Visual walkthrough — Uniform, Exponential, and Beta distributions

This page builds the Beta distribution from nothing but a coin you are unsure about. We will not write a single symbol before we have drawn it — so the famous formula is deliberately withheld until Step 5, once every piece ($x$, the two exponents, and the normaliser) has been earned.

If any word here feels new, the parent note is your home base. We also lean on Continuous Random Variables and PDFs (what a density even is) and Bayesian Inference and Conjugate Priors (why "belief" is a distribution).

Roadmap of the eight figures (each step points at one, so you can always say "look at Figure N"):

• Figure 1 — the $[0,1]$ segment of possible biases.
• Figure 2 — evidence carves a raw hump.
• Figure 3 — relabelling with $\alpha ,\beta$; the flat "know-nothing" curve.
• Figure 4 — normalising by the area $B(\alpha ,\beta )$.
• Figure 5 — the finished, valid Beta densities.
• Figure 6 — the mean as a balance point.
• Figure 7 — every shape case, including the one-sided singular ones.
• Figure 8 — the one-picture summary.

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Step 1 — The thing we are uncertain about lives on a line from 0 to 1

WHAT. A coin's bias is a number $x$ with $0\le x\le 1$. Here $x=0$ means "always tails," $x=1$ means "always heads," and $x=0.5$ means "fair." That's the whole playground: the segment from 0 to 1.

WHY. A probability can never be negative and never exceed 1. So whatever describes our belief about $x$ must be zero outside $[0,1]$ — it lives only on that segment. This is the first constraint the Beta shape must obey.

PICTURE — Figure 1. The red segment below is the entire universe of possible biases. A belief is a way of shading this segment: taller shading = "more believable."

Figure 1 — the bias lives on the red segment $[0,1]$.

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Step 2 — Evidence reshapes the segment: each head tilts it right, each tail tilts it left

WHAT. Suppose the true (but hidden) bias is $x$. The chance of seeing one head is exactly $x$. The chance of seeing one tail is $1-x$. If we flip and get $H$ heads and $T$ tails, and the flips are independent, the chance of that exact sequence is

$\underset{\text{per head}}{\underbrace{x}}\times x\times \dots \times \underset{\text{per tail}}{\underbrace{(1-x)}}\times (1-x)\dots =x^H(1-x{)}^T.$

Every symbol earns its place: $x^H$ is "$x$ multiplied by itself once per head," and $(1-x{)}^T$ is "$(1-x)$ once per tail."

WHY multiply? Independent events combine by multiplication — that's what independence means. Each head demands a factor $x$; each tail demands a factor $1-x$.

PICTURE — Figure 2. The plain function $x^H(1-x{)}^T$ is drawn for a few $(H,T)$. More heads → the hump slides toward $x=1$. More tails → toward $x=0$. This hump is not yet a probability distribution (its area isn't 1) — it's the raw shape evidence carves.

Figure 2 — each head pulls the hump right, each tail pulls it left.

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Step 3 — Rename the exponents so the curve stands alone: meet $\alpha$ and $\beta$

WHAT. We rename the counting so the formula works even before any flips. Define

$\alpha =H+1,\beta =T+1.$

Then the exponent of $x$ becomes $H=\alpha -1$, and the exponent of $(1-x)$ becomes $T=\beta -1$. The raw shape is now

$x^{\alpha -1}(1-x{)}^{\beta -1}.$

WHY the "+1"? With zero data ($H=T=0$) we want a flat curve — total ignorance, every bias equally believable. Setting $\alpha -1=0$ and $\beta -1=0$ (i.e. $\alpha =\beta =1$) makes both exponents zero, so the shape is $x^0(1-x{)}^0=1$: perfectly flat. The "+1" is exactly the bookkeeping that turns "no data" into "flat." So $\alpha$ and $\beta$ are pseudo-counts: heads-plus-one and tails-plus-one.

PICTURE — Figure 3. Same humps as Figure 2, relabelled with $\alpha ,\beta$, plus the flat $\alpha =\beta =1$ line highlighted in red — the "I know nothing" belief.

Figure 3 — $\alpha =H+1,\beta =T+1$; the flat $\alpha =\beta =1$ curve is total ignorance.

Recall Why

$\alpha =\beta =1$ is Uniform(0,1) With $\alpha =\beta =1$ the shape is $x^0(1-x{)}^0=1$ on $[0,1]$ ::: a constant, i.e. the flat Uniform density — the same "no preference" idea from the Uniform section.

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Step 4 — The area is wrong, so we divide by it: the Beta function is born

WHAT. The raw shape $x^{\alpha -1}(1-x{)}^{\beta -1}$ almost never encloses area 1, so it is not yet a valid density. We measure its area and divide by it. Call that area

$B(\alpha ,\beta )={\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}dx.$

WHY divide? A density must integrate to 1 (that's the law from Continuous Random Variables and PDFs). If a curve has area $B$, then curve$/B$ has area $B/B=1$. Dividing by the total area is the only rescaling that fixes the area without changing the shape.

WHY the Gamma closed form? We do not have to leave $B$ as an unevaluated integral — it collapses to a ratio of Gamma functions. Here is the honest reason, in plain steps. The Gamma function is defined by $\Gamma (\alpha )={\int }_0^{\infty }{t}^{\alpha -1}{e}^{-t}dt$, the smooth continuation of the factorial ($\Gamma (n)=(n-1)!$ for whole $n$). If you write down the product $\Gamma (\alpha )\Gamma (\beta )$ as a double integral over two independent variables $s,t\ge 0$ and then switch to the new coordinates "total $u=s+t$" and "fraction $x=s/(s+t)$," the double integral factorises cleanly: the total $u$ part reassembles into $\Gamma (\alpha +\beta )$, and the fraction $x$ part becomes exactly our integral ${\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}dx=B(\alpha ,\beta )$. Reading that factorisation as an equation gives $\Gamma (\alpha )\Gamma (\beta )=\Gamma (\alpha +\beta )B(\alpha ,\beta )$, i.e. the identity below. (The change of variables is the reason the "total" and "fraction" separate — the same split that later makes Beta the natural belief over a fraction.)

PICTURE — Figure 4. Left: the raw hump with its (wrong) grey area labelled $B(\alpha ,\beta )$. Right: the same hump squashed vertically by $1/B$ so the red area is exactly 1. Same shape, correct total.

Figure 4 — normalise by the total area $B(\alpha ,\beta )$ so the red area becomes 1.

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Step 5 — Assemble the Beta distribution

WHAT. Put the pieces together: the shape from Step 3 divided by the area from Step 4.

$f(x;\alpha ,\beta )=\frac{1}{B(\alpha ,\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1},0\le x\le 1,\alpha >0,\beta >0.$

WHY this is exactly right. Each part answers one demand:

• $x^{\alpha -1}$ — the heads-pull (Step 2–3),
• $(1-x{)}^{\beta -1}$ — the tails-pull,
• $\frac{1}{B(\alpha ,\beta )}$ — the divide-by-area fix so total probability is 1 (Step 4).

Nothing is decorative; drop any factor and either the shape or the area breaks.

PICTURE — Figure 5. Four finished Beta curves side by side — flat, right-skewed, left-skewed, and sharp — each a valid density (red area = 1).

Figure 5 — four valid Beta densities, each enclosing area 1.

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Step 6 — Where is the balance point? Reading off the mean

WHAT. The mean (centre of mass) is

$E[X]=\frac{\alpha }{\alpha +\beta }=\frac{H+1}{(H+1)+(T+1)}.$

WHY — the one-line derivation. By definition the mean multiplies each $x$ by its density and integrates:

$E[X]={\int }_0^{1}x\cdot \frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{B(\alpha ,\beta )}dx=\frac{1}{B(\alpha ,\beta )}{\int }_0^1{x}^{\alpha }(1-x{)}^{\beta -1}dx.$

Look at that last integral: multiplying by $x$ turned the exponent $\alpha -1$ into $\alpha$ — i.e. it is itself a Beta integral, but with $\alpha$ shifted up by one. So it equals $B(\alpha +1,\beta )$ (the shift property: raising $\alpha$'s exponent by one just bumps $\alpha \to \alpha +1$ inside $B$). Therefore

$E[X]=\frac{B(\alpha +1,\beta )}{B(\alpha ,\beta )}=\frac{\Gamma (\alpha +1)\Gamma (\beta )/\Gamma (\alpha +\beta +1)}{\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )}=\frac{\Gamma (\alpha +1)}{\Gamma (\alpha )}\cdot \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha +\beta +1)}.$

Now use the single fact $\Gamma (z+1)=z\Gamma (z)$: the first ratio is $\alpha$, the second is $\frac{1}{\alpha +\beta }$. Multiplying gives $\frac{\alpha }{\alpha +\beta }$. Done.

WHY it makes sense. It is (heads+1) over (total flips + 2) — almost the plain fraction of heads, gently nudged toward $1/2$ by the two phantom flips baked into the "+1"s. More heads ⇒ mean slides toward 1; more tails ⇒ toward 0. Balanced counts ⇒ mean at $1/2$.

PICTURE — Figure 6. A Beta curve with a red triangle marking the balance point at $x=\alpha /(\alpha +\beta )$; arrows show it sliding right as $\alpha$ grows.

Figure 6 — the mean is the centre of mass at $\alpha /(\alpha +\beta )$.

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Step 7 — Every case: peaks, U-shapes, one-sided spikes, and the corners

WHAT & WHY, case by case — the two exponents $\alpha -1$ and $\beta -1$ decide the behaviour at the edges $x=0$ and $x=1$. A positive exponent pins that end to zero; a negative exponent ($0<\alpha <1$ or $0<\beta <1$) sends that end to $+\infty$; a zero exponent leaves it finite and non-zero.

• Both $\alpha ,\beta >1$ (both exponents $>0$). $x^{\alpha -1}\to 0$ at $x=0$ and $(1-x{)}^{\beta -1}\to 0$ at $x=1$. Pinned to zero at both ends ⇒ a single interior hump. "Data points to a middling rate."
• Both $\alpha ,\beta <1$ (both exponents $<0$). $x^{\alpha -1}\to \infty$ as $x\to 0$ and $(1-x{)}^{\beta -1}\to \infty$ as $x\to 1$ ⇒ a U-shape (infinite at both ends): "probably near-always-heads or near-always-tails, unsure which."
• One-sided singular: $0<\alpha <1$ and $\beta >1$ (so $\alpha -1<0$ but $\beta -1>0$). The left end $x=0$ blows up to $+\infty$, while the right end $x=1$ is pinned to zero ⇒ a "J" that is infinite at 0, zero at 1, sliding down across the interval. Belief concentrated near 0 (mostly tails, but not certain).
• One-sided singular: $\alpha >1$ and $0<\beta <1$ (mirror image). Now $x=1$ blows up to $+\infty$ and $x=0$ is pinned to zero ⇒ a reversed "J", infinite at 1, zero at 0. Belief concentrated near 1 (mostly heads).
• $\alpha =\beta =1$. Both exponents zero ⇒ flat ⇒ Uniform (Step 3).
• Border $\alpha =1,\beta >1$ ⇒ straight-ish slope, finite non-zero at $x=0$, zero at $x=1$; $\alpha >1,\beta =1$ ⇒ the mirror, finite at $x=1$, zero at $x=0$.
• Degenerate limit, $\alpha ,\beta \to \infty$ with fixed ratio ⇒ the hump narrows to a spike at $\alpha /(\alpha +\beta )$: overwhelming evidence, near-certainty.

PICTURE — Figure 7. A gallery: hump, U-shape, the two one-sided singular J-shapes, flat, and a narrow spike — the red curve is the one named in each panel's title.

Figure 7 — the exponents decide every edge behaviour, including one-sided singular shapes.

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The one-picture summary

Everything above, in a single frame: start with the flat "know-nothing" line, let heads and tails carve a raw hump, divide by its area to make it a legal density, and read the balance point.

Figure 8 — flat → carve → normalise → read the mean.

Recall Feynman retelling — say it like you'd explain to a friend

A coin has a secret heads-chance somewhere between 0 and 1, but you don't know it — so instead of guessing one number you keep a curve over all the possibilities. Before any flips the curve is flat: every bias is equally believable. Then you flip. Each head multiplies the curve by $x$ (which is small near 0, so heads push the curve away from 0), and each tail multiplies by $1-x$ (small near 1, pushing away from 1). After $H$ heads and $T$ tails the curve looks like $x^H(1-x{)}^T$ — a hump leaning toward whichever you saw more of. But that hump doesn't have area 1, so it's not a real probability curve; you shrink it uniformly by dividing by its total area, a number we name $B(\alpha ,\beta )$ — which itself equals $\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )$ once you split a product of Gammas into "total times fraction." Rename the exponents with $\alpha =H+1,\beta =T+1$ (always keeping $\alpha ,\beta >0$) and you've got the Beta distribution. Its balance point $\alpha /(\alpha +\beta )$ — found by nudging one exponent up by one inside $B$ and cancelling — is basically your heads-fraction, softened toward $\frac{1}{2}$. And the two exponents alone tell the whole shape story: both above 1 → a tidy hump, both below 1 → a two-horned U, one below and one above 1 → a one-sided spike infinite at just one end, both exactly 1 → flat ignorance, both huge → a confident spike.