More steps, and what the tree converges to

Chapter 4 ended with a wrong answer and a diagnosis: two outcomes is not enough of a world. The repair is mechanical.

Two steps, in full

Same contract — index 24,000, strike 24,000, thirty days, 6.5% interest, 14% volatility — split into two periods of 15 days.

Recalibrate for the shorter step. $\Delta t = 0.041096$ years, so

$u = e^{0.14\sqrt{0.041096}} = 1.028788, \qquad d = 0.972018, \qquad q = 0.540022$

The lattice. Because $d = 1/u$, an up followed by a down returns to the start, so three terminal nodes rather than four:

	$t = 0$	$t = 15$ days	$t = 30$ days
up-up		24,690.90	25,401.69
up-down	24,000.00		24,000.00
down-down		23,328.43	22,675.66

The payoffs. Call payoffs at the three terminal nodes: ₹1,401.69, ₹0, ₹0.

Work backwards one layer. At the upper node after 15 days the option can still go to ₹1,401.69 or to ₹0:

$c_u = e^{-r\Delta t}\left[0.540022 \times 1{,}401.69 + 0.459978 \times 0\right] = ₹754.93$

At the lower node both outcomes pay nothing, so $c_d = ₹0$.

And back to the root.

$c_0 = e^{-r\Delta t}\left[0.540022 \times 754.93 + 0.459978 \times 0\right] = ₹406.59$

That is the procedure in full, and it is all there ever is. The source notes drily that the original paper leaves the justification of this step implicit, with the authors stating they "now have a recursive procedure for finding the value of a call with any number of periods to go." Chapter 3's two-state argument, applied at every node, is that justification.

The hedge is no longer a fixed holding

Something important changed between chapter 4 and here, and it is easy to miss inside the arithmetic.

Compute the replicating holding at each node.

Node	Underlying	$\Delta$
Root, $t=0$	24,000.00	0.5541
Up node, $t=15$	24,690.90	1.0000
Down node, $t=15$	23,328.43	0.0000

The holding changes. Start with 0.554 units; if the index rises, buy up to a full unit; if it falls, sell the lot. The option is still being replicated exactly, but by a portfolio that must be traded rather than held.

This is the step from static to dynamic hedging, and three consequences follow that run through the rest of the subject.

The hedge buys high and sells low. Up moves require buying more, down moves require selling. That is not a flaw in the method — it is the cost of manufacturing the option, and chapter 9 shows it is exactly what the option's time decay is paying for.

The replication needs no transaction costs and no gaps to work. It assumes you can trade at every node, at the node's price. Chapter 7 is about what happens when you cannot.

And the tree does not know which option it is pricing. As Derman and Kani put it, the lattice "is the same for all options on that stock, irrespective of their strike level or time to expiration" — because "the stock tree cannot 'know' about which option we are valuing on it." One tree, built from the underlying's volatility, prices every contract written on it. Chapter 11 is about what happens when the market disagrees with that.

Convergence

Now just keep subdividing. The same contract, priced with more and more steps:

Steps	Value	Error against the limit
1	₹544.23	+₹93.66
2	₹406.59	−₹43.98
4	₹427.37	−₹23.20
8	₹438.71	−₹11.86
16	₹444.58	−₹5.99
32	₹447.56	−₹3.01
64	₹449.07	−₹1.51
128	₹449.82	−₹0.75
256	₹450.20	−₹0.38
1,024	₹450.48	−₹0.09
4,096	₹450.55	−₹0.02

Read the error column rather than the value column. From 4 steps onwards it halves every time the number of steps doubles. That is first-order convergence, and it means the sequence has a limit the values are closing in on — about ₹450.57.

Two features worth naming.

The first two rows are erratic. One step overshoots by ₹94; two steps undershoot by ₹44. With $d = 1/u$ and an at-the-money strike, the two-step tree puts a terminal node exactly at the strike, where the payoff is zero — so it throws away the entire middle of the distribution. Coarse lattices have artefacts of this kind, which is why nobody uses four-step trees for anything.

The convergence is to a specific number, and that number is the point. The source describes the lattice as approximating Black-Scholes-Merton prices "with a very rapid rate of convergence as the number of time steps grows." Chapter 6 is about what sits at the end of that column.

The CRR formula, which is the table in closed form

You do not have to walk a lattice node by node. Collecting all the paths that reach each terminal node gives the binomial pricing formula, which the source states as

$C_f(0) = \frac{1}{(1+r)^T}\sum_{x=0}^{T} \binom{T}{x} q^x (1-q)^{T-x}\, f\!\left(S_0(1+u)^x(1+d)^{T-x}\right)$

where $T$ is the number of steps and $f$ the payoff at maturity.

It is worth seeing what this is. $\binom{T}{x}q^x(1-q)^{T-x}$ is the binomial weight on ending with $x$ up-moves; $f(\cdot)$ is the payoff there; the sum is a weighted average; the leading factor discounts it. A discounted risk-neutral average of payoffs — chapter 3's formula, with more states.

And it explains the shape of the limit. A binomial distribution with many trials approaches a normal one. So the limiting price is a discounted average of payoffs over a normal distribution of $\ln S_T$ — which is to say a lognormal distribution of $S_T$. That observation is the whole of chapter 6.

What the tree can do that a formula cannot

The lattice is not merely a stepping stone to a closed form. It handles something the formula cannot, and chapter 1's European restriction is the reason.

At every node, an American option offers a choice: hold, or exercise now. So the backward step gains a comparison:

$f_i = \max\!\Big(\underbrace{e^{-r\Delta t}\left[q f_u + (1-q) f_d\right]}_{\text{hold}},\; \underbrace{\text{intrinsic value at this node}}_{\text{exercise}}\Big)$

One line added to the recursion, and American options are priced. No formula does this as simply, which is why trees remain in use.

Working the problem

A 26,000 put on an index at 24,000, thirty days, 6.5%, 14% volatility. European: ₹1,873.68. American: ₹2,000.00.

Where the ₹126.32 comes from — and notice what ₹2,000.00 is. It is exactly $26{,}000 - 24{,}000$, the intrinsic value. The American put is worth precisely what exercising it right now pays. The tree is telling you that immediate exercise is optimal, and the gap over the European price is the value of being allowed to do that.

Why holding is worse here. Compare the two courses of action.

Exercise now. You receive ₹2,000 today. Over the next thirty days at 6.5% that ₹2,000 earns about ₹10.70, and more to the point the whole ₹26,000 strike is in your hands earning interest rather than sitting in a contract.

Hold. You keep the chance that the index falls further, which would pay more. But a put's upside is capped — chapter 1 showed it cannot exceed $Ke^{-rT}$, and at 24,000 the index would have to collapse for much more to arrive. Meanwhile the discounting works against you: the European bound $Ke^{-rT} - S = ₹1{,}861.47$ is below the ₹2,000 you could have today.

So the deciding quantity is interest on the strike. For a deep in-the-money put, waiting means deferring receipt of a large, nearly certain sum. That is a cost, it grows with the rate and with how deep in the money the option is, and at some point it swamps the remaining time value.

What an American holder should do on the day they buy it: exercise. Which raises the obvious question of why they bought it, and the honest answer is that a position whose optimal action is immediate exercise should not have been opened — it is a long position in cash dressed as an option.

The asymmetry with calls, stated carefully. On a non-dividend-paying underlying, an American call should never be exercised early: exercising pays the strike sooner, which is a cost, and destroys the remaining time value, which is another. Chapter 1's bound $C \ge S - Ke^{-rT} > S - K$ is the compact version — the call is always worth more alive than exercised. Dividends can break that, and chapter 12 covers how Indian contracts handle them.

And the Indian relevance. Index options here are European, so none of this applies to them. It matters for stock options, where SEBI permits exchanges to introduce "Premium Settled American / European Style Stock Options" — so the exercise style is a contract term to check rather than assume.

The point

Subdivide the period, recalibrate $u$ and $d$ for the shorter step, and work backwards node by node: thirty days in two steps gives ₹406.59, and the error halves with every doubling of steps towards about ₹450.57. Two things change along the way. The replicating holding stops being fixed and must be rebalanced at every node, which makes the hedge a trading strategy with costs rather than a portfolio — and the recursion, with one $\max$ added, prices early exercise, which no closed formula does so easily. Collected into a single sum, the lattice is a discounted binomial average of payoffs, and a binomial average with many steps becomes a normal one.