Quantitative Finance — the Mathematics of Markets · chapter 12

12The Distribution Family Tree

Chapter 11 built one machine and pointed it in four directions. Feed E[·] the value and you get the mean. Feed it squared distance from the mean and you get the variance. It worked beautifully, and there is a quiet problem sitting underneath it. Every distribution we ran that machine on was something I made up for the paragraph it appeared in. A die that pays −2, +1 and +6. A triangle density on the interval from 0 to 1. Real problems do not arrive with a hand-built pmf attached to them. They arrive as questions like how many of my twenty quotes get filled, or how long until the first one does, and the shapes that answer those questions are not arbitrary at all. There is a small family of them, and the usual way to learn it is a table of a dozen rows you are quietly expected to memorise. We are going to do the exact opposite. We start with one brick, a single yes-or-no trial, and then we ask it a different question each time. Count them. Wait for them. Let them go rare. Spread them out. Add them up. Multiply them. Believe something about them. Each question grows a branch, and every branch carries its own formula, mean and variance rebuilt from the root rather than recalled. By the end there is no zoo — there is one tree, and you can regrow any part of it from memory of a single brick.

Before the first coin lands, check what this page is standing on. Chapter 11 handed us three tools that do all the heavy lifting here: E[X] as a weighted average, the identity Var = E[X²] − E[X]², and linearity of expectation, which splits a sum without asking how the pieces are related. Chapter 2 handed us the infinite geometric series and the constant e. Chapter 1 handed us C(n,k). Nothing else is needed.

The seam, and the tree we are about to grow — tap a branch to reveal its question, tap a tool to light the line it hands you.

The tools this page stands on — tap one to light the line it hands you:

Every branch asks the one coin a different question — tap a branch to hear its question (the formula stays dark until the chapter builds it).

tap a branch — watch it grow

What you're looking at — one coin, asked nine ways, on tools you already hold

E[X] = weighted average → coin's mean is p

Var = E[X²]−E[X]² → coin's var p(1−p)

linearity → Binomial mean np, for free

geometric series (Ch 2) sums the waiting

(1+x/n)ⁿ→eₓ turns Binomial into Poisson

C(n,k) — Ch 1 counting, unchanged

Fig. 1. The whole zoo, on one page — and it was never a zoo. At the root sits a single coin flip; every branch is that same flip asked a different question (count them, wait for one, let them go rare), so tapping a branch reveals only the question — the formula stays dark until the chapter builds it. The six chips are the entire toolkit the page needs, and you already hold every one: E[X], the variance identity and linearity from Ch 11, the geometric series and e-limit from Ch 2, and C(n,k) from Ch 1. The dim chips downstream (Ch 13, Ch 26, Ch 37) are debts this chapter names but pays later. One brick, nine questions — learn the root and the branches regenerate.

Come back to that panel later, because the branches are dark now and light up as we build them. Three roads lead out of it. Chapter 13 takes this whole tree and collapses it onto one branch. Chapter 26 derives the shape that prices live on. Chapter 37 models order arrivals with the rare-event branch we grow in the middle of this page.

01The brick everything is made of

Start with the smallest random thing that can happen to you. You post one order at the top of the book, and one of exactly two things follows. It fills, or it does not. There is no third outcome and no partial credit.

I want to be honest that this looks far too small to deserve a name. It is one flip. It is one customer who either buys or walks. It is one click on a Geiger counter in a given tick of the clock. Every instinct says to skip past it and get to the real distributions, and that instinct is the single most expensive mistake in this chapter.

Give the outcome a number, exactly the way Chapter 11 taught us. Write X = 1 when the order fills and X = 0 when it does not. Call p the probability of a fill, which leaves 1 − p for no fill. Those two cases exhaust everything that can happen. That two-row table is a random variable, and it is called a Bernoulli trial with success probability p.

Now point last chapter's machine at it and notice how little there is to do. The expectation is a probability-weighted sum over two values, so E[X] = 1·p + 0·(1−p) = p. For the variance we need E[X²] first, and here something convenient happens. Squaring does nothing at all to 0 and 1, so E[X²] = p as well, and the identity gives Var(X) = p − p² = p(1−p).

The atom of the whole tree — one trial, and Chapter 11's machine finishes it in two lines

pick the instance — the picture changes, the maths won't

success probability p — drag →

E[X] = 0.30 · Var = 0.21

both lines land on p — that's why this brick needs no new theory: Chapter 11's machine finishes it.

What you're looking at — the smallest brick, and the two lines that finish it for free

blue = the one trial (a coin, still unresolved) — the whole tree grows from this

green = the success X=1, carrying weight p

red = the failure X=0, carrying weight 1−p

gold = the numbers the machine returns: E[X], E[X²] and Var

Point Chapter 11's E[X]=Σ x·P(x) at a single trial and it is done in two lines: E[X]=p, and since squaring can't touch 0 or 1, E[X²]=p too — so Var = p − p² = p(1−p). No new formula, no memorising: every mean and variance later on this page is built from these two lines.

Fig. 2. The atom of the whole distribution tree: one trial that lands X=1 with probability p or X=0 with probability 1−p. Pick the concrete instance — a posted order, a Geiger tick, a customer — and the picture changes while the arithmetic does not. Now drag p: Chapter 11's machine (E[X]=Σ x·P(x)) finishes the whole thing in two lines that recompute live — E[X]=p, and because squaring can't touch 0 or 1, E[X²]=p as well, so Var = p − p² = p(1−p). Tap why is E[X²]=E[X]? to see the square vanish. That's why the smallest brick is free: it needs no new theory at all.

That is the whole brick, and nothing in it is new. It uses the definition of expectation and the variance identity, both of which you already own. With a fill probability of p = 0.3 the atom has mean 0.3000 and variance 0.2100, and I would like you to hold both numbers for a moment rather than skim them.

Here is the promise this section is really making. Every named distribution on the rest of this page is built out of copies of this brick. Not analogous to it, not inspired by it. Literally made of it, the way a wall is made of bricks. So the two lines above are not a warm-up we discard — they are the source that every later mean and variance on this page is derived from.

One thing about p(1−p) is worth staring at before we move, because it does real work when the rare-event branch arrives. It is not a straight line in p. It is a hill.

The uncertainty hill — drag p and watch variance rise, peak, and die

p — the coin's bias (drag the dot, or slide)

Var = p(1−p) = 0.2500

Var / mean = (1−p) = 0.5000

fair coin — the noisiest brick

What you're looking at — variance is a hill over the coin's bias p, not a straight line

the hill is the variance p(1−p) — 0 at each certain end, peaking at ¼ for the fair coin

the two bars are the coin's pmf; the heads bar height = p = the mean, and it collapses to one certain bar at each end

gold zone: once p ≤ 0.05, Var/mean = 1−p ≈ 1, so variance ≈ mean — the Poisson fingerprint, five sections early

Fig. 3. The uncertainty hill: variance p(1−p) is 0 at both certain ends and peaks at ¼ for the fair coin — and once p goes tiny, (1−p)≈1 makes variance ≈ mean, the fingerprint the Poisson branch will wear.

Read the two ends first. At p = 0 the event never happens and at p = 1 it always does, and in both cases the variance is exactly 0. That is not a quirk of the algebra — a certain outcome has nothing to be uncertain about, so there is no spread to measure. The peak sits at p = 0.5 with variance 0.2500, which says the fair coin is the noisiest brick you can build.

Now walk to the left end slowly and watch the two factors separate. When p is genuinely small, say 0.01, the factor (1 − p) is 0.99, which is very nearly 1. So the variance 0.0099 is almost exactly the mean 0.0100. Rare bricks have a variance that all but equals their mean. Park that. It comes back wearing a Greek letter.

02★★ Count them: the Binomial

One order is not a trading session. You post n = 20 quotes over the morning, each with the same fill probability p = 0.3, and the quotes do not influence one another. The question that actually matters to you is not whether quote number seven filled. It is how many filled in total.

That total is a new random variable, and it is the first branch of the tree. Before we can average it we need its distribution, so let's build that from the ground rather than quote it. Shrink the problem to five fair coin flips, because the arithmetic stays readable.

Take one specific sequence, say H H T H T. The flips are independent, so the multiplication principle from Chapter 1 prices the whole sequence as a product: ½ × ½ × ½ × ½ × ½ = 0.03125. Nothing subtle has happened yet. Now ask a sharper question. How many other sequences also give exactly three heads?

Price ONE sequence — then just count the rest

Every reorder of the same flips multiplies the same factors — identical price. So the only thing left to know is: how many orderings are there?

flip coins — the price won’t move

What you're looking at — the price is fixed, only the count grows

blue = one ordering (the H/T sequence of flips you built)

gold = the price — (½)⁵; every reorder pays the identical tag

cyan = the count — C(n,k) is just how many orderings there are

Fig. 4. One specific run of five fair flips costs ½·½·½·½·½ = 0.03125 by the multiplication principle. Flip any coin: the pattern changes but the price never moves — the same five factors, multiplied in a new order, give the same product. So press SHOW MY SIBLINGS and watch every sequence with the same head count slide in, each wearing that identical tag, while the counter climbs to C(5,3) = 10 and the tags stack to 10 × 0.03125 = 0.3125. Switch to the biased case (n=20, p=0.3, k=6) and the machinery is unchanged: one ordering costs 0.3⁶·0.7¹⁴ = 4.9442e−6, there are C(20,6) = 38,760 of them (far too many to draw), and they total 0.1916. That is where the choose function comes from — not decree, just a count of equally-priced orderings.

This is the step that demystifies the whole formula. Every sequence with three heads costs the identical 0.03125, because each one multiplies the same five factors in a different order. So the only thing left to know is how many such sequences exist, and choosing which 3 of the 5 slots hold a head is precisely C(5,3) = 10. Ten equally-priced orderings, stacked, give P(3 heads) = 10 × 0.03125 = 0.3125.

Write that in general and every factor now has a job. If X counts successes in n independent trials each with probability p, then:

P(X = k) = C(n,k) · p^k · (1−p)^(n−k)

Read it left to right as three separate things. p^k pays for the k successes and (1−p)^(n−k) pays for the failures, while C(n,k) counts the orderings that all cost exactly the same. That is a Binomial distribution, written Binomial(n, p), and for our session it says the chance of exactly six fills is 38760 × 0.3^6 × 0.7^14 = 0.1916.

Now the question the rest of the course depends on, and I want you to answer it in your head before reading on. Flip a fair coin 100 times. On average, how many heads?

Your gut said 50 and your gut is correct. But we are doing this from first principles, so prove it. The definition of expectation says multiply each value by its probability and add, which for this variable means:

E[X] = Σ_k=0..100 k · C(100,k) · (½)¹⁰⁰

Look at that honestly for a second — it is one hundred and one terms, each carrying a factorial ratio in the hundreds of digits. It is genuinely horrible, it is the route the definition hands you, and almost nobody wants to walk it. So let's not.

The keystone — ask for the average heads in 100 flips, then race the ugly definition against one line of linearity.

1 · drag your guess 2 · Lock it in 3 · race the two routes

your guess for the average 50

Commit a guess, then race the routes.

What you're looking at — one mean E[X], reached two ways.

Route A obeys the definition literally: 101 terms, each a factorial monster. It crawls — and early terms add almost nothing, so it sits near zero for ages.

Route B splits X into 100 atoms worth 0.5 each; linearity adds them in one line: 100 × 0.5 = 50.0000. Same answer, no pmf touched.

Linearity never asked how the atoms relate — that's why E[Binomial] = np falls out free, and it is the one move every branch of the tree is built on.

Fig. 5. Ask for the average heads in 100 fair flips — your gut says 50, so commit it: drag the dial and Lock it in. Then race. Route A obeys the definition, stepping the 101-term sum Σ k·C(100,k)(½)¹⁰⁰ — hold Grind and watch it inch, stuck near zero for ages. Route B shatters X into 100 atoms worth 0.5 each; linearity — which never cared how the atoms relate — adds them in one line: 100 × 0.5 = 50.0000. Both print the same number; only one is walkable, and that single move is why E[Binomial] = np and why you never touch the pmf. Switch to the 20-quote session and it regenerates: 20 × 0.3 = 6.0000.

Here is the move, and it is the engine of this entire chapter. That count is not a mysterious new object — it is a sum of bricks. Let X_i be the atom for flip i, worth 1 for a head and 0 for a tail. Adding up a hundred zeros and ones is exactly the same thing as counting the heads, so:

X = X₁ + X₂ + … + X_n

And Chapter 11 proved that expectation splits across a sum no matter what, with no independence required anywhere. Each atom has mean p, and there are n of them, so E[X] = p + p + … + p = np. For the hundred flips that is 100 × 0.5 = 50.0000, in one line, with the monstrous sum never touched.

Sit with what just happened, because it is the most valuable paragraph on this page. The gut answer was right, and now it is rigorous. The rigour cost one line instead of a hundred and one terms. The mean of a Binomial is not a fact to memorise. It is a consequence of the count being a pile of bricks, and of linearity being blind to how those bricks relate.

For our twenty quotes at a 30% fill rate, the expected number of fills is 20 × 0.3 = 6.0000. Notice you never needed the pmf to get there.

The shape lab — drag n and p; the balance point is always np

n how many coins p each coin's chance

n=10 · p=0.50 · np=5.00

press CHECK to weigh the bars

What you're looking at — the same mean, built two ways

each blue brick is one coin — n of them, every one worth p; laid end to end they reach np.

the gold fulcrum sits at np — the balance point of the bars, dead under where the bricks end.

CHECK weighs the bars: Σ k·P(X=k) equals np to four decimals — a keystone, not a coincidence.

Fig. 6. The Binomial's shape swings wildly with n and p — a lopsided lump near zero when p is small, a symmetric mound at p = 0.5 — yet the balance point is np every single time, because the count is n copies of one brick no matter what the bars look like.

Drag n and p around in that panel and one thing refuses to break. The shape moves a great deal, from a lopsided lump hugging zero when p is small, to a symmetric mound when p sits at a half. The balance point, though, lands on np every single time, because it is always just n copies of one brick added together.

03★ When spreads are allowed to add

The mean fell out so cleanly that the obvious next move is to try the same trick on the variance. Write the count as a sum of atoms, add up each atom's p(1−p), and claim Var(X) = np(1−p). That answer is correct. The reasoning, as stated, is not, and this is the exact inch where a hidden assumption gets smuggled into risk calculations.

Here is the trap. Chapter 11 made a point of saying that linearity of expectation ignores dependence entirely, and that lesson lands so hard that readers quietly extend it. If E[X+Y] = E[X] + E[Y] always holds, surely Var(X+Y) = Var(X) + Var(Y) does too. It does not. Not always.

The counterexample is embarrassingly small, so let's do it before the algebra. Take one fair atom with variance 0.2500. If Y is a second independent atom, the pair has variance 0.5000. Now weld them together and set Y = X, so the sum is 2X and its variance is 4 × 0.25 = 1.0000, double the naive answer. Mirror them instead, with Y = 1 − X, and the sum is always exactly 1, so the variance is 0.0000. All three pairs have the same mean of 1.0000.

A six-beat derivation: does variance add? Watch the exact inch where independence enters.

Beat 1 / 6

independence — live on beat 4

The claim: does the spread add?

What you're looking at — variance adds only when independence pays for it.

2E[XY] — the joint term

2E[X]E[Y] — product of the means

the mean — same 1.0000 no matter what

The asymmetry nobody flags: the mean never cared about dependence and never will — the variance always does. That leftover cross term when independence fails is not an error, it is covariance.

Fig. 7. 7. Does variance add? Step through the derivation. Expectation added for free in Ch 11, so the spread surely does too — until you open Var(X+Y) and two cross terms fall out, 2E[XY] against 2E[X]E[Y]. On beat 4 flip the independence switch: E[XY] factorises, the two annihilate on screen, and Var(X)+Var(Y) survives. Flip it off and three pairs on one fair coin print the identical mean 1.0000 with spreads of 0.5, 1.0 and 0.0 — dependence, not the mean, moved them. Finally 20 independent atoms rebuild 20 × 0.3 × 0.7 = 4.2000. The mean never cared about dependence; the variance always does, and the leftover is covariance.

Step through the derivation and the condition appears on its own. Expand Var(X+Y) = E[(X+Y)²] − (E[X+Y])², and both squares produce a cross term. On the left you get 2·E[XY]. On the right you get 2·E[X]E[Y]. Those two are different in general, and the whole question is when they cancel.

They cancel when E[XY] = E[X]E[Y], and Chapter 10's independence is exactly what buys you that. Independent variables factorise inside the expectation, the cross terms annihilate, and what is left is Var(X) + Var(Y). So variances add for independent pieces, and only for independent pieces.

Our Binomial atoms genuinely are independent, since the quotes do not influence each other, so we are safe. Adding n copies of p(1−p) gives:

Var(Binomial) = np(1−p)

For the trading session that is 20 × 0.3 × 0.7 = 4.2000, and the standard deviation is √4.2 = 2.0494 fills. So a typical morning brings six fills, give or take about two — and both of those numbers came out of the brick rather than out of a table.

One honest flag before we leave. When the pieces are not independent, the cross terms do not vanish, and what survives has a name and a chapter of its own. It is covariance, it is Chapter 18, and every correlated-portfolio disaster in finance lives in that leftover term.

n against √n — the same fair coin, counted as a raw count and as a proportion

flips (drag → left bell widens, right bell narrows)

n = 100 flips of a fair coin (p = ½)

mean np50

σ √(np(1−p))5.0000

σ/n spread of proportion0.0500

×100 flips buys only ×10 spread — the √ is doing that.

np ∝ n · σ ∝ √n · σ/n ∝ 1/√n

What you're looking at — one coin flipped n times, drawn two ways: the count spreads out while the proportion tightens.

COUNT of heads. Its spread is σ = √(np(1−p)) = ½√n. More flips → the bell gets wider (n 100→10000 grows σ only 5→50).

PROPORTION heads÷n. Its spread is σ/n = ½/√n. More flips → the bell narrows onto 0.5 (0.05→0.005). A bridge to Ch 13.

Fig. 8. One knob — n, the number of times you flip a fair coin — feeding the same experiment into two pictures. On the left it is a raw count of heads, whose spread is σ = √(np(1−p)) = ½√n; push n up and the bell gets wider (n from 100 to 10,000 grows σ only from 5 to 50 — a hundred times the flips buys ten times the spread, because the square root is sitting right there). On the right it is the proportion heads÷n, whose spread is σ/n = ½/√n; the identical experiment now narrows onto 0.5, tightening tenfold. That is why a bigger sample is a better sample — arithmetic, not faith. The bridge to Ch 13's Central Limit Theorem.

One consequence of np(1−p) is the seed of the next chapter. The mean grows like n, but the standard deviation only grows like √n, since the square root is sitting right there. Flip a fair coin 100 times and you get 50 heads with σ = 5.0000. Multiply the flips by a hundred and the spread multiplies by only ten.

Divide through and the point lands. The spread of the proportion is 0.0500 at a hundred flips, and by ten thousand flips it has fallen to 0.0050. That collapse is why a large sample is trustworthy, and Chapter 13 turns it into the law of large numbers.

04Wait for the first one

The Binomial answers one question about the bricks: how many succeed in a fixed number of tries. Change the question and you get a different branch off the same root. Instead of fixing n and counting successes, fix the success and count the tries. How many quotes do you post before one finally fills?

One piece of bookkeeping first. Two different variables live here wearing almost the same name, and mixing them up quietly ruins answers. One counts trials up to and including the fill, so its smallest value is 1. The other counts failures before the fill, so its smallest value is 0.

They differ by exactly one, always, and their means differ by exactly one too. I am pinning the first convention for this whole chapter, so our variable starts at k = 1 and counts the successful trial itself. When you meet the other convention in a textbook, the only fix you need is to add or subtract a 1.

Pin the Geometric's convention, then close its infinite tail — two counts of one wait, welded exactly 1 apart, and a sum that reaches 1.

Tap Step ▸ to reveal quotes. Watch K and F march together through misses — then the fill splits them by exactly one.

Step ▸ to reveal the first quote.

What you're looking at — one wait, counted two ways, and its tail.

K counts trials up to and including the fill — floor 1, mean 1/p = 3.3333. It is the answer to "which quote filled?"

F counts the failures before it — floor 0, mean (1−p)/p = 2.3333. Same event, near-identical name, off by one.

The fill is a trial but not a failure, so K = F + 1 every run — the mean gap is welded at exactly 1.0000. Pick the wrong one and every answer is off by one.

Close the tail: the running total of the pmf climbs 0.8319→0.9718→0.9992 and Ch 2's series pays it out — p/(1−(1−p)) = 1.

Fig. 9. The Geometric wears two names, and mixing them silently wrecks answers. K counts trials up to and including the fill — its floor is 1, its mean 1/p = 3.3333; F counts the failures before it — floor 0, mean (1−p)/p = 2.3333. Tap Step ▸ and they march together through every miss, then the fill — a trial, but not a failure — ticks K alone: K = F + 1 in every run, so the mean gap is welded at 1.0000. Run ×25 to watch the means settle onto their targets. Then Close the tail: the pmf (1−p)^(k−1)p gives bars 0.3000, 0.2100, 0.1470, … whose running total climbs 0.8319 → 0.9718 → 0.9992, and Ch 2's geometric series pays the debt — p/(1−(1−p)) = 1.0000, the ceiling it can never cross.

The pmf writes itself from independence. To have the first fill land on trial k, the first k − 1 quotes must all miss and the k-th must hit:

P(X = k) = (1−p)^(k−1) · p for k = 1, 2, 3, …

That is the Geometric distribution, and the name is the one honest bit of naming in the family. With p = 0.3 the probabilities run 0.3000, 0.2100, 0.1470, 0.1029, and so on forever, each one 0.7 times the last — a geometric sequence, which is exactly where the name came from.

There are infinitely many terms, so a fair question is whether they even add to 1. Chapter 2 already answered it. The sum p + p(1−p) + p(1−p)² + … is a geometric series with first term p and ratio (1−p), so it totals p / (1 − (1−p)) = p/p = 1.0000. Watch the running total in that panel crawl up: five terms reach 0.8319, ten reach 0.9718, twenty reach 0.9992.

Now the property that makes this distribution genuinely strange, and I want a real commitment from you first. You have posted ten quotes and every one of them missed. On average, how many more quotes until you finally get a fill? More than when you started, fewer, or exactly the same?

The overlay that does not move — a losing run buys you nothing; the wait resets to fresh

1 · ten quotes just missed. The remaining wait is now…

3 · push the miss-run locked until reveal

Predict, then reveal the truth

What you're looking at — the wait to a fill, drawn twice, landing as one

fresh wait — quotes-till-a-fill from a cold start (geometric, p = 0.3)

remaining wait after the losing run — lands on the blue bar-for-bar (μ = 3.3333)

the register that would make a fill “due” — it is nowhere in the mechanism

Nothing is ever due: the spent trials aren't stored anywhere, so they can't be owed back.

Fig. 10. Predict the remaining wait after ten straight misses, then reveal: the conditional distribution (gold) lands exactly on the fresh one (blue), bar for bar, both means 3.3333. Push the miss-run as far as you like — the overlay never moves, because no register anywhere holds the spent trials.

Nearly everybody says fewer, and that feeling has a name from Chapter 10. It is the gambler's fallacy, the sense that a run of misses has built up a debt the market now owes you. The correct answer is that your remaining wait is exactly the same as it was before the first quote went out, and the overlay in that panel is the proof you can see: the conditional distribution after ten failures lands precisely on top of the fresh one.

Say the mechanism in plain hardware terms and the mystery disappears. Each quote is an independent trial — there is no register anywhere in the system storing how many times you have missed, and no counter for the market to consult. The ten spent trials are simply gone. That property is called memorylessness, and among the discrete distributions the Geometric is the only one that has it.

Memorylessness is not just a curiosity, either — it hands us the mean without a single infinite sum.

The loop that closes — the Geometric's mean, with no infinite sum

Drag p, then press break — watch the loop's term vanish.

the exit chance p = 0.30

1/p = 3.33 · pmf balances ✓

What you're looking at — one trial, then a fork: end now, or land on a start identical to the first.

exit now, chance p — the run is over

loop back, chance 1−p — an identical start (memorylessness)

so E = 1 + (1−p)E → E = 1/p, no infinite sum

Fig. 11. Spend one trial. With chance p you exit; with chance 1−p you loop back to a start identical to the first — that sameness is memorylessness. It lets you write E on both sides: E = 1 + (1−p)E, so pE = 1 and E = 1/p — drag p and all three lines rebuild, and the mini-distribution below balances at the very same mean. Now hit break memorylessness: the return arrow greys out, the (1−p)E term vanishes, and the equation loses its right-hand side. That is the property the derivation was spending — no series, no calculus, just a loop that could only close because the restart was the same.

Follow the loop. You post one quote, which costs you one trial no matter what. With probability p it fills and you are finished. With probability 1 − p it misses, and because of memorylessness you are now standing in a situation identical to the one you started in. So the expected wait E satisfies:

E = 1 + (1−p)·E

One equation, one unknown, and school algebra finishes it. Move the term across to get p·E = 1, so E = 1/p. At a 30% fill rate you wait 1/0.3 = 3.3333 quotes on average. If one in three lands, you expect to wait about three.

Notice what carried that derivation. The recursion is only legal because the restart is genuinely identical, so memorylessness is not a side note about this distribution. It is the thing that lets us compute with it.

05Wait for the r-th one

Waiting for one fill is useful. Waiting for a target is more useful. You need r = 5 fills before you can close out the position, so the real question is how many quotes that takes.

Let me deal with the name first, because it actively fights the concept. This is called the Negative Binomial distribution, and nothing about it is negative. There are no minus signs to hunt for and no subtraction hiding anywhere. The name is a leftover from how the formula was first written, using a series expansion with negative exponents — so mentally rename it wait for the r-th success and file the etymology as trivia.

Five waits, stacked — the Negative Binomial is r Geometric waits laid end to end, so its mean is r×(1/p) for free

r — the target success drag → stack more waits r = 5 successes p — success chance per trial drag → rebuild every leg p = 0.30

mean = r/p = 16.6667

What you're looking at — r fresh Geometric waits, laid end to end on one timeline

each blue leg = one Geometric wait (“wait for one success”), expected length 1/p

the ↻ re-seed at each leg's start = a fresh, memoryless coin — the past is forgotten

each green dot = a success landing; the last one is the r-th success you waited for

the gold mark = the running mean, +1/p per leg, landing exactly on r/p

Its terrible name is pure algebra — tap the name (or “why negative?”) to defuse it.

Fig. 12. The Negative Binomial earns its scary name and then hands you the mean for free. Reaching the r-th success is just doing the "wait for one" job r times over, and by memorylessness each leg starts a fresh Geometric wait — a brand-new coin with no memory of the failures behind it (the ↻ re-seed). So the total is a sum of independent waits, T = G₁ + … + Gᵣ, and linearity — the keystone from Ch11 — adds their means with no new algebra: each leg contributes 1/p, and r of them land the running mean exactly on r/p (r=5, p=0.3 → 16.6667). Drag r to stack more waits; drag p to rebuild every leg at once and watch r/p move as a whole. The name was never a new object to memorise — it's r bricks you already own.

Now watch how little this costs you, because the tree is about to regenerate a branch for free. Getting to the fifth fill is just doing the "wait for one" job five times in a row. The wait to the first fill is a Geometric. Then, by memorylessness, the wait from the first to the second is a fresh independent Geometric with the same p. And so on, five times.

T = G₁ + G₂ + … + G_r

That is the keystone move again, with Geometrics as the bricks instead of Bernoullis. Linearity splits the sum, each piece contributes 1/p, and the mean is r/p. Five fills at a 30% rate takes 5 / 0.3 = 16.6667 quotes on average, and you got that without meeting a single new formula.

This is the moment the chapter's thesis becomes something you can feel rather than something I asserted. A distribution with an intimidating name arrived, and you built it out of parts you already had, instead of looking it up.

06★ Let the events go rare

Here is a question the Binomial cannot answer as it stands. How many trade prints land on the tape in the next second? There is no obvious n here. Nobody handed you a fixed number of trials, because a print can arrive at any instant in a continuous window.

So build the n yourself. Chop the second into n tiny slots, thin enough that at most one print can fall inside any one of them. Each slot is then a Bernoulli brick, and we are back on familiar ground with a Binomial(n, p).

Now pin the thing you actually know. You do not know n or p separately, but you do know the average count in the window, which we call λ. Since the mean is np, holding the average fixed at λ means setting p = λ/n. Chop finer and the slots multiply while each one's chance shrinks in exact proportion.

The slot chopper — pin the mean at λ = 3, chop the window into n slots, watch Binomial crawl onto Poisson

chop finer — slot count n n = 10 · p = λ/n = 0.3000

0.028248 → e⁻³ 0.049787

drag to slice the window into more, tinier trials — the mean np stays pinned at 3.

What you're looking at — the same average, chopped into more and more slots

Binomial(n, 3/n) — n coin-flips, each with success chance p = 3/n, so the mean np stays pinned at 3

Poisson(3) — the fixed ghost the bars settle onto; the gauge's target is e⁻³ = 0.049787

(1−3/n)ⁿ — that's the P(zero events) bar itself, crawling up toward e⁻³

e⁻³ isn't a constant somebody chose: (1−λ/n)ⁿ → e⁻ᵤ is Chapter 2's definition of e — walking in as the probability of no events once the trials go infinitely many and infinitely rare.

Fig. 13. Pin the mean at λ = 3 and chop the window into n Bernoulli slots, each with p = 3/n. As n climbs, the blue Binomial bars crawl onto the fixed gold Poisson ghost, the k = 3 bar walks 0.266828 → 0.224042, and the isolated factor (1−3/n)ⁿ — which is just the P(zero events) bar — chases e⁻³ = 0.049787: not a chosen constant, but Chapter 2's e-limit walking into a probability.

Drive the slot slider with λ = 3 and keep your eye on the factor (1 − 3/n)^n, which is the "no print in this slot" term raised to the number of slots. At four slots it reads 0.003906. At ten it is 0.028248. At a hundred it is 0.047554, and at two thousand it is 0.049676. It is crawling toward something.

You have met that something already. Chapter 2 defined e by exactly this limit, e^x = lim (1 + x/n)^n, so with x = −λ the factor converges to e^(−λ). For λ = 3 that is 0.049787, sitting right where the numbers were heading. The rest of the pmf limits just as cleanly, and the whole thing collapses to:

P(X = k) = e^(−λ) · λ^k / k!

That is the Poisson distribution, the count of rare events in a fixed window. So the e in front is not decoration and not a normalising constant somebody chose. It is Chapter 2's limit, walking into a probability formula because we made the trials infinitely many and each one infinitely unlikely.

Now the fingerprint, and it genuinely surprises people. Every distribution so far had a mean and a variance that were different numbers. Poisson has them equal. We do not have to assert that, because we can just take limits of the two things we already built.

The factor that dies — one knob, twin gauges: watch np(1−p) climb to meet np

keep the mean at λ = 3, spread it over more trials drag → each event gets rarer (p = λ/n falls)

n = 100 trials · p = λ/n = 0.030

variance climbing toward λ

What you're looking at — the same λ = 3, counted two ways, and the one factor that separates them

mean = np, pinned at λ by construction — it never moves

variance = np(1−p) = λ(1−λ/n), climbing as trials get rarer

the (1−p) toll λ·p — the wedge that dies as p→0, closing the gap

Fig. 14. One knob, twin gauges. The mean reads np and stays welded at λ = 3.0000 — construction pins it there. The variance reads np(1−p) = λ(1 − λ/n); drag n up and the (1−p) toll — the red wedge λ·p — evaporates, so the variance climbs through 2.9100 at n = 100 and 2.9910 at n = 1000 toward 3. Equal mean and variance is not a coincidence to memorise — it is what rare looks like. Hit break it and p jumps large: the toll returns and the gauges split, showing which assumption carried the equality. The field-test chip flips it into a tool: a real count whose variance runs far above its mean is telling you the events cluster — they are not independent.

The mean is easy. np = λ was true by construction, and it stays true in the limit. The variance is np(1−p), which is λ(1 − λ/n), and as n grows that trailing factor climbs toward 1. At a hundred slots it gives 2.9100, at a thousand 2.9910, and in the limit exactly 3.0000.

So look back at the uncertainty hill we drew for the atom, because this is the payoff on the thing I asked you to park. Rareness is precisely what kills the (1 − p) factor. Mean equal to variance is not a coincidence to memorise — it is what rare looks like, and it gives you a genuinely useful field test. If you measure a count whose variance is much larger than its mean, the events are not independent, and something is clustering them.

07★ Crossing into the continuous

Everything so far has been countable. Fills, quotes, prints, waits. Now the tree crosses to the other side, where the outcome can be any real number in a range, and one specific thing about that crossing breaks nearly everyone's intuition.

Take the plainest possible continuous variable. A bus arrives at some point in the next 20 minutes, and no minute is more likely than any other. That is a continuous Uniform on the interval from 0 to 20, and its density is flat at 1/20 = 0.0500 everywhere.

Asking for the chance the bus comes between minute 5 and minute 10 is now a matter of shading. The favourable stretch is 5 minutes long out of 20, so the probability is 5/20 = 0.2500. This is geometric probability, and it is as simple as it looks — a length over a length.

Squeeze the window, then go 2D — probability is the shaded length, never the height

drag the two ▸◂ edges · then hit SQUEEZE

probability = length

0.2500

height (welded)

0.0500

window width 5.00 min · P = width × 0.05

Kill the width and P dies to 0 — but the height won't flinch. A point has no length to shade.

length 0.2500 · height 0.0500

What you're looking at — probability lives in the shaded amount, never in the height.

Shaded length (then area) is the probability. Width 5→10 min gives 0.25; squeeze it and P dies while the height sits still.

Height 0.05 is the density, not a probability — it can even exceed 1 (a Uniform on a half-minute support stands at 2).

A single point has zero length → P(X = x) = 0. That is why, in the continuous world, you only ever ask about a range.

Fig. 15. The pmf-to-pdf tear, met head-on. A bus arrives Uniform on [0,20], so its density is flat at 0.05. Probability is the shaded length times that height: five-to-ten minutes shades a width of 5, giving P = 0.25. Drag either edge and the length — and the probability — move, while the height stays welded at 0.05 (it is a density, not a probability: on a narrow enough support it would climb past 1). Hit SQUEEZE and watch the window collapse through width 0.1 (P = 0.005) and 0.001 (P = 0.00005) down to a single point, where P(X = x) = 0 — a point has no length to shade — and the height never so much as flinches. Toggle to 2D and the same idea climbs a dimension: two arrival times fill a 60×60 square, probability becomes area ÷ area, and a patience window of 15 minutes lands on 1 − (45/60)² = 0.4375. This is why, in the continuous world, you only ever ask about a range.

Now shrink the window and watch the two quantities behave completely differently. A window of width 0.1 carries probability 0.0050. Width 0.001 carries 0.00005. The height, meanwhile, has not moved at all — it sits welded at 0.0500 the entire time.

Push it to the end and the shock arrives. What is the probability the bus comes at exactly minute 7.0000000…, on the nose forever? The answer is 0. A single point has no length, so there is no area to collect, so there is no probability there.

People read that as an insult to reality, since the bus obviously arrives at some instant. But the resolution is a change of question, not a paradox. In the continuous world you never ask about points. You ask about ranges, and the answer is always the area you shade, which Chapter 8's integral computes for a density of any shape.

Chapter 11 defined the density, and the gut keeps overriding it anyway. So say the rule plainly. The height f(x) is not a probability. The sliver f(x)dx is, because a height times a width is an area.

Flip the panel to its second view and the same idea moves up a dimension. Two people agree to meet within the hour, each arriving uniformly at random, and each willing to wait 15 minutes. Now the pair of arrival times is a point in a 60-by-60 square, the meeting condition shades a band through the middle, and the probability is area over area. The miss region is two corner triangles forming a square of side 45, so the answer is 1 − (45/60)² = 0.4375.

08The shape that sums take

Now the most famous shape in the subject, and I want to handle it in an unusual order. Most treatments open by putting the density on the board and asking you to accept the √(2π). That approach buries the only two things you actually need.

So let's ask where the bell comes from instead. Go back to the keystone. A Binomial is a sum of many independent bricks, and when you slid n upward in the shape lab, the bars stopped looking lumpy and started looking like a smooth symmetric mound. That was not an accident of coin flips.

The adder — take any ugly shape, add copies of it, and watch a bell appear from nowhere

pick an ingredient

copies added slide 1 → 12

k = 1 copy · the sum is just one copy

k=1 · bell match 61%

What you're looking at — the SAME shape, added to itself. Exact arithmetic, not a simulation.

the blue bars on the left are your ingredient; on the right, the exact sum of k copies of it — and the faint blue outline is where it started (k=1).

the gold bell is the shape the sum settles into. By four or five copies the bars leave the origin and climb into it — the ingredient is forgotten.

Why it settles there at all is Chapter 13's Central Limit Theorem — owed, not yet paid. This panel only does the adding, exactly.

Fig. 16. Take any ugly ingredient — a flat block, a lopsided spike, a two‑hump mess — and add independent copies of it. By four or five copies the sum is already a mound; by ten you cannot tell what you started from. The ingredient stops mattering, which is why the same bell turns up in places that have nothing to do with each other. Everything here is exact arithmetic (the discrete convolution, rescaled to a common width), not a simulation — the proof that sums must go this way is Chapter 13's Central Limit Theorem.

Pick any ugly ingredient in that panel. A flat block, a lopsided spike, a two-humped mess. Then add a few independent copies of it together and watch the result. By four or five copies the sum is already a mound, and by ten you would struggle to tell which ingredient you started from.

That is the Normal distribution: the shape a sum of many independent small contributions settles into. The ingredient stops mattering, which is exactly why it turns up everywhere. Measurement error is a sum of many small errors, and a daily return is a sum of many small trades.

I owe you a flag here, and I would rather state it than let it pass. That panel is exact arithmetic on the sums, not a proof, and it does not show you why the ingredient washes out. The theorem that guarantees it is the Central Limit Theorem, it has real conditions attached, and it is the whole of Chapter 13. This is a promise, not a hand-wave.

What you need now is how to read one, and that needs no formula at all. Every Normal has the same shape, stretched by its standard deviation, so the useful ruler is measured in σ rather than in units.

The sigma ruler — sweep a band in σ, then stretch σ and watch the % refuse to move

band width ±k σ (sweep it outward) stretch σ (daily £) (widen the world, watch the %)

inside ±2.00σ → 95.45%

drag either slider · or tap a σ button. The % depends on k alone — never on how wide σ makes the curve.

What you're looking at — a ruler, not a table. A distance measured in σ is already a probability.

the blue curve is the Normal — every Normal is this one shape, only stretched by σ.

the gold band ±kσ and its area: 68.27% at 1σ, 95.45% at 2σ, 99.73% at 3σ.

the cyan σ-ruler is glued to the curve; stretch σ and it spreads too, so ±kσ holds the same area to the digit.

a fixed £40k loss means nothing raw — divide by σ. At σ=£10k it is 4σ, a ~1-in-31,500-day shock; double σ and the same loss is only 2σ.

Fig. 17. The Normal's real gift is not the formula e^−x²/2/√(2π) — it is a ruler. Drag band width outward and the shaded area prints continuously, locking onto the three numbers worth memorising: 68.27% inside 1σ, 95.45% inside 2σ, 99.73% inside 3σ. Now the proof it is a ruler and not a lookup table: drag stretch σ and the bell widens dramatically — the cyan σ-ruler spreads with it — yet those percentages do not shift a single digit, because they depend on k alone. That is why a raw number is meaningless until you divide by σ: a £40k loss is 4σ when the daily σ is £10k (a ~1-in-31,500-day shock), but only 2σ if σ doubles. A distance in sigmas is a probability. (Fat tails make the real Normal a poor fit for market returns — taken seriously in a later chapter.)

Sweep the band outward and three numbers do all the work in practice. Inside one sigma of the mean sits 68.27% of the mass. Inside two sigma, 95.45%. Inside three, 99.73%. That is the 68-95-99.7 rule, and it converts a distance into a probability on sight.

Its real use is turning a raw number into a surprise measure. A daily loss of £40,000 means nothing on its own. A loss of 4σ is one the model says should turn up about once in thirty thousand days, which is a statement you can argue with. Chapter 17 formalises that habit as the z-score, and Chapter 31 leans on it for value at risk.

09What multiplying gives you instead

The Normal comes from adding many independent things. That immediately raises the sharper question, because money does not add. It compounds.

A price does not go up by £3 and then down by £2. It goes up 5% and then down 3%, which means today's price is yesterday's multiplied by a factor. A year of returns is a long product of such factors, not a long sum.

Two lanes, one shock: adding builds a bell that crosses zero · multiplying builds a price that can't

☝ fire a shock — both lanes move together

straighten the multiply lane:

fire a shock — both lanes move

What you're looking at — the SAME shocks feed two lanes, and only one can be a price

ADD lane: a running sum of +0.2 / −0.2. Sums pile into a symmetric bell that happily crosses 0 — which a price cannot do.

MULTIPLY lane: a running product of ×1.2 / ×0.8. Money compounds, so a year of returns is a long product — right-skewed, pinned above 0 because exp never returns a negative.

Log axis turns the product into a sum, straightening it onto the add lane: ln 1.2 = +0.18232, ln 0.8 = −0.22314, sum −0.04082 → exp = 0.9600. The down move is genuinely bigger in log space, so an equal up-and-down pair still costs 4%.

Fig. 18. 18. One shock, two fates. Fire the same up-and-down shocks into both lanes: the ADD lane keeps a running sum of +0.2 and −0.2 — it piles into a symmetric bell and cheerfully crosses 0, which a price can never do. The MULTIPLY lane keeps a running product of ×1.2 and ×0.8 — money compounds, so a year of returns is a long product, and its shape is right-skewed and pinned above zero because exp never returns a negative. Hit ▲ then ▼: the add lane returns to 0, yet the product lands at exactly 0.9600 — a 4% loss from two “cancelling” moves. Flip to log axis and the mystery dissolves: logs turn the product into a sum, straightening the multiply lane onto the add lane, and ln 1.2 = +0.18232 against ln 0.8 = −0.22314 sum to −0.04082, which exponentiates back to 0.9600. The down move is genuinely bigger in log space — that's why equal-looking percentage moves don't cancel, and why prices live on the multiplying lane. (If ln X is Normal, then X = exp(Normal) is the lognormal; the full shape is Ch 26.)

Chapter 5 gave us the tool that converts one into the other. A logarithm turns a product into a sum. So take logs of all those growth factors, and the product becomes a sum of many independent log-returns, which is precisely the setup that produces a Normal.

Read that backwards and the new branch appears. If log(X) is Normal, then X is e raised to a Normal, and that variable is called lognormal. Two properties follow immediately from what e^x does. It is never negative, whatever you feed it, so a lognormal price cannot cross zero. And it stretches the upper half of the bell far more than the lower half, so the shape leans right.

The two lanes in that panel make the difference physical. On the adding lane, shocks of +0.2 and −0.2 cancel and you end where you began. On the multiplying lane, up 20% then down 20% gives 1.2 × 0.8 = 0.9600, so a pair of equal-and-opposite moves has cost you 4%.

That asymmetry is not a trick of arithmetic, and it is the reason this shape earns a place in the family. In logs it is perfectly balanced, since ln 1.2 = 0.18232 against ln 0.8 = −0.22314. The down move is simply bigger in log space, and the sum −0.04082 exponentiates back to exactly 0.96.

I am naming this branch here and deliberately not deriving its density. That derivation belongs to Chapter 26, where the price path is built from Brownian motion and the lognormal falls out of the model rather than being asserted. What you should carry from here is the placement: sums give Normal, products give lognormal.

10★ A distribution over a probability

One branch left, and it requires a shift in what a distribution is about. Every shape so far has described an outcome. How many fills, how long a wait, what price. In each of them p was a fixed parameter you were handed.

But where did p come from? In a real market nobody hands you the fill probability. You estimate it from what you have seen, which means the honest object to carry is not a single number but a whole spread of plausible values, sharpening as evidence lands.

So we need a distribution whose outcomes are the possible values of p itself. It lives on the interval from 0 to 1, because that is where probabilities live. And it is continuous, because p can be anything in between.

Start from total ignorance. You have a coin of unknown bias and no data at all, so every value of p is equally plausible. That is the flat density on [0,1], which is the continuous Uniform we already built, and in this family it has a second name: Beta(1,1).

Belief over the coin's true head-rate p — start flat, then tap H or T and watch it sharpen

tap H or T — each flip tilts the belief & adds 1 to a count

the Bayes ledger — the update IS the count

α = 1 + 0 heads = 1

β = 1 + 0 tails = 1

mean α∕(α+β)

0.5000

width (sd)

0.2887

peak

1.00

Conjugate: a Beta prior + coin data gives a Beta posterior — a full recompute collapses into +1 to a count.

Beta(1,1) — the flat Uniform start

What you're looking at — a distribution over the PARAMETER p, not over an outcome; and it sharpens as data lands.

The belief curve — a Beta over p. It begins flat (Beta(1,1), the same Uniform), and the fading ghost is the belief one flip ago.

Every head adds 1 to α and tilts the mass right; every tail adds 1 to β and tilts it left — that bookkeeping is Bayes.

The mean α∕(α+β) is your best guess for p, and the width shrinks as evidence piles up — the belief gets sure.

Fig. 19. A distribution over the parameter itself — not over an outcome, but over the coin's unknown head-rate p. Total ignorance is the flat density on [0,1]: Beta(1,1), which is exactly the Uniform from the continuous section. Now feed it data. Each head adds 1 to the shape parameter α and tilts the curve right; each tail adds 1 to β and tilts it left — the fading ghost is the belief one flip ago, so you see it sharpen rather than take it on faith. The presets jump to Beta(2,2) (mean 0.5000, density 1.5000 at p = 0.5) and Beta(9,3) (mean 0.7500, peak near 0.8), and the width readout shrinks as evidence lands — the belief grows sure. This is what conjugate prior actually buys, once the jargon is stripped off: the shape survives its own Bayes update, so a full posterior recomputation collapses into adding 1 to a count. That closes the loop back to Chapter 10 — Bayes, made continuous.

Now flip the coin and run Chapter 10's machinery, which said to shrink your world to where the evidence held and renormalise. A head makes higher values of p more plausible, so the curve tilts right. A tail tilts it left. Click through the panel and watch the flat line grow a bump, then a hump, then a spike.

Here is the part that makes this family the one everybody uses. Multiply a Beta prior by the Bernoulli likelihood of what you saw, and the answer is another Beta. Not a new shape needing a new name — the same shape, with its two counts updated:

posterior = Beta(1 + heads, 1 + tails)

That property is called being a conjugate prior, and the jargon hides something almost disappointingly simple. It means the shape survives its own update, so a full Bayesian revision collapses into bookkeeping. Saw a head, add one to the left count. Saw a tail, add one to the right count.

Run it once to see the whole loop. One head and one tail takes you from flat to Beta(2,2), a gentle hump centred at 0.5000, which says you now lean toward fairness but weakly. Eight heads and two tails takes you to Beta(9,3), with mean 9/12 = 0.7500 and a clear peak near 0.8000. The curve got both taller and narrower, and narrower is the visual signature of confidence.

Notice what this closes. Chapter 10's Bayes updated a handful of discrete hypotheses in a table. This is that identical move on a continuum of hypotheses, and it is how a live model keeps a running belief about a parameter instead of a single brittle guess.

11The tree, whole

Step back and look at what we actually did. We never learned a dozen distributions. We learned one brick and then asked it a different question each time, and every question grew a branch with its own formula, mean and variance attached.

The distribution family tree — one coin flip, nine questions. Click a branch to re-derive it; then Locate three cold prompts.

Tap any branch to begin →

its mean re-derives from the brick

— not looked up, derived

one coin flip grows the whole tree

Then hit Locate — I hand you an unseen problem, you find the branch.

Explore — click any branch

The skill that outlives the chapter: you don't recall a distribution — you locate it. Name the question, find the branch, re-derive the mean straight from one coin flip.

the brick — Bernoulli, one flip; every mean below is re-derived from it

counting · how many? Bernoulli→Binomial→Poisson (n huge, p tiny)

waiting · how long? Geometric→Neg-Binomial

continuous · area, not counts: Uniform→Normal→Lognormal

belief · about p itself: Beta (prior + data → posterior)

Fig. 20. There is only one tree, and it grows from a single coin flip — the Bernoulli brick, whose mean is p in one line. Every other distribution is that brick asked a new question: count n of them and the mean is np; let them go rare and Binomial becomes Poisson, where mean = variance; ask how long instead of how many and you get Geometric, whose coin has no memory. So don't memorise twelve unrelated objects — learn the root and the branches regenerate. Then the real test: three problems you've never seen, and you locate each one from its mechanism rather than recall it. (Why sums go Normal is Ch13's CLT; Lognormal's shape is Ch26 — owed, not yet paid.)

Read the branches by their questions rather than their names. Count the successes in n tries and you get the Binomial. Let the tries go to infinity while the successes go rare and it becomes Poisson. Wait for the first success and you get the Geometric, and waiting for the r-th just stacks r of those.

The continuous side is the same logic with lengths instead of counts. Spread the chance evenly and you get the Uniform. Add many independent pieces and you get the Normal. Multiply them instead and you get the lognormal. And off to one side, ask about p itself rather than about outcomes, and you get the Beta.

This is the skill you keep, and it outlasts any single branch. When you meet a distribution you have never seen, you do not try to recall it — you locate it, and the tree panel carries three cold prompts to practise on.

Typos on a printed page — many characters, each with a tiny independent chance of going wrong, so it is a rare count in a fixed window. That is Poisson, and if the average is 2 per page then the variance is 2 as well, without looking anything up.

Trades until your first fill at a 25% hit rate — that is a wait for the first success, so it is Geometric with mean 1/0.25 = 4. The sum of 300 daily P&Ls is a sum of many roughly independent pieces, so it leans Normal, with all the caveats Chapter 13 is about to attach.

Carry three things out of this page. A distribution is a question asked of a brick, so name the question and the branch follows. When a variable is a sum of pieces, linearity hands you the mean for free, and if the pieces are independent it hands you the variance too. And the limits between branches are not trivia, because Poisson is Binomial gone rare and the Normal is what sums become.

That last one is the crack the next chapter drives a wedge into. We keep saying sums of many independent things go Normal, and so far you have watched it happen without being told why it must. Chapter 13 proves it, states the conditions honestly, and shows you the case where it fails.

‹ prevRandom Variables, Expectation & Variance next ›The Law of Large Numbers & the CLT

iolinked.com

Written by Ajai Raj