Backporch

Pillars · deep dive

The distribution foundry

The densities in the catalog look conjured — $Γ (k)$ here, $Γ (α) Γ (β) /Γ (α + β)$ there, as if a crystal ball were required. It wasn't. Every $Γ$ is one of three bricks: an ordering counted, a sphere measured, or a latent scale integrated out. Each brick below assembles live from raw draws.

Companions: the t & F forge (change of variables), generating functions (the same laws as transforms), and the transforms table (Γ is Mellin's creature: $Γ (s) = \int_{0}^{\infty} x^{s - 1} e^{- x} d x$ , the Mellin transform of $e^{- x}$ ).

1 · The ordering brick — waits convolve into Gamma

Wait for the $k$ -th arrival of a rate- $λ$ Poisson process. The $k - 1$ arrivals before time $x$ must land in order — and the volume of that ordered region is the whole mystery:

k (arrivals)0 / 6,000

Given T, the two earlier arrivals are uniform on the triangle t₁ < t₂ < T. Its area is T²/2! — the (k−1)! is a simplex volume.

Wait for the 3rd arrival, 6,000 runs — against the exact Gamma(3, 1) density.

(λ e^{- λ t})^{* k} (x) = λ^{k} e^{- λ x} x^{k - 1} / (k - 1)! vol {0 < t_{1} < \dots < t_{k - 1} < x} = \frac{λ ^{k} x ^{k - 1}}{Γ ( k )} e^{- λ x}

Brick I. $Γ (k) = (k - 1)!$ continued off the integers — a simplex volume, i.e. a count of orderings. The same brick puts the $k!$ in Poisson's pmf.

2 · The sphere brick — Gaussians square-sum into χ²

Square-sum $ν$ standard normals. The Gaussian cloud has no preferred direction, so its integral only sees the radius — and pays a sphere-area factor for each shell:

ν (dimensions)0 / 6,000

Two of the ν coordinates. The cloud has no preferred direction — so the integral only sees the radius, sphere by sphere.

‖Z‖² = Z₁² + ⋯ + Z₃², 6,000 draws — against the exact χ²(3) density.

\int_{R^{ν}} f (∥ z ∥) d z = \int_{0}^{\infty} S_{ν - 1} r^{ν - 1} f (r) d r, S_{ν - 1} = \frac{2 π ^{ν /2}}{Γ ( ν /2 )} ⟹ f_{χ_{ν}^{2}} (x) = \frac{x ^{ν /2 - 1} e ^{- x /2}}{2 ^{ν /2} Γ ( ν /2 )}

Brick II. $Γ$ is the sphere-area function. It haunts everything born from Gaussians — χ², Rayleigh, Maxwell — because polar coordinates always route through it. And in high dimensions that same factor concentrates the Gaussian onto a thin shell — a sphere persistent homology can read: the Gaussian is a sphere →

3 · The ratio brick — Gammas trade to Beta

Take independent $X \sim Γ (α)$ , $Y \sim Γ (β)$ and keep only the share $W = X / (X + Y)$ . In the quadrant, $W$ is the ray and $S = X + Y$ is the radius — and they separate:

αβ0 / 6,000

Color = the ratio w = x/(x+y). Every ray from the origin is one color: the ray forgets the radius.

W = X/(X+Y), 6,000 draws — against the exact Beta(2, 5) density.

x^{α - 1} e^{- x} y^{β - 1} e^{- y} d x d y x = s w, y = s (1 - w), ∣ J ∣ = s \int = Γ (α + β) s^{α + β - 1} e^{- s} d s \cdot w^{α - 1} (1 - w)^{β - 1} d w

⟹ B (α, β) = \frac{Γ ( α ) Γ ( β )}{Γ ( α + β )}

Brick III. The famous identity is not clever — it is forced: the radius integral factors out as $Γ (α + β)$ because ray ⟂ radius. Order statistics of uniforms land on the same density by counting who falls left and right.

4 · The mixture finale — a Normal with a random scale is t

Student's $t$ is bricks II and III composed: a Normal whose variance is itself random ( $χ_{ν}^{2}$ -distributed). Integrate the latent scale out and the Γ-ratio is the residue:

ν (df)0 / 6,000

T = Z / √(V/ν) with V ~ χ²(3) — solid: exact t(3); dashed: N(0,1). Small ν = fat tails; ν → ∞ recovers the normal. Built by hand in the forge; here the point is where the Γ-ratio comes from.

f_{ν} (x) = \int_{0}^{\infty} φ (x 0, \frac{ν}{v}) f_{χ_{ν}^{2}} (v) d v = \frac{Γ ( \frac{ν + 1}{2} )}{ν π Γ ( \frac{ν}{2} )} (1 + \frac{x ^{2}}{ν})^{- \frac{ν + 1}{2}}

The mixture brick. Every “magic” Γ-ratio — $t$ , $F$ , Negative Binomial with real $r$ — is a latent variable someone integrated out. The constants are receipts, not incantations.

5 · The three bricks

Brick	Move	Where Γ comes from	Lands on
Ordering	convolve waits	simplex volume $x^{k - 1} / (k - 1)!$	Gamma, Erlang, Poisson's $k!$
Sphere	square-sum normals	$S_{ν - 1} = 2 π^{ν /2} /Γ (ν /2)$	χ², Rayleigh, Maxwell
Mixture / ratio	integrate a latent scale	the $s$ -integral = $Γ (α + β)$	Beta, t, F, NegBin

One function, three doors — and all three are the same door from higher up: $Γ$ is the Mellin transform of $e^{- x}$ , the normalizer of the multiplicative world. That is the door the Mellin bridge walks through toward $ζ$ and modular forms: $π^{- s /2} Γ (s /2)$ — the factor in the functional equation — is the Mellin transform of the Gaussian.

6 · The missing brick — dependence is its own object

The three bricks forge one-variable laws — and margins never determine the joint. Strip any continuous margin to uniform (the probability integral transform, $U = F_{X} (X) \sim Unif (0, 1)$ ) and what remains is the copula: dependence, isolated. Five different joints below — with the identical margin every time:

copula0 / 6,000

Elliptical dependence, no tail clustering: λ = 0 for every ρ < 1.

The margin, X = −log(1−U), against the exact Exp(1) density — identical under every preset. Margins are blind to the joint.

F (x, y) = C (F_{X} (x), F_{Y} (y)) (Sklar — margins in, dependence separate)

C_{θ} (u, v) = (u^{- θ} + v^{- θ} - 1)^{- 1/ θ} from a shared Γ (1/ θ) frailty, λ_{L} = 2^{- 1/ θ}

The missing brick. The right panel never moves — margins are blind to the joint. And tail dependence is structure a correlation coefficient cannot carry: Gaussian ρ has $λ = 0$ , while Clayton's shared-Gamma frailty (the mixture brick, again) keeps $λ_{L} = 2^{- 1/ θ} > 0$ . Pricing joint defaults with a Gaussian copula is how that lesson got learned in 2008.

7 · The workbench — diagnose a dataset, build a law

Two directions through one machine. Diagnose runs the derivation checklist on a dataset: support → moment ladder → variance scan → moment fit → KS verdict. Build runs the three moves on atoms: tilt (which generates the exponential family), convolve (which walks to the bell), mix (which inflates variance — NegBin and t are mixtures caught in the act). Every curve is a closed form, not a simulation.

dataset

Support. 400 values on ℤ≥0 → count laws (min 0.00, max 9.00)
Ladder. x̄ = 1.84 · s² = 3.03 · skew g₁ = 1.28 · excess kurtosis g₂ = 1.81
Variance scan. D = s²/x̄ = 1.64 > 1 — overdispersed → mixture (NegBin)
Fit (moments). NegBin(r=2.86, p=0.61) vs Poisson(λ=1.84)
Verify (KS). NegBin: 0.029 · Poisson: 0.090 → NegBin wins

Solid: NegBin(r=2.86, p=0.61) (KS 0.029) · dashed: Poisson(λ=1.84) (KS 0.090).

PP plot for the winner — model CDF vs empirical. On the diagonal = the story holds.

Samples are synthetic and seeded (reproducible); every step runs in your browser. Verdicts read “consistent with”, never “is” — the checklist narrows the shelf, the KS number keeps it honest.

f_{θ} (x) = e^{θ x - ψ (θ)} f (x) (tilt — the natural exponential family of the atom f)

V (μ) = σ^{2} \Rightarrow Normal, μ \Rightarrow Poisson, \frac{μ ^{2}}{k} \Rightarrow Gamma, μ + \frac{μ ^{2}}{r} \Rightarrow NegBin, μ - \frac{μ ^{2}}{n} \Rightarrow Binomial

The skill, named. Morris's theorem: a natural exponential family with variance quadratic in the mean must be one of six laws — so the variance scan in the checklist isn't a heuristic, it's reading off a classification. Pick the shelf by support, read $V (μ)$ , tilt/mix to fit, verify by KS. That is “deriving a distribution on demand.” (v1 uses built-in seeded samples; pasting your own data arrives next.)

8 · Where probability lives — the Fisher–Rao sphere

Write $q_{i} = p_{i}$ and “probability requirements satisfied” becomes the unit-sphere equation $\sum q_{i}^{2} = 1$ . That is Rao's theorem: the Fisher information metric on distributions is the round sphere metric in these coordinates — so Hellinger distance is the chord and the Bhattacharyya coefficient is the cosine of the angle at the origin. Drag $p$ and read the same contours in both charts:

p =p = (0.45, 0.33, 0.22)

$d_{FR}$ arc = 0.286 $H$ chord = 0.143 $BC$ cosine = 0.990

Arc, chord, and cosine are three readings of the same separation from p to the uniform center (grey). The flat chart distorts them toward the boundary; in √p coordinates the geometry is exactly the round unit sphere — “probability requirements satisfied” is $\sum q_{i}^{2} = 1$ .

d s_{FR}^{2} = i \sum \frac{d p _{i}^{2}}{p _{i}} = 4 i \sum d q_{i}^{2}, d_{FR} (p, r) = 2 arccos (\sum_{i} p_{i} r_{i}), H^{2} (p, r) = \sum_{i} (p_{i} - r_{i})^{2}

The sphere brick, globalized. §2 met the sphere locally (Gaussian in polar coordinates); here the sphere is the space of distributions itself. The same square-root map is the Born rule read backwards — $p = ∣ ψ ∣^{2}$ , quantum amplitudes as the unit $L^{2}$ sphere (the QBism door) — and high-dimensional uniform sphere measure projects to the Gaussian (Poincaré–Borel).

9 · Where the mean dies — compactness vs central tendency

On a compact space the mean must be intrinsic: the Fréchet mean, the minimizer of expected squared distance. Von Mises data on the circle shows its engine — the resultant vector — and its death: as concentration $κ \to 0$ the resultant collapses and the bootstrap spray of mean directions widens from an arc to the whole rim.

κ (concentration)

resultant length |R̄| = 0.838

circular variance 1 − |R̄| = 0.162

The arrow points at the mean direction; the violet spray is its sampling uncertainty, still a tight arc.

\overset{ˉ}{R} = \frac{1}{n} i \sum (cos θ_{i}, sin θ_{i}), \overset{μ}{^} = μ arg min i \sum d (θ_{i}, μ)^{2}, κ = 0 \Rightarrow every μ minimizes

Limits of central tendency, dispersion, and convergence — named honestly. This death is compactness and curvature, not the Hairy Ball: the circle combs perfectly (see §10) yet its mean still dies at maximal dispersion. Near uniformity even the convergence rate degrades — the smeary CLT of Eltzner–Huckemann (2019) replaces $n^{- 1/2}$ with $n^{- 1/6}$ — which is what the widening violet spray is showing you.

10 · Comb the sphere — every flow has a still point

The Hairy Ball theorem is about vector fields, not measures: a continuous tangent field on $S^{2}$ must vanish, and Poincaré–Hopf says precisely how much — the indices of its zeros always sum to $χ (S^{2}) = 2$ . Comb it, spin it, or hoard the winding in one point: the budget of 2 is conserved.

field

Poincaré–Hopf: Σ indices = +1 +1 = 2 = χ(S²)

Comb everything toward +e₁: the hair must part somewhere. Two zeros, +1 each.

The gold rings are where the flow stops — no continuous field on S² avoids them, however you comb. The budget of 2 can move and merge, never vanish.

v (x) = 0 \sum ind_{x} (v) = χ (S^{2}) = 2, p \cdot z (p) = 0 (Walras) \Rightarrow z is a tangent field on the price sphere

Decision flows. Read $v$ as a preference gradient or an excess-demand field: Walras's law makes demand tangent to the price sphere, so “an equilibrium exists” is “this field has a zero” — the same index-theory family as the ball you can't comb. That, and not the mean's death in §9, is where the Hairy Ball genuinely constrains decisions.

Next in this dive: paste your own data into the workbench (v2).