Quant Finance Foundations, Part 6. Previously: What an Option Really Is.
Binomial tree option pricing is the most important topic in this series. Everything else in derivatives — Black-Scholes, Monte Carlo, and the whole apparatus of a modern trading desk — rests on the single idea it demonstrates. Moreover, that idea needs no calculus at all. In fact, you can follow the entire argument with arithmetic a secondary school student would recognise.
The idea is replication. In other words, you build a portfolio that reproduces the option’s payoff in every possible future state. Consequently, the option must cost exactly what that portfolio costs.
This post assumes you have read what an option really is and, ideally, the introduction to quantitative finance that opens the series.
The setup for binomial tree option pricing
Radical simplification comes first. We allow one period only, and the stock can end at exactly two prices.
- Stock today: S = 100
- In one year it goes up to 120 or down to 90. Nothing else.
- Risk-free rate: 5% annually
- We price a European call, strike K = 105
At expiry the option is worth 15 in the up state (120 − 105) and 0 in the down state.
Admittedly, the two-state assumption looks crude. However, it is not a limitation. As you will see at the end, adding more steps converges to the continuous answer, and the logic never changes along the way.
Here is the whole method in outline before we work through it:
- Solve for the hedge ratio that matches the option in both states.
- Solve for the cash position that completes the match.
- Price the resulting portfolio today.
- Verify the payoff in every state.
- Conclude by no-arbitrage.
Step 1: find the hedge ratio
Construct a portfolio of Δ shares plus a cash amount B, and require it to match the option in both states.
- Up state: 120Δ + 1.05B = 15
- Down state: 90Δ + 1.05B = 0
Subtract the second from the first. The cash term cancels:
30Δ = 15, so Δ = 0.5
There is a shortcut worth memorising: Δ = (payoff up − payoff down) / (price up − price down) = (15 − 0) / (120 − 90) = 0.5. This is the hedge ratio, and it says the option behaves like half a share over this move.
Step 2: find the cash position
Substitute Δ = 0.5 into the down-state equation:
90(0.5) + 1.05B = 0 → 45 + 1.05B = 0 → B = −42.857
Negative means borrowing. The replicating portfolio is: hold half a share, borrow 42.857.
Step 3: price the portfolio today
Cost = 0.5 × 100 − 42.857 = 50 − 42.857 = 7.143
Step 4: verify it in both states
Never skip this step. In one year the borrowing has grown to 42.857 × 1.05 = 45.
| State | Value of 0.5 share | Debt repaid | Net | Option payoff |
|---|---|---|---|---|
| Up (120) | 60.00 | 45.00 | 15.00 | 15.00 |
| Down (90) | 45.00 | 45.00 | 0.00 | 0.00 |
The portfolio reproduces the option’s payoff in every possible state of the world. It is not an approximation — it is exact within the model.
Step 5: conclude by no-arbitrage
Two things with identical payoffs must have identical prices. The call is worth 7.143.
If it traded at 8, you would sell the option, buy the replicating portfolio for 7.143, and keep 0.857 with zero residual risk. If it traded at 6, do the reverse. Either way the market would close the gap.
The result that should unsettle you
Look back over those five steps. Then try to find where we used the probability of the stock going up.
It is not there. Therefore the price 7.143 holds whether the up move has a 10% chance or a 90% chance. Two traders with opposed views on the stock must nevertheless agree on the option’s value. After all, either of them can construct the replicating portfolio, and its cost does not care about their opinions.
What does matter is the size of the two possible moves, since that is what determines Δ. The spread between the up and down states is volatility in its rawest form. This is the same conclusion as the previous post, now derived rather than asserted.
Risk-neutral probabilities
There is a second route to the same number. Solve for the probability p that makes the stock’s expected return equal the risk-free rate:
100 × 1.05 = p(120) + (1 − p)(90)
105 = 90 + 30p → p = 0.5
Now take the expected option payoff under that probability and discount it:
[0.5(15) + 0.5(0)] / 1.05 = 7.5 / 1.05 = 7.143
The same answer, by a different road. This p is the risk-neutral probability, and it is important to be clear about what it is not: it is not anybody’s forecast, and it does not become the real probability just because we computed it. It is a mathematical device that encodes the replication argument in a form that is far easier to extend to complex payoffs.
The phrase “risk-neutral” causes endless confusion. Nobody assumes investors are indifferent to risk. The point is that under this artificial measure, the replication argument lets you price as though they were, and the answer comes out right.
Scaling binomial tree option pricing to a real market
One period with two outcomes is obviously not a market. Fortunately, the fix is mechanical. You chop the year into n shorter periods, each with its own small up or down move. Then you work backwards from expiry through the resulting tree, applying the identical five steps at every node.
Convergence is quick in practice:
| Steps in the tree | Behaviour |
|---|---|
| 1 | Crude, but the logic is already correct |
| 50 | Price is close to stable |
| Several hundred | Converges to the Black-Scholes value |
In other words, the continuous formula is simply the limiting case of this tree as the steps become infinitesimal. That is why understanding the tree first makes Black-Scholes an outcome rather than a magic incantation.
Trees also handle things closed-form solutions cannot. American options need an early-exercise check at each node — compare the continuation value against immediate exercise and take the larger. That single extra line of logic is why binomial trees remain in production use on real desks decades after more sophisticated methods arrived.
Try it yourself
Build the one-period example in a spreadsheet from scratch. Then change the down state from 90 to 80 and re-solve. The option gets more expensive despite the stock’s starting price being unchanged — wider possible outcomes, more valuable optionality, exactly as the previous post predicted.
Then extend to two periods: four terminal nodes, work backwards, and confirm the delta changes at each node rather than staying fixed. That changing delta is what forces a real hedger to rebalance continuously, and it is the subject of a later post in this series.
Frequently asked questions
Why does binomial tree option pricing ignore the real probability of an up move?
Because the option is priced by replication rather than by forecasting. The replicating portfolio matches the payoff in every state, so its cost is fixed regardless of how likely each state is.
Are risk-neutral probabilities a prediction?
No. They are a computational device that encodes the replication argument. They do not become real probabilities simply because we calculated them.
Is the binomial method still used in practice?
Yes. It remains in production because it handles American early exercise with one extra line of logic, which closed-form solutions cannot do.
How many steps are enough?
Around 50 steps gives a stable price for most European options. Several hundred brings it in line with Black-Scholes.
Next in this series: random walks, Brownian motion and Itô’s lemma — the continuous-time picture, without the measure theory.