The Binomial Model in Full

Last lesson you built an option out of stock and cash, watched the real-world probabilities cancel out of the algebra, and saw a strange new number — a risk-neutral probability $q$ — appear from nowhere to do the pricing. That was the magic trick. This lesson is the instruction manual: we’ll write $q$ down as a clean formula, roll it across a multi-step tree, calibrate the whole thing to a real stock’s volatility, and watch it quietly turn into Black–Scholes as we slice time finer. By the end you’ll be able to price an option with a pen, a single exponential, and some patient arithmetic.

First, a sanity check on what you carried over from lesson 1.

The one-step formula, stated cleanly

Set up the simplest possible future: one period of length $T$, at the end of which the stock has either gone up to $S_0 u$ or down to $S_0 d$. Here $u$ is the up-factor and $d$ is the down-factor, with $d < 1 < u$ (down means shrink, up means grow). The risk-free rate over the step is $r$, so a dollar in the bank becomes $e^{rT}$.

The option is worth $V_u$ if the stock goes up and $V_d$ if it goes down. Lesson 1’s replication argument collapses to two lines. Define the risk-neutral up-probability:

$q = \frac{e^{rT} - d}{u - d}$

Then today’s fair price of the option is the discounted, $q$-weighted average of the two outcomes:

$V = e^{-rT}\,\big[\,q\,V_u + (1 - q)\,V_d\,\big]$

Read each piece slowly, because every later step is just this sentence repeated:

• $q\,V_u + (1-q)\,V_d$ is an expected payoff — but weighted by $q$, not by the real odds.
• $e^{-rT}$ discounts that expected payoff back to today’s money (a dollar at expiry is worth less than a dollar now).
• $q$ itself is built only from $u$, $d$, and the risk-free growth $e^{rT}$. Notice what’s missing: the real probability of an up-move, the stock’s expected return, your opinion. None of it is in the formula.

The no-arbitrage bound

For $q$ to behave like a probability it must sit between 0 and 1, and that forces a condition on the factors:

$d < e^{rT} < u$

This isn’t a technicality — it’s the law that keeps the market honest. Suppose it broke. If $e^{rT} \le d$, then even the down outcome ($S_0 d$) grows at least as fast as cash: the stock can only beat or tie the bank, never lose to it. Borrow money, buy the stock, and you’ve got a guaranteed free lunch — arbitrage. If $e^{rT} \ge u$, the stock can never even keep up with the bank, so you short it, park the proceeds at the risk-free rate, and again pocket a riskless profit. Only when the risk-free growth sits strictly between the down and up outcomes does the stock genuinely have an upside and a downside — and only then is $q$ a number between 0 and 1.

Why q is not the real probability

This is the idea people trip over for years, so let’s nail it. Rearrange the definition of $q$ and ask what it makes true. Plug $q$ back into the expected stock price next period:

$q\,(S_0 u) + (1 - q)\,(S_0 d) = S_0\,e^{rT}$

In words: $q$ is precisely the probability that makes the stock’s own expected growth equal the risk-free rate. Under the real odds, a risky stock should be expected to out-grow cash (that extra return is your reward for stomaching the risk). Under $q$, that risk premium is surgically removed — every asset, risky or not, is “expected” to grow at exactly $r$. That’s why $q$ is called the risk-neutral probability: it describes a make-believe world where nobody demands extra return for taking risk.

So $q$ is a pricing device, not a forecast. If you genuinely believe the stock will rise 70% of the time, that’s a fine trading opinion — but it does not go into the option price. The price uses $q$, and $q$ is fixed by $u$, $d$, and $r$ alone. The market prices the option “as if” everyone were risk-neutral, not because they are, but because that’s the only weighting under which the replicating recipe and the option must cost the same. Confuse $q$ with your forecast and you’ll mis-price every contract you touch.

A fully worked two-step example

Numbers make it concrete. Take a stock at $S_0 = 100$, with $u = 1.2$ and $d = 0.8$ over each step, a European call struck at $K = 100$, and — to keep the arithmetic clean — a risk-free rate of essentially zero for now, so $e^{rT} = 1$. We’ll run two steps.

Step A — grow the tree forward. Each node either multiplies by 1.2 or by 0.8:

So the three terminal prices are 144, 96, 64. The call’s payoff at expiry is $\max(S - K, 0)$ with $K = 100$:

• $V_{uu} = \max(144 - 100, 0) = 44$
• $V_{ud} = \max(96 - 100, 0) = 0$
• $V_{dd} = \max(64 - 100, 0) = 0$

Step B — compute $q$. With $e^{rT} = 1$:

$q = \frac{1 - 0.8}{1.2 - 0.8} = \frac{0.2}{0.4} = 0.5$

A tidy fifty-fifty. (That’s a coincidence of our zero-rate, symmetric-ish factors — don’t expect 0.5 in general.)

Step C — backward-induct, node by node. Now we walk the tree backwards, applying $V = e^{-rT}[q V_u + (1-q) V_d]$ at every interior node. With $e^{-rT} = 1$ and $q = 0.5$ it’s just the average of the two children.

The upper node after one up-move (stock at 120) has children 44 and 0:

$V_u = 0.5 \times 44 + 0.5 \times 0 = 22$

The lower node after one down-move (stock at 80) has children 0 and 0:

$V_d = 0.5 \times 0 + 0.5 \times 0 = 0$

The root (today) has children $V_u = 22$ and $V_d = 0$:

$V_0 = 0.5 \times 22 + 0.5 \times 0 = 11$

The call is worth $11 today. Not the average payoff under your gut feeling — the discounted $q$-weighted roll-back of the whole tree. Every interior node is the same formula; we just applied it three times, right-to-left.

Now drive the tree yourself. This island uses the same spot and factors but a small positive rate, so $q$ won’t be a clean 0.5 — watch the readout. Move the sliders and the values roll back from the terminal column to the root in real time.

Calibrating u and d to a real stock

So far $u = 1.2$ and $d = 0.8$ were plucked from thin air. To price a real option you need the tree to match the real stock’s wobbliness — its volatility $\sigma$ (the annualised standard deviation of its returns). The Cox–Ross–Rubinstein (CRR) recipe sets the factors directly from $\sigma$ and the length of one time step $\Delta t$:

$u = e^{\sigma \sqrt{\Delta t}}, \qquad d = \frac{1}{u} = e^{-\sigma \sqrt{\Delta t}}$

Two things to notice. First, $d = 1/u$, so an up-move followed by a down-move lands you exactly back where you started — the tree recombines, which keeps the number of nodes small (a 50-step tree has 51 terminal prices, not $2^{50}$). Second, the size of each move scales with $\sqrt{\Delta t}$: chop a year into more steps, each $\Delta t$ shrinks, and each up/down jump gets smaller. A jittery, high-$\sigma$ stock gets a wide tree (big $u$); a sleepy bond-like stock gets a narrow one.

Worked calibration. Say $\sigma = 20\%$ per year and you want monthly steps, so $\Delta t = 1/12$. Then $\sqrt{\Delta t} = \sqrt{0.0833} \approx 0.2887$, and

$u = e^{0.20 \times 0.2887} = e^{0.0577} \approx 1.0594, \qquad d = 1/u \approx 0.9439$

Each month the stock either rises about 5.9% or falls about 5.6% in this model — a believable monthly wobble for a 20%-vol stock. More steps means finer time slices, smaller jumps, and a more realistic spread of outcomes.

Convergence: the tree is Black–Scholes in slow motion

Here’s the payoff for all that calibration. Take a fixed option and price it with a 2-step tree, then 10 steps, then 50, then 500. The prices don’t wander off — they converge, homing in on a single limiting number. That limit is the Black–Scholes price (the subject of the next lesson). As the number of steps $\to \infty$, each move shrinks toward zero, the jagged binomial distribution of final prices smooths into the bell-shaped lognormal distribution Black–Scholes assumes, and the discrete roll-back becomes the continuous formula.

So the binomial model isn’t a rough cousin of Black–Scholes — it’s Black–Scholes in slow motion, frozen into discrete frames you can actually compute by hand. The convergence isn’t perfectly smooth, either: as you add steps the binomial price tends to oscillate around the limit (even step counts and odd step counts approach from slightly different sides) before settling down. Doubling the steps roughly halves the error. A few hundred steps is usually plenty for a price you’d trade on.

American options: the one thing the tree can do that the formula can’t

Everything so far priced a European option (exercise only at expiry). Trees earn their keep on American options, where the holder may exercise any time. The plain Black–Scholes formula can’t handle that early-exercise choice; the tree handles it with a one-line tweak.

At every node of the tree, instead of just taking the rolled-back continuation value, you take the larger of two things:

$V_{\text{node}} = \max\Big(\underbrace{e^{-rT}[q V_u + (1-q)V_d]}_{\text{hold / continuation value}},\ \underbrace{\text{payoff}(S_{\text{node}})}_{\text{exercise now}}\Big)$

At each step backward you ask: is exercising right now worth more than holding on? If yes, you exercise (and the node takes the immediate payoff); if no, you hold (and it keeps the continuation value). That single $\max$ is the entire difference between American and European pricing in a tree — and it’s exactly why an American option is worth at least as much as its European twin: the extra right to bail early can only help.

When does that early-exercise right actually have value? Almost always for American puts: deep in the money, a put’s payoff $K - S$ is capped (the stock can’t fall below zero), so it can be worth grabbing the cash now and earning interest on it rather than waiting. For American calls on a non-dividend stock, early exercise is never optimal — you’d throw away time value and the interest on the strike — so they price the same as European calls. The exception is calls on a dividend-paying stock, where exercising just before a dividend can pay off.

Putting it together

You can now price an option from scratch: grow a recombining tree with $u = e^{\sigma\sqrt{\Delta t}}$ and $d = 1/u$, stamp the payoff on every terminal node, compute $q = (e^{rT} - d)/(u - d)$, and roll backward with $V = e^{-rT}[qV_u + (1-q)V_d]$ — taking a $\max$ against immediate exercise at every node if the option is American. Add steps and the price marches toward Black–Scholes.

Next up — The Black–Scholes Formula — we let the steps go to infinity. The tree’s discrete roll-back smooths into a single closed-form equation, the lognormal bell curve replaces the jagged terminal distribution, and that mysterious $q$ reappears wearing a new costume: $N(d_2)$, a probability hiding inside the most famous formula in finance.