Taylor Expansion — Cheatsheet

The idea in one sentence

Stand at one spot on a curve, measure everything you can there, and use those measurements to guess the height somewhere else.

You’d do this when the curve is hard to compute everywhere but easy to measure at one point, or when you want to see what drives a change. A bond’s price is a familiar case: you know its yield and duration today and want to know what happens if rates move.

The guess is built in layers. Each layer uses one more thing you measured at the anchor. The first layer is a straight line. The second bends it. The third bends the bend. You stop when the layers stop mattering, or when you run out of measurements.

Worked example: guess x³ at x = 1.2 using only what you know at x = 1

The curve is y = x³. You are standing at x = 1. The true answer is 1.2³ = 1.728, but pretend you can’t compute it.

What you know at x = 1

Thing	How you get it	Value at x = 1
Height	x³	1
Slope	derivative, 3x²	3
How fast the slope changes	derivative of that, 6x	6
How fast that changes	derivative of that	6
Anything further	derivative of a constant	0

The walk. Destination minus start: 1.2 − 1 = 0.2. Call it h.

Layer 1: pretend the slope stays 3. Guess = height + slope × walk = 1 + 3 × 0.2 = 1.6. Off by 0.128.

Layer 2: the slope drifts. It goes up at 6 per unit, so over the walk it rises from 3 to 3 + 6 × 0.2 = 4.2. Use the average slope, 3.6, instead of 3. Guess = 1 + 3.6 × 0.2 = 1.72. Off by 0.008.

Written as a separate correction: the new piece is ½ × 6 × 0.2² = 0.12, added to the 1.6.

Layer 3: the drift rate drifts. Same move one level deeper. The correction is 6 × 0.2³ ÷ 6 = 0.008. Guess = 1.728. Off by exactly 0.

Layer 4 and beyond: the measurement is 0, so every further correction is 0.

The guess is now perfect at x = 1.2, and it’s perfect at every other x too. x³ only has three pieces of information in it. Use all three and you have rebuilt the function.

Try it at x = 3 (walk h = 2): 1 + 3(2) + 3(2)² + (2)³ = 1 + 6 + 12 + 8 = 27 = 3³. Still exact, even two units from the anchor.

Layer 1 is the tangent line. Layer 2 bends it into a parabola that hugs the curve near x = 1 but misses on both sides. Layer 3 (dashed) lies on top of the black x³ curve at every x, which is the whole point: for a polynomial, enough layers is exactly the function.

Why you divide by 2, then 6, then 24

Each layer’s measurement gets divided before it’s used: layer 1 by 1, layer 2 by 2, layer 3 by 6, layer 4 by 24. Those are 1!, 2!, 3!, 4!. Two ways to see why.

The averaging picture. Layer 2 used the average slope over the walk. The slope was a ramp going from 3 to 4.2, and the average of a ramp is halfway: that’s the ÷2. Layer 3 needs the average of something that grows like a parabola, and a parabola from zero spends most of the walk being small, so its average is only a third of its end value: another ÷3. Stack them: 2 × 3 = 6. One more level and the average of a cubic is a quarter: 2 × 3 × 4 = 24.

Check the parabola claim with numbers. Sample t² at t = 0.1, 0.2, …, 1.0: you get 0.01, 0.04, 0.09, 0.16, 0.25, 0.36, 0.49, 0.64, 0.81, 1.00. They add to 3.85, average 0.385, and with finer sampling it settles to ⅓.

The power-raising picture. Differentiating hⁿ gives n × hⁿ⁻¹: lowering a power by one multiplies by that power. So raising a power by one divides by it. To turn a constant measurement into a term in h³ you raise the power three times, paying ÷1, ÷2, ÷3 along the way. That product is 3! = 6.

Layer	Measurement (x³ at 1)	Divide by	Term
1	3	1	3h
2	6	2	3h²
3	6	6	h³
4	0	24	0

Worked example: ln x, where the corrections stop helping

Same game, anchor at x = 1. Height is ln 1 = 0. Slope is 1/x, so 1 at the anchor.

What you know at x = 1

Layer	Derivative	Value at 1	Divide by	Term
1	1/x	1	1	h
2	−1/x²	−1	2	−h²/2
3	2/x³	2	6	h³/3
4	−6/x⁴	−6	24	−h⁴/4
5	24/x⁵	24	120	h⁵/5
6	−120/x⁶	−120	720	−h⁶/6

The measurements never hit zero. They alternate sign and grow. So there is always another correction, forever. Whether the corrections help depends on how far you walk.

Short walk: x = 1.5, so h = 0.5. True value ln 1.5 = 0.4055.

Layers used	Guess	Gap
1	0.5	0.0945
2	0.375	0.0305
3	0.4167	0.0112
4	0.4010	0.0044
5	0.4073	0.0018
6	0.4047	0.0008

Each layer roughly halves the gap. Keep going and it heads to zero.

Long walk: x = 2.5, so h = 1.5. True value ln 2.5 = 0.9163.

Layers used	Guess	Gap
1	1.5	0.5837
2	0.375	0.5413
3	1.5	0.5837
4	0.2344	0.6819
5	1.7531	0.8368
6	−0.1453	1.0616

The gap gets worse. Each term is ±hᵏ/k, and with h = 1.5 the hᵏ grows faster than the k can shrink it: 1.5, 1.125, 1.125, 1.27, 1.52, 1.90, … The guess swings wider and wider around the truth.

Inside the band the colored curves pile onto the black one, each layer tighter than the last. Right of x = 2 they fan out: layer 3 shoots up, layer 4 dives, layer 5 shoots higher, layer 6 dives harder. Every extra layer swings further from ln x instead of closer.

The rule. For ln x about 1, the corrections help when |h| < 1 and hurt when |h| > 1. That distance, 1, is the radius of convergence. It’s set by where the function itself breaks: ln x blows up at x = 0, exactly one unit left of the anchor, and the series can’t reach further right than it can reach left.

x³ had no such limit because its corrections ran out before they could misbehave. That is the difference between a polynomial and everything else.

The formula, decoded

Everything above, written the standard way:

Pₙ(x) = Σₖ₌₀ⁿ f⁽ᵏ⁾(x₀) / k! · (x − x₀)ᵏ

Spelled out for the first few terms:

Pₙ(x) = f(x₀) + f′(x₀)(x − x₀) + f″(x₀)/2 · (x − x₀)² + f‴(x₀)/6 · (x − x₀)³ + ⋯

Every symbol, in the order it appears:

Symbol	Read it as	What it is	In the x³ example
f	“the function”	the curve you’re guessing; f(x) is its height at x	x³
x	“x”	the destination, any point on the horizontal axis	1.2
x₀	“x-nought”	the anchor, where you stood and took measurements	1
x − x₀	“the walk”	destination minus start, also written h	0.2
P	“the polynomial”	the guess; a polynomial because it’s a sum of powers of the walk	1 + 3h + 3h² + h³
n	“n”	how many layers you used; the highest power in the guess	3
Pₙ(x)	“P-n of x”	the guess using n layers, evaluated at x	P₃(1.2) = 1.728
k	“k”	the counter: which layer you’re on, running 0, 1, 2, … up to n	0, 1, 2, 3
Σₖ₌₀ⁿ	“sum from k = 0 to n”	add up the term for every k from 0 through n	four terms
f′, f″, f‴	“f-prime, double-prime, triple-prime”	first, second, third derivative: slope, rate of slope, rate of that	3x², 6x, 6
f⁽ᵏ⁾	“f-k”	the k-th derivative; f⁽⁰⁾ is f itself, f⁽¹⁾ is f′, and so on	f⁽²⁾ = 6x
f⁽ᵏ⁾(x₀)	“f-k at x-nought”	the k-th derivative evaluated at the anchor, a plain number	1, 3, 6, 6
k!	“k factorial”	1 × 2 × … × k, the divide-by column; 0! = 1	1, 1, 2, 6
(x − x₀)ᵏ	“the walk to the k”	the walk raised to the layer number	1, 0.2, 0.04, 0.008
f(x) − Pₙ(x)	“the gap”	true height minus guess	0
ξ	“xi” (Greek letter)	some unknown point between x₀ and x; used only in the error formula	somewhere in [1, 1.2]

So the k = 2 term of the sum is: take the second derivative (6x), evaluate at the anchor (6), divide by 2! (3), multiply by the walk squared (0.04). That’s 0.12, the layer-2 correction from the worked example.

How big is the gap? It’s controlled by the next measurement you didn’t use, taken somewhere along the walk:

f(x) − Pₙ(x) = f⁽ⁿ⁺¹⁾(ξ) / (n+1)! · (x − x₀)ⁿ⁺¹ for some ξ between x₀ and x

Read it as: the gap is small when the walk is short (the hⁿ⁺¹ is tiny), when the next derivative is tame, or when n is large enough that (n+1)! dominates. The gap is large when the walk is long and the higher derivatives are big, which is exactly what happened to ln x at x = 2.5.

Real-world uses

Four places you’ve met this without the name. Each is worked with numbers.

Bond prices: duration and convexity. A 10-year zero at a 4% yield is priced 100/1.04¹⁰ = 67.56. The anchor is 4%. The two measurements are duration (layer 1, slope) = 10/1.04 = 9.62 and convexity (layer 2) = 10 × 11/1.04² = 101.7.

Yield rises 1%, so the walk is 0.01:

Layers	Guess	True price	Gap
1 (duration only)	67.56 × (1 − 9.62 × 0.01) = 61.06	61.39	0.33
2 (add convexity)	61.06 + 67.56 × ½ × 101.7 × 0.01² = 61.40	61.39	0.01

Yield rises 3%, walk 0.03: duration alone says 48.07, convexity pulls it to 51.16, true is 50.83. The gap is 30 times larger than for the 1% move. That’s the long walk. Traders quote duration and convexity for exactly the reason your tool quotes delta and gamma.

How a calculator computes sin. There’s no sin key inside the chip; it sums the series about 0: x − x³/6 + x⁵/120 − x⁷/5040 + ….

sin(0.5): 0.5 − 0.0208 + 0.0003 = 0.4794. True value 0.4794. Three terms.

sin(3): 3 − 4.5 + 2.025 − 0.434 + 0.050 − 0.004 = 0.137. True value 0.141. Six terms and still off in the third decimal. The series converges everywhere, unlike ln x, but a long walk needs many more layers. Calculators dodge this by folding the input back to a small angle first, which is the “walk less” strategy.

Pendulum clocks. The textbook period T = 2π√(L/g) comes from replacing sin θ with θ, which is layer 1 of the sine series. The next layer says the true period is longer by a factor of about 1 + θ₀²/16, where θ₀ is the swing in radians.

Swing	θ₀ in radians	Correction	Period error if you ignore it
5°	0.087	1.0005	0.05%
20°	0.349	1.0076	0.8%
60°	1.047	1.069	7%

A clock built on layer 1 keeps time at small swings and drifts at big ones. Same story: the anchor is θ = 0 and the walk is the amplitude.

Compound growth. (1 + r)ⁿ about r = 0 is 1 + nr + n(n−1)r²/2 + …. For 5% over 10 years: layer 1 says 1.50, layer 2 says 1.50 + 45 × 0.0025 = 1.61, true is 1.63. Layer 1 alone is the “simple interest” mental shortcut, and the layer-2 term is exactly how much compounding beats it.

All four break the same way ln x did. Past some size of move the layers you kept stop describing the function, and the fix is either more layers, a shorter walk, or computing the real thing.

Quick reference

Recipe for any function about any anchor

Pick the anchor x₀. Compute the walk h = x − x₀.
Take derivatives of f until you have as many as you want, and evaluate each at x₀.
Divide the k-th one by k!, multiply by hᵏ.
Add them up. That’s the guess. The leftover is the gap.
If the derivatives hit zero, the guess becomes exact. If they don’t, check whether the terms are shrinking; if not, you walked too far.

The two examples side by side

	x³ about 1	ln x about 1
Derivatives at anchor	1, 3, 6, 6, 0, 0, …	0, 1, −1, 2, −6, 24, …
k-th term	3h, 3h², h³, then 0	(−1)ᵏ⁺¹ hᵏ / k
Exact after	3 layers	never
Works for	every x	0 < x < 2 only
Why	polynomial: information runs out	ln x breaks at 0, one unit from the anchor

Factorials

k	k!	Average of tᵏ over [0, h]
1	1	h/2
2	2	h²/3
3	6	h³/4
4	24	h⁴/5
5	120	h⁵/6

Common series about 0, for reference

Function	Series	Converges for
eˣ	1 + x + x²/2 + x³/6 + …	all x
sin x	x − x³/6 + x⁵/120 − …	all x
cos x	1 − x²/2 + x⁴/24 − …	all x
1/(1−x)	1 + x + x² + x³ + …	\|x\| < 1
ln(1+x)	x − x²/2 + x³/3 − …	−1 < x ≤ 1

The last two have a radius because the function breaks at x = 1 or x = −1. The first three never break, so the series works everywhere even though it never terminates.

Key Insights

What	Why it matters
Anchor and walk	Everything is measured at one point; the guess only ever knows about that point
Layers = derivatives	Each derivative at the anchor buys one more correction; that is all the information you have
Divide by k!	Raising a power costs a division each time; averaging a ramp, parabola, cubic costs ÷2, ÷3, ÷4
Polynomials terminate	Derivatives hit zero, so finitely many layers rebuild the function exactly, everywhere
Radius of convergence	For everything else, past the distance to the nearest breakdown the layers make the guess worse
The gap formula	The error is the first term you dropped, evaluated somewhere on the walk

soulcraft

My second favorite thing after a piece of art, writing, music, movie, and sport that I love is watching someone else gush over things they love. It’s why the Lost in Vegas guys are so endearing, especially when the channel first started. You were watching hip-hop heads not only discover but unpeel the layers in Tool’s music.

One of my dad friends recently bought a 1979 Jeep to restore with his son and asked if my son Zak (13) would like to help. Zak’s “hell yea” beat the sound of my last syllable when I asked him. He started watching YouTube, and we landed on a channel where these obsessed details comb the U.S. looking for “barn finds” to clean for the owners. For free! It’s free because the channel has almost 2 million subscribers clamoring to watch an impossible cleaning.

I watched a few videos. I get it. Renewal of beauty, just like the objects of renewal never goes out of style. The process is as timelessly rewarding as the result.

Caitlin’s tweet is an appeal to our spirit of obsession and craft.

I like ‘em thick (Adam Mastroianni)

This essay not only feels good but is an important lens on the fast-approaching experiment of whether infinite electric monkeys will pound out Shakespeare.

Normally I might apologize for extensive excerpting but they are the point. I can do no better than this.

It opens:

I owe an apology to every English teacher I ever had. I always assumed that so-called “great” literature was a hoax, a punishment inflicted upon adolescents for the crime of being young. These books did not have anything special about them, and covering up that fact was simply a make-work exercise for former English majors, a sort of “jobs for snobs” program.

I was wrong about this. There is such thing as greatness. More specifically, there is such thing as thickness. Great works of fiction—for that matter, great works of any art—unfurl in response to your attention. The more time you spend with them, the more you get out of them. That kind of responsiveness is so addicting that it can lead people to do crazy things, like try to teach literature to high schoolers.

But thickness is tricky, because rewarding the careful reader often means repelling the casual one. And this is where I would like an apology in return from my English teachers, because while this might have been obvious to them, they never made it obvious to me.

I was presented with art and literature as if it was self-explanatory, and that everything wonderful about it was plainly visible from the outside. But those works were much more like dark, winding caves with treasure stashed inside of them. My teachers were like, “Right, well, into the cave you go!” and I was like “But there’s nothing in there” and they were like “Entering the cave is 30% of your grade” and so I took a few steps into the darkness and I was like “Just as I suspected: an empty cave” and then I came trudging back out and pretended that I saw something.

It goes on to describe spectacular examples of “thickness”. You won’t want to miss the The Garden of Earthly Delights painting and its unintended invitation to hear “butt music”.

Adam presents 4 qualities that make something thick with examples from a children’s book, the “ears” in Hamlet, and Penn (of Penn and Teller fame) eating fire.

There’s an amazing takedown of what passes for popular non-fiction.:

Reading a book like this feels like wandering through a Potemkin village. Touch any of the ideas, and they tip over.

He contrasts gilded examples of non-fiction with solid gold:

Thickness comes from surfacing a few facts well, and in such a way that you realize the existence of entire universes of additional facts that could be known….[Jane Jacobs book] is pointing out a fact that millions of people observe every day, but almost none of them notice.

He addresses an easily anticipated objection which if you’ll be familiar with if you have read any of deBoer on “poptimism”:

If you allow for the existence of secret treasures that can only be accessed with effort and analysis, then you empower the elitists and the snobs. “Buddy, don’t even talk to me until you’ve been in the cave!”

But look around. The snobs are in full retreat. We have swung the pendulum so far toward poptimism, toward the blinkered idea that all art is equal because all humans are equal, toward the ethos that guilty pleasures are simply pleasures, that I’m not sure if we can ever swing it back.

And finally, Adam addresses the elephant:

Erasing the line between the thick and the thin has left us defenseless against slop at the exact moment of its onslaught. Everyone can sense there’s something amiss with the prose that comes out of the machines, but we lack the language to talk about it, and so we’ve converged on the idea that slop simply means using too many em dashes, bullet points, and line breaks.

What separates substance from slop is thickness. Slop holds no secrets; it signifies nothing. Under scrutiny, it evaporates. All it can offer is bottomlessness—sure, there’s nothing on, but at least there are infinite channels!

That’s why I’m neither surprised nor dismayed when studies find that people prefer AI art to human art. Of course they do! In the short term, thinness prevails. When people are making snap judgments, they want pretty flowers, poems that rhyme, pleasing pablum, the simulacrum of thought. But none of these last…

About once a week, I get a pitch from some AI startup that wants to automate some part of my writing. The most recent one says it’s “built for credible thinkers who have a book’s worth of ideas but not the time it typically takes to write one”.

I’m sorry, but if you’re building or using a tool like this, then you’ve got slop for brains. There is no such thing as having a “book’s worth of ideas” that are all ready to go except for the small matter of choosing the right words and putting them in the right order.

I know exactly the feeling that these slop-trepeneurs are preying on, because I feel it all the time: I’ve got these thoughts in my head, and boy oh boy they’re good ones, all-timers, really, and it’s so annoying that I have to spend all this time making the words sound good, when the ideas behind the words are already so good!

But this is an illusion. The ideas are not already good. They need to be thickened. I understand why it’s tempting to force a machine do the hard part for you, but it can’t, and the hard part is the only part worth doing anyway.

Moontower #329

In this issue:

soulcraft
grab a coffee and derive this with me

Friends,

I watched a few videos. I get it. Renewal of beauty, just like the objects of renewal never goes out of style. The process is as timelessly rewarding as the result.

Caitlin’s tweet is an appeal to our spirit of obsession and craft.

Article content — https://x.com/KrisAbdelmessih/status/2102392847676637227?s=20

Which brings me to a luxurious reading experience:

I like ‘em thick (Adam Mastroianni)

This essay not only feels good but is an important lens on the fast-approaching experiment of whether infinite electric monkeys will pound out Shakespeare.

Normally I might apologize for extensive excerpting but they are the point. I can do no better than this.

It opens:

It goes on to describe spectacular examples of “thickness”. You won’t want to miss the The Garden of Earthly Delights painting and its unintended invitation to hear “butt music”.

Adam presents 4 qualities that make something thick with examples from a children’s book, the “ears” in Hamlet, and Penn (of Penn and Teller fame) eating fire.

There’s an amazing takedown of what passes for popular non-fiction.:

Reading a book like this feels like wandering through a Potemkin village. Touch any of the ideas, and they tip over.

He contrasts gilded examples of non-fiction with solid gold:

He addresses an easily anticipated objection, which you’ll be familiar with if you have read any of deBoer on “poptimism”:

But look around. The snobs are in full retreat. We have swung the pendulum so far toward poptimism, toward the blinkered idea that all art is equal because all humans are equal, toward the ethos that guilty pleasures are simply pleasures, that I’m not sure if we can ever swing it back.

And finally, Adam addresses the elephant:

Money Angle

People seemed to like the post I wrote 2 weeks ago teaching readers how to compute Kelly optimal bet sizes in their heads. A lot of people reached out saying that despite learning Kelly in the past, this treatment not only made it clearer but also helped them appreciate how its approach informs risk-taking in wider contexts.

Panoptica graciously asked me to republish it under their own banner:

After this post you will be sizing bets in your head (Panoptica)

Just to put a bow on it, I’ll restate what I think are the most crucial lessons without dwelling on the formula:

Even educated people are terrible at sizing bets. It’s not because it’s so complex, but I guess it’s like squatting. It seems like you should just know how to do it, but it’s actually something you need to learn the mechanics of.
Overbetting is incinerating money. This is something that’s hard to appreciate until you see the math. The reason you size smaller is because of the asymmetry of being wrong on your edge. If you underbet, you slow your growth rate but slow risk even faster. At least you’re exchanging lower returns for a better risk/reward. But if you overbet, you lower your growth rate AND increase your risk even faster than you reduce your reward. Both the numerator and denominator of your risk/reward move in the wrong directions!

If interested, Matt does this neat meta series “Notes on Notes”. It’s a short chat about how and why a particular article comes together in the first place. You can watch it here:

via Matt Zeigler@CultishCreative

A short yet broad-ranging talk with @KrisAbdelmessih on @Panoptica_ai for his @EpsilonTheory: Unplugged essay, “After this post you will be sizing bets in your head” Kelly bets, using AI to learn math, creativity… NEW Notes on Notes!

To round out this Kelly sprint, I have 2 more bits that are again overtures to curious learners who may find math intimidating.

1) Slides

The first is a condensed slide version, which I hope makes this accessible. It was born out of teaching Kelly to my 8th grader at breakfast this past Tuesday. “Zak, I wanna see if I can teach you something neat in 5 minutes.” He rolled his eyes, but at least he humored me while scarfing down his cereal.

📊Moontower Kelly Slides

2) An empowering derivation

The derivation of the Kelly formula is so fun because as it rolls downhill, a number of concepts we talk about in this letter stick to it, so when we get to the end it feels like something grand, but it’s so damn compact.

Another teaching experiment. Let’s see if I can narrate the derivation in a way so that as you follow along it never feels “hard”. I want to prove that this is fun to do and while I don’t expect to convert everyone, I do think there’s a bunch of you who’d like to be able to learn this but feel blocked because you have gaps in your foundations or can’t remember HS math, or just lack some confidence.

Screw all that, I’ll lay my jacket over the puddles so you can see that it’s not so bad out here. Just come along.

Money Angle For Masochists

Deriving Kelly From Scratch

Stating the question

You have a bet. You win with probability p and lose with probability q = 1 − p. If you win, you get paid B times what you risked. If you lose, you lose what you risked. B is also just a return. So if you double your money on a bet, B=1=100% return.

You’re going to bet the same fraction f of your bankroll every time and let it ride.

What’s the f that yields the highest compounded return?

Step 1: What one bet does to your wealth

If your wealth is W and you bet the fraction f:

Win: W → W(1 + Bf) Lose: W → W(1 − f)

Example:

Your starting wealth is $100 and you bet 50% of it on a coin flip. Remember B =1 because when you win you make 100%. When you lose you always lose f which is your bet size.

Win: 100 → 100(1 + 1*.50) = $150 Lose: 100 → 100(1 − .50) = $50

You keep the part you didn’t bet no matter what. A win adds B times your stake. A loss takes your stake away.

Step 2: What many bets do to your wealth

Play n bets. You win h of them and lose the other n − h. Each bet multiplies whatever the last one left you, so the multipliers stack:

Wₙ = W₀(1 + Bf)ʰ(1 − f)ⁿ⁻ʰ

The order of wins and losses doesn’t matter. Only the count does.

Example:

I bet 50% of my wealth each turn on the coin. I play 4 times and I win 3 of them.

100(1+1*.5)3(1-.5)1= 168.75

Order doesn’t matter. If I lost 50% on the first flip, then won 3 in a row:

Start: 100 Lose (×0.5): 50 Win (×1.5): 75 Win (×1.5): 112.50 Win (×1.5): 168.75

Step 3: Turn total growth into a growth rate

Wₙ is where you end up. We want a per-bet growth rate.

You already know how to do this. If an investment grew by a factor of (1 + 8%)¹⁰ over 10 years, you take the 10th root to get the CAGR back. Same move here: take the nth root of this equation: Wₙ = W₀(1 + Bf)ʰ(1 − f)ⁿ⁻ʰ

Note that W₀ can be divided out, just as if your starting identity was 150 = 100(1+8%)¹⁰ and you turned that into 1.5 = 1.08¹⁰ before taking the 10th root to get to annual growth rate.

This equation took our simple total growth equation and turned it into a growth rate equation:

Step 4: Let probabilities take over

We can clean up our new growth equation with several handy notation substitutions.

Ps and Qs

h/n is the share of bets won

(n-h)/n is the share of bets lost

Over a long run, the fraction of bets you win settles down to your win probability:

h/n → p and (n − h)/n → q

Wₙ/W₀ is just a wealth multiple. If you made 200% on your investment portfolio over a decade, your wealth multiple is 3 because you started with $1 and ended up with $3.

We’ll call that wealth multiple G (for “gross multiple”)

So the per-bet growth factor becomes:

G = (1 + Bf)ᵖ(1 − f)ᑫ

G = 1.05 means your bankroll grows 5% per bet on a compounded basis. In traditional investing, we’d substitute the word “year” for “bet”. Each flip is like a year.

Our job is to find the f that makes G (the gross multiple of our wealth) as big as possible.

Step 5: Take the log to make the math easy

Maximizing G means taking its derivative with respect to f and setting it to zero.

It’s 2026, you don’t need to know how to actually differentiate an equation. You just have to awaken that part of your brain that knows:

a) a derivative is the slope of a function at a given point on a curve

b) when the slope of a curve is 0 this is a maximum or minimum

c) intuitively, we can reason that this is a growth curve is a hill with no bumps. Its slope starts positive and only ever gets smaller as you bet more, so it can hit zero exactly once, and when it does, you’re at the top. (the jargon version: this growth curve is concave: it bends downward everywhere from the max, with no inflection points)

We are still here:

G = (1 + Bf)ᵖ(1 − f)ᑫ

But G is a product of powers, which is nasty to differentiate.

But remember from simple pleasures, logs fix this. They are the inverse of exponentiation, allowing them to turn multiplication into addition!

Push the cobwebs away to recall that this works:

log(100) = log(10 × 10) = log(10) + log(10) = 1 + 1 = 2

That’s not all logs do.

They pull exponents down in front: log(xᵖ) = p · log(x).

We can prep that equation for differentiation by taking the log of both sides and using both of those sexy log features:

log(G) = p · log(1 + Bf) + q · log(1 − f)

Maximizing log(G) gives the same f as maximizing G, because log always rises when its input rises.

There’s another little bonus.

Log(G) is the log return per bet, the same quantity as ln(S₁/S₀).

It re-expresses a compounded, multiplicative growth factor as a continuously compounded rate, and rates add cleanly.

Step 6: Differentiate and set to zero

Once you know this is a derivative problem because you are trying to maximize then we can rely on crutches (Claude) for thing that’s hard to remember.

Namely that:

the derivative of log(x) is 1/x.
If there’s something inside the log, also multiply by the derivative of that inside piece (the chain rule)

So we have our equation:

log(G) = p · log(1 + Bf) + q · log(1 − f)

then we differentiate our terms to be added with respect to f:

p · log(1 + Bf) → pB/(1 + Bf) (the inside, 1 + Bf, has derivative B)
q · log(1 − f) → −q/(1 − f) (the inside, 1 − f, has derivative −1)

Set the sum equal to zero, which is where the growth curve is flat at its peak:

pB/(1 + Bf) − q/(1 − f) = 0

Step 7: Solve for f

Move the second term across and cross-multiply:

pB(1 − f) = q(1 + Bf)

pB − pBf = q + qBf

pB − q = pBf + qBf

pB − q = Bf(p + q)

But don’t turn your pattern recognition noggin off!

Since p + q = 1, this collapses to:

f = (pB − q) / B

f = p − q / B

Can you reproduce this right now on a blank sheet of paper?* Probably not. But go through it once the way you used to trace when you learned the motions for drawing comic book characters or flowers or in my case TMNT. After a single reproduction by hand, I assure you some neural pathways will re-open that have had construction signs in front of them for years.

And to tie this back to the beginning, that is the joy of thickness.

[Get your mind back over here, this is a family letter.]

*You will be able to reproduce it on your own after 2 or 3 attempts. I don’t know why the motion of the pencil is a 10x better instructor than reading it a bunch of times (generation effect maybe?). But this is one of those learning principles I take seriously and impress on my kids.

The world will seduce you with ease where you’d be better served by friction. It is a 21st-century skill to have a point of view on the difference.

Kelly Criterion — Cheatsheet Derivation

Step 1 One Period Expectancy

E = pB − q

where p = win probability, q = 1−p = loss probability, B = net odds (win B per unit staked, lose 1).

Why: Weighted average of outcomes. Win B with probability p, lose 1 with probability q.

Step 2 Per-Flip Wealth Multipliers

Bet fraction f of current wealth W:

Win: W₁ = W(1+Bf)
Lose: W₁ = W(1−f)

Why: You keep the unbet portion W(1−f) regardless. On a win you collect B times your stake Wf on top. On a loss your stake Wf is gone.

Step 3 Wealth After n Flips

After h wins and (n−h) losses:

Wₙ = W₀(1+Bf)^h(1−f)^n−h

Why: The flips compound — each one rescales whatever the previous left. That means multiply, not add. Order doesn’t matter, only h and n−h.

Step 4 Per-Flip Growth Rate G

Take the nth root of total growth to extract the per-period rate:

G = (Wₙ/W₀)^1/n = (1+Bf)^h/n · (1−f)^(n−h)/n

As n → ∞, law of large numbers: h/n → p, (n−h)/n → q.

G = (1+Bf)^p(1−f)^q

Why nth root: Same logic as extracting r from (1+r)ⁿ = total growth. Geometric mean, not arithmetic, because the process is multiplicative.

Step 5 Take ln Before Differentiating

Define g = ln(G). Since ln is monotonically increasing, maximizing g gives the same f* as maximizing G.

Apply two log rules:

ln(AB) = ln(A) + ln(B) → product becomes sum
ln(A^p) = p·ln(A) → exponent drops to coefficient

g = p·ln(1+Bf) + q·ln(1−f)

Why: Differentiating a product of powers is a mess. A sum of logs is trivial. Valid because ln is monotone — same maximum, easier math.

Step 6 Differentiate and Set to Zero

Rule: ddx[ln(x)] = 1x. Chain rule: multiply by derivative of the inside.

ddf[p·ln(1+Bf)] = pB1+Bf ← chain rule gives B from inside (1+Bf)
ddf[q·ln(1−f)] = −q1−f ← chain rule gives −1 from inside (1−f)

Set dg/df = 0:

pB1+Bf − q1−f = 0

Step 7 Solve for f*

Cross-multiply:

pB(1−f) = q(1+Bf)

Expand:

pB − pBf = q + qBf

Collect f terms:

pB − q = pBf + qBf = Bf(p+q)

Since p+q = 1:

f* = pB−qB = p − qB

The Answer

f* = pB − qB

Read as: edge / odds

Numerator pB−q is your expected profit per unit bet
Denominator B scales it by the odds

Special case B=1 (even money): f* = p−q

Your optimal bet equals your raw edge.

Key Insights

What	Why it matters
Multiplicative wealth function	One bad bet can’t be offset by other bets — sizing matters
Geometric mean not arithmetic	Compounding processes need per-period rates, not averages
ln transform	Turns product into sum without moving the maximum
Chain rule on ln(1−f)	The −1 derivative is what creates a finite optimum
p+q=1	The simplification that closes the algebra cleanly

“end behavior”

Alex is an options trader you should follow in case he ever tweets a lot. Because he doesn’t, when he posted the question below a year ago, it got few responses. I took the liberty of posting it myself this week.

This was fun because it led to a lot of discussion on the timeline and DMs. I was told it sparked a bunch of quant debate on one trader’s desk.

The most popular answer, which was still less than 1/3 of the responses, was the correct answer.

Why?

The maximum value of a put is the strike. The maximum value of a call is the stock price.

Straddle is C + P so $100+$100 = $200

Notice how this means all call spreads go to zero since the calls are worth the same — the stock price. All put spreads go to their max value— the distance between strikes because the puts themselves are worth the strikes.

Logic for delta:

Delta is the change in option price per change in stock.

But the put’s strike is fixed, so the value of the put doesn’t depend on the stock price. The put has zero delta. It’s always worth $100. Which means it has no gamma either

The call is $100 because the max value of the call is the stock price. The call value moves 1-to-1 with the stock, so it has a delta of 1 or 100%

The max value of a straddle is therefore the stock price plus the strike price.

If you sell the straddle or either option at max value and hedge on its delta one time (this is known as a static hedge in contrast to dynamic hedging where you would rebalance as your hedge ratio changes), you cannot lose. It is that simple fact of arbitrage that makes it the upper bound.

To address the second most popular response in the poll, those who said the straddle is $100 (wrong) and has a 1.00 delta (correct), we will demonstrate why this is incorrect.

What’s your p/l if you sell 1 straddle at $100 and buy 100 shares against, if the stock goes to $300?

The straddle will be worth $400, so you lose $300 but make $200 on your long share.

Hmm, maybe I’m just underhedged. Fine, what if I hedge on a 200 delta?

In that case, you actually make money; you win $400 on your 2 shares more than offsetting the $300 straddle loss. But what if the stock went to zero?

Your straddle p/l is unchanged, but you lost $200 on the long stock position. Arbitrage max value means you cannot lose if you sell at that price. Since we found a losing scenario, the price is not the maximum arbitrage bound. If you sell the straddle at $200 and buy a single share of stock, there’s no scenario where you lose. It is the lowest straddle value for which this no-lose scenario is true, thus it’s the arbitrage bound.

Of course, this is but a toy problem where the call and put go to their maximum values because it’s a degenerate case of infinite time or vol. But learning how a function (an option price is just a function) behaves by observing its boundaries is good for intuition. You did this in 9th grade. Khan Academy can jog your memory:

In the real world, you can fleetingly find options that trade beyond their arbitrage values:

Earlier in the week, @DeepDishEnjoyer aka p4 wrote a thread about a dividend mispricing.

It led to some back and forth with passersbys who use options but appear to have large gaps in the fundamentals.

Between the maximum value poll and p4’s dividend lesson, it’s worth saying it:

In a proper option education, you spend a lot of time on arbitrage relationships, cost of carry, and synthetics before you ever hear the word “volatility”.

I didn’t study formal math but I imagine there’s a lot in common with the process of proofs. Arbitrages rest heavily on assumptions. So to understand the relationships, you are forced into an intimate familiarity with the assumptions. And in the extremes of everything, it’s the failure to examine assumptions that leads to being blindsided. But also, when things get extreme, to go on the attack means asking yourself, “Who’s on autopilot? Is this price resting on a stale assumption?” The arbitrage relationships give you the highest conceptual ROI that derivatives offer, you never learn the most useful thing derivatives can teach…passage over the “bridge of asses”.

If you want to see more examples of why option basics are so key to understanding assumptions and opportunities when things get weird:

blindsided

My cousin Nicole just had her first child in the past year and is now compiling resources for new parents in light of where her attention has obviously been.

In our family chat, she asked:

When you became a new parent, what blindsided you the most?

I’ll share my answer, which I qualify with both the awareness that having a child is a gamble on many levels and that conception itself is a miracle and should never be taken for granted. What was I blindsided by?

[9:16 AM, 9/16/2026] Kris Abdelmessih: that they were gonna be so awesome

[9:17 AM, 9/16/2026] Kris Abdelmessih: that last one is important when we live in times where people have less kids and talk about it as though it’s a chore (it is) but not the amazing upside

The most transcendent single moment of my life thus far was to hear my son’s voice the day he came into the world. I don’t know if that’s every parent’s experience, but immediately I felt the joy and clarity of purpose. Before that cry, I don’t think I would have said I had no purpose, but the moment revealed that I didn’t believe I did. The speed and intensity of this rush of belief was a novel feeling. Thus, irreparably blindsided.

Share your own answers in the comments and I’ll share them with Nicole. Thank you!

I offered a couple of less serious answers to her question as well.

On that last one, I wrote about that 2 years ago in our minds love to betray us:

When I was at the Sphere with my family over Spring Break, I wouldn’t ride the long, exposed escalators. I took the elevator where I found the rest of the scaredy-cats.

If there’s anything good about having a phobia, it’s empathy for the range of what can go on in people’s minds and bodies.

I’m watching this poor guy thinking, don’t do it man, it’s not worth and all he wants is a glimpse:

Man crawls through his phobia of heights to get a view of the Atlantic Ocean from the edge of a cliff.

1:44 PM · Sep 12, 2026 · 36.9M Views

2.99K Replies · 5.3K Reposts · 131K Likes

The comment section understands, and based on the number of “likes”, many others do too.

Anyway, I blame my kids for my embarrassment when I go to the Sphere to see Metallica with a group of guys next month and have to explain that I’ll meet them at the seats.

Moontower #328

In this issue:

blindsided
“end behavior”

Friends,

Blindsided

My cousin Nicole just had her first child in the past year and is now compiling resources for new parents in light of where her attention has obviously been.

In our family chat, she asked:

When you became a new parent, what blindsided you the most?

[9:16 AM, 9/16/2026] Kris Abdelmessih: that they were gonna be so awesome

[9:17 AM, 9/16/2026] Kris Abdelmessih: that last one is important when we live in times where people have less kids and talk about it as though it’s a chore (it is) but not the amazing upside

Share your own answers in the comments and I’ll share them with Nicole. Thank you!

I offered a couple of less serious answers to her question as well.

On that last one, I wrote about that 2 years ago in our minds love to betray us:

When I was at the Sphere with my family over Spring Break, I wouldn’t ride the long, exposed escalators. I took the elevator where I found the rest of the scaredy-cats.

If there’s anything good about having a phobia, it’s empathy for the range of what can go on in people’s minds and bodies.

I’m watching this poor guy thinking, don’t do it man, it’s not worth and all he wants is a glimpse:

Man crawls through his phobia of heights to get a view of the Atlantic Ocean from the edge of a cliff. 😅

1:44 PM · Sep 12, 2026 · 36.9M Views

2.99K Replies · 5.3K Reposts · 131K Likes

The comment section understands, and based on the number of “likes”, many others do too.

Anyway, I blame my kids for my embarrassment when I go to the Sphere to see Metallica with a group of guys next month and have to explain that I’ll meet them at the seats.

Money Angle

A couple of Option Trench episodes to share:

📺Terminal vs Path-Dependent Value Explained Using Collars | 39 min

📺The (Not So) Efficient Market Hypothesis? | 59 min

The first one will be useful for anyone wanting to learn more about option collars, which I’ve been writing a lot about. A video might be a gentler format so check that out.

The second one applies to investing broadly. I also use the Paradox of Provable Alpha at the end to answer a good question Erik asks.

Money Angle For Masochists

This was fun because it led to a lot of discussion on the timeline and DMs. I was told it sparked a bunch of quant debate on one trader’s desk.

The most popular answer, which was still less than 1/3 of the responses, was the correct answer.

Why?

The maximum value of a put is the strike. The maximum value of a call is the stock price.

Straddle is C + P so $100+$100 = $200

Logic for delta:

Delta is the change in option price per change in stock.

But the put’s strike is fixed, so the value of the put doesn’t depend on the stock price. The put has zero delta. It’s always worth $100. Which means it has no gamma either 🙂