From the Binomial Tree to Black-Scholes

Black–Scholes looks like magic until you price an option on a one-step tree by hand. The tree contains every idea in the formula — replication, risk-neutral probability, why drift drops out — in arithmetic a middle-schooler could check. This derivation is also, verbatim, a common interview question.

One step, one option, no model

A stock trades at \100. In one period it will be either \120 or \80 (that's the entire model). The risk-free rate is 0. Price a call struck at \100 — payoff \20 in the up state, \0 in the down state.

Replicate it. Hold $\Delta$ shares and $B$ in bonds, and match the payoff in both states:

$$120\Delta + B = 20, \qquad 80\Delta + B = 0$$

Subtracting: $\Delta = \frac{20 - 0}{120 - 80} = \frac{1}{2}$, then $B = -40$. So: buy half a share, borrow $40. That portfolio pays exactly the call in both states, so today the call is worth

$$C = \tfrac{1}{2}(100) - 40 = \mathbf{\$10}.$$

Notice what we never used: the probability of the up move. Whether the stock rises with probability 0.9 or 0.1, the call is worth $10, because the price is the cost of manufacturing the payoff, not a forecast of it. Interviewers set this trap constantly — offering "the stock goes up 70% of the time" and watching who reaches for it.

Risk-neutral probability

Rearranging the replication algebra, the price can be written as a discounted expectation under artificial probabilities:

$$C = e^{-rT}\left[\, q \cdot C_{up} + (1-q)\cdot C_{down} \,\right], \qquad q = \frac{e^{rT} - d}{u - d}$$

where $u, d$ are the up/down gross returns. Here $q = \frac{1 - 0.8}{1.2 - 0.8} = \frac12$, and indeed $\frac12(20) + \frac12(0) = 10$. ✓

$q$ is the risk-neutral probability: the unique weighting that makes the stock itself earn the risk-free rate. It is not anyone's belief about the world — it's a bookkeeping device that encodes replication cost as an expectation. That one sentence — "risk-neutral probabilities are prices, not forecasts" — resolves most of the conceptual confusion candidates display about derivatives.

Many steps, then the limit

Chain steps into a tree of $n$ periods, sized so the tree matches a volatility $\sigma$ (e.g. $u = e^{\sigma\sqrt{\Delta t}}$). Price by backward induction: option values at the leaves, then discounted $q$-expectations back to the root — re-deriving $\Delta$ at every node, which is dynamic hedging. As $n \to \infty$ the binomial distribution converges (CLT!) to a lognormal, and the price converges to the Black–Scholes formula:

$$C = S_0\, N(d_1) - K e^{-rT} N(d_2), \qquad d_{1,2} = \frac{\ln(S_0/K) + (r \pm \sigma^2/2)T}{\sigma\sqrt{T}}$$

Read it structurally, the way you'd explain it aloud: it's the one-step answer grown up — $N(d_1)$ is the hedge ratio $\Delta$ (the "half a share"), and $Ke^{-rT}N(d_2)$ is the borrowing, with $N(d_2)$ interpretable as the risk-neutral probability of finishing in the money. Same recipe: option = Δ shares minus a loan.

What the model assumes (and what breaks)

Continuous frictionless hedging, constant volatility, lognormal moves. Reality violates all three — hedging is discrete and costly, volatility moves, and returns jump — which is why traders quote the implied vol surface (the market's corrections to Black–Scholes, strike by strike) rather than believing the flat-σ model. The formula survives not because it's true but because it's a robust interpolation language: everyone converts prices to implied vols and negotiates in those units.

The interview version

The one-step replication above is the question, at every level of seniority. Variations: different numbers; "price the put" (parity or re-replicate: $\Delta = -\tfrac12$); "now the stock is 100/110/90 — why can't you replicate with three states and two instruments?" (incomplete markets — the honest answer is a price range, and it's a strong-signal answer to know it).