oscillating does not.

4. The sine, where it finally looks beautiful

On the whole line, \( {}_{-\infty}D_x^{\alpha} e^{ix} = i^{\alpha}e^{ix} = e^{i\alpha\pi/2}e^{ix} \), and taking imaginary parts gives something lovely:

\[ {}_{-\infty}D_x^{\,\alpha} \sin x \;=\; \sin\!\left(x + \frac{\alpha\pi}{2}\right) \]

Fractional differentiation of a sine wave is just a phase shift. A whole derivative shifts by 90 degrees, turning sine into cosine, which you already knew without thinking of it that way. Half a derivative shifts by 45 degrees:

\[ {}_{-\infty}D_x^{1/2}\sin x = \sin\!\left(x+\tfrac{\pi}{4}\right) = \frac{\sin x + \cos x}{\sqrt{2}} \]

Do it twice, 45 plus 45, and you're at cosine. The order of the derivative is a dial that rotates the wave continuously, and the integers we normally use are just four evenly spaced clicks on it.

The property that actually matters: memory

Look back at the definition. An ordinary derivative at \( x \) only needs an infinitesimal neighbourhood of \( x \). It's local. It has no idea what the function did an hour ago.

A fractional derivative integrates from \( a \) all the way to \( x \), weighting the past by \( (x-t)^{-\alpha} \). That weight decays like a power law, not an exponential, so the distant past never fully stops counting. The operator remembers.

That single structural fact is why fractional calculus escaped pure mathematics. Viscoelastic materials remember being stretched. Charge in some batteries remembers its cycling history. Anomalous diffusion, where a particle spreads out faster or slower than the usual square root of time, comes out naturally.

And insurance claims, it turns out, remember too.

Ruin theory, and the assumption everyone quietly makes

Here's the classical setup, which is about as compact as risk models get. An insurer starts with capital \( u \), collects premium at a steady rate \( c \), and pays claims as they arrive:

\[ U(t) \;=\; u + ct - \sum_{i=1}^{N(t)} X_i \]

\( N(t) \) counts claims, \( X_i \) are their sizes. Ruin is the event that \( U(t) \) ever drops below zero, and the quantity everyone wants is \( \psi(u) \), the probability that it eventually does.

This is the Cramer-Lundberg model, and it's the piece of the syllabus the whole subject hangs off. Its famous result is Lundberg's inequality, a clean upper bound:

\[ \psi(u) \;\le\; e^{-Ru} \]

where \( R \), the adjustment coefficient, falls out of the claim distribution and the premium loading. It's a genuinely lovely theorem. Ruin probability decays exponentially in your starting capital, so every extra dollar of reserve buys you a fixed multiplicative reduction in risk. That exponential is doing a lot of work in how capital requirements get argued about.

But it rests on an assumption that's easy to miss. Taking \( N(t) \) to be a Poisson process means the waiting times between claims are exponential, and the exponential distribution is memoryless. Having waited an hour for the next claim tells you precisely nothing about how much longer you'll wait. The process has no history.

Which is a strange thing to assume about catastrophes. Hurricanes come in seasons. Earthquakes have aftershocks. A hailstorm generates a thousand motor claims in one afternoon and then nothing for a year. Claims arrive in bursts, and bursts are exactly what a memoryless model cannot produce.

What happens when you let the arrivals remember

This is where the two halves of this article meet. Replace the Poisson process with a fractional Poisson process, where the waiting time between claims follows a Mittag-Leffler distribution instead of an exponential. The Mittag-Leffler function

\[ E_{\alpha}(z) \;=\; \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma(\alpha k + 1)} \]

is what the exponential turns into when you solve a fractional differential equation instead of an ordinary one. At \( \alpha = 1 \) the Gamma reduces to a factorial and you get \( e^z \) back, so the classical model is the special case sitting inside the general one, which is the sort of thing that makes a framework worth taking seriously.

Its tail is heavier than exponential, and the resulting counting process is non-Markovian with long-range dependence. Claims cluster. The past leaks into the future.

So what does that do to the answer? Four results I found genuinely surprising:

• The long run is tougher than you'd think. Biard and Saussereau established the long-range dependence of the fractional Poisson process and worked out ruin probabilities under it, for both light and heavy claim sizes.
• The stress shows up early, not late. Kumar, Leonenko and Pichler found that the fractional model puts what they call initial stress on the surplus process, while the average capital needed to recover after ruin does not change when you switch to the fractional regime. Clustering hurts you near the start of the horizon, and some long-run quantities stay put. If you only ever look at the ultimate ruin probability, you'd miss the entire effect.
• The Lundberg bound has been rebuilt. Leonenko, Pepelyshev, Pichler and coauthors published a Cramer-Lundberg inequality for fractional risk processes in TEST in 2026, along with closed-form finite time ruin probabilities for exponential claims. Finite time is the practically important case and the classical theory is mostly silent on it.
• Catastrophe clustering has its own machinery now. Hu, Rachev, Sayit, Yang and Yildirim built a model on repeatedly time-changed Poisson processes specifically to capture clustered catastrophic arrivals, and derived a closed-form scale function so ruin probabilities stay computable. They aim it directly at solvency assessment, reinsurance pricing and capital reserving.

The "ceiling or floor" question is the interesting one. Lundberg's exponential bound is a ceiling on ruin probability, and its exponential shape is what makes holding capital feel so effective. Once arrivals cluster and dependence stretches across time, that shape is not guaranteed. A bound that decays more slowly in \( u \) means each additional dollar of capital buys less safety than the classical model promised, and the gap widens exactly where you care most, out in the tail where the catastrophes live.

That is the practical worry in one line. Not that the classical answer is wrong, but that it may be optimistic in precisely the scenario you bought the capital for.

An honest caveat

I should be careful here, because it would be easy to oversell this. As far as I can tell this is research-stage mathematics, not something your insurer is running tonight. Regulatory capital frameworks are conservative by design and they do not adopt a new arrival process because a paper is elegant. Most of the applied work still handles clustering with tools that are already accepted, things like compound distributions and catastrophe models built on simulated event sets.

What I find compelling is the direction. The memory that makes fractional calculus awkward is the same memory that insurance risk actually has, and it's rare for a piece of abstract machinery to fit a real problem that snugly.

Why I think this is worth a student's time

Every step here was an extension of something small. Factorials to the Gamma function. One integral to \( n \) of them to \( \alpha \) of them. A memoryless waiting time to one that remembers. Nobody had to invent a new universe, they just refused to accept that a parameter had to be a whole number.

That's the move, and you can start practising it now. When you meet a rule stated for integers, ask what it would mean in between. Most of the time nothing useful happens. Occasionally you get three hundred years of mathematics and a new way to think about whether an insurance company survives a bad decade.

And if you ever ask your teacher what half a derivative is, I hope they do better than I did.

Sources

Everything above is drawn from the following. The arXiv links are free to read in full.

• Kumar, A., Leonenko, N., Pichler, A. Fractional risk process in insurance, Mathematics and Financial Economics. Preprint: arXiv:1808.07950
• Biard, R., Saussereau, B. Fractional Poisson process: long-range dependence and applications in ruin theory, Journal of Applied Probability 51(3), 727-740. Cambridge Core
• Leonenko, N., Pepelyshev, A., Pichler, A. et al. Probability of ruin within finite time and Cramer-Lundberg inequality for fractional risk processes, TEST 35, 23-48 (2026). Springer
• Hu, D., Rachev, S. T., Sayit, H., Yang, H., Yildirim, Y. Iterated Poisson processes for catastrophic risk modeling in ruin theory (2025). arXiv:2501.11322
• Background on the classical model: Ruin theory and Fractional calculus.

The four worked derivatives were checked numerically to twenty five decimal places, and the logarithm case was confirmed analytically after the first numerical attempt disagreed. If you spot an error, I would genuinely like to hear about it.