Calibrating the tree

Chapter 3's two outcomes, 25,200 and 22,800, were invented. That is the only thing wrong with it, and fixing it is the whole of this chapter.

The general one-period result

First, write chapter 3's arithmetic in symbols so it can be reused. The underlying is $S$; after one period of length $\Delta t$ it is either $Su$ or $Sd$, with $u > 1 > d$. The derivative pays $f_u$ or $f_d$.

The replicating holding of the underlying is

$\Delta = \frac{f_u - f_d}{Su - Sd}$

and the value today is

$f = e^{-r\Delta t}\left[q f_u + (1-q) f_d\right], \qquad q = \frac{e^{r \Delta t} - d}{u - d}$

Three things to notice before moving on.

$\Delta$ is a ratio of spreads — how much the payoff moves divided by how much the underlying moves. That is the definition chapter 8 generalises into delta.

$q$ depends only on $u$, $d$, $r$ and $\Delta t$. It is manufactured from the tree, not supplied by a view.

The formula values any payoff — call, put, or anything else written on the two outcomes. Nothing in it is specific to options.

Where $u$ and $d$ have to come from

So the model needs $u$ and $d$. They cannot be invented, and they cannot come from a forecast of the level — chapter 3 showed the level forecast does not belong in a price.

They come from volatility. Volatility is the standard deviation of returns, and the Quantitative methods subject's treatment of dispersion is the relevant background. The standard choice is

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

Why that specific form, in three parts.

Why exponential. Prices are modelled multiplicatively, not additively — a move is a percentage, so the natural variable is $\ln S$. Working in logs, an up step adds $\sigma\sqrt{\Delta t}$ and a down step subtracts it. Derman and Kani describe exactly this geometry: the stock evolves "with constant logarithmic stock price spacing, corresponding to constant volatility."

Why $\sqrt{\Delta t}$. Variance accumulates with time, so standard deviation accumulates with the square root of time. This is the same $\sqrt{t}$ that the Measuring your return subject uses to annualise a daily standard deviation, applied in reverse to split an annual volatility across a short step.

Why $d = 1/u$. It makes the tree symmetric in logs, so that an up followed by a down returns exactly to the start. That keeps the lattice tidy when chapter 5 adds steps, and it is a convention rather than a necessity.

Pricing a real contract, once

Now a contract with Indian dimensions rather than invented ones. An index at 24,000, a 24,000 strike, thirty days to expiry, 6.5% interest, and 14% annualised volatility.

Step 1 — the step size. $\Delta t = T = 30/365 = 0.082192$ years, and $\sqrt{\Delta t} = 0.286691$.

Step 2 — the factors.

$u = e^{0.14 \times 0.286691} = 1.040953, \qquad d = 0.960658$

Step 3 — the two outcomes.

$Su = 24{,}982.87, \qquad Sd = 23{,}055.79$

Step 4 — the payoffs. The call pays $24{,}982.87 - 24{,}000 = ₹982.87$ if the index rises, and ₹0 if it falls.

Step 5 — the risk-neutral weight.

$q = \frac{e^{0.065 \times 0.082192} - 0.960658}{1.040953 - 0.960658} = 0.556681$

Step 6 — the value.

$C = e^{-0.065 \times 0.082192}\left[0.556681 \times 982.87\right] = ₹544.23$

And the replication, to confirm it is the same thing. $\Delta = 982.87 / 1{,}927.08 = 0.510033$ units of the index, funded by ₹11,696.56 of borrowing: $0.510033 \times 24{,}000 - 11{,}696.56 = ₹544.23$.

The put, from the same tree. It pays ₹0 up and ₹944.21 down, so

$P = e^{-r\Delta t}\left[0.443319 \times 944.21\right] = ₹416.35$

Chapter 2's test, as a check on the arithmetic. $C - P = 544.23 - 416.35 = ₹127.88$, and $S - Ke^{-rT} = ₹127.88$. Parity holds exactly — as it must, since both prices came from the same two states.

Now the bad news

₹544.23 is wrong. Not approximately right — wrong by about 21%, as chapter 5 will show the correct figure to be about ₹450.

It is worth being precise about why, because the reason is not the volatility or the rate.

A one-step tree says the index does exactly one thing. It ends at 24,982.87 or 23,055.79, and nothing else is possible. The call therefore pays either ₹982.87 or ₹0.

But the real contract pays every value in between. An index ending at 24,100 pays ₹100; one ending at 24,900 pays ₹900. The one-step model has assigned those outcomes zero probability and replaced them with a coarse pair.

So the model's error is not in its inputs but in its state space. It has the right volatility and the wrong world. And notice the direction of the error: lumping all the upside into a single large outcome overstates the value of a kinked payoff, which is why ₹544 is too high rather than too low.

This is the useful failure in the subject. Everything in chapters 5 and 6 is the single repair of adding states — and the repair converges, which is the subject's central technical fact.

Working the problem

Hold everything fixed and change only $\sigma$.

$\sigma$	$u$	Up outcome	Down outcome	$q$	Call value
10%	1.0291	24,698	23,322	0.5862	₹407.03
14%	1.0410	24,983	23,056	0.5567	₹544.23
20%	1.0590	25,416	22,663	0.5324	₹749.98
30%	1.0898	26,156	22,022	0.5096	₹1,092.64

What I trust in this table — the direction and roughly the proportionality. More volatility means a wider pair of outcomes, a larger payoff in the up state, and a higher value. The relationship is close to linear in $\sigma$ over this range: trebling volatility from 10% to 30% multiplies the price by about 2.7. Chapter 10 shows that near-linearity is a real property of at-the-money options, not an artefact of the coarse tree.

Notice also what happens to $q$. It falls as volatility rises, from 0.586 to 0.510. The risk-neutral weight on the up state goes down while the option gets dearer — which would be incoherent if $q$ were a probability forecast. It is not. It is the number that keeps the underlying's risk-neutral growth equal to the interest rate, and as the outcomes spread apart, less weight on the higher one is needed to hold that average fixed. If you ever find yourself reasoning from $q$ as a likelihood, this row is the correction.

What I do not trust — the levels. Every figure in that column is too high, for the state-space reason above. The ₹544.23 at 14% should be about ₹450.57. The errors are large and they are systematic, so the table is useful for showing how the price responds and useless for telling you what to pay.

And the general lesson about models. A model can be right about comparative statics and wrong about levels, and the two kinds of trust are separate. Reporting the ₹544.23 as a price would be a mistake; reporting "volatility is the dominant input and the price rises roughly in proportion to it" would not be.

The point

In one period, $\Delta = (f_u - f_d)/(Su - Sd)$ and $f = e^{-r\Delta t}[qf_u + (1-q)f_d]$ with $q = (e^{r\Delta t} - d)/(u-d)$ — and the only inputs that are not observable are $u$ and $d$, which come from volatility as $u = e^{\sigma\sqrt{\Delta t}}$ and $d = 1/u$ because variance accumulates with time and prices move multiplicatively. Applied once to a thirty-day at-the-money index call at 14% volatility, that gives ₹544.23, which is about 21% too high — not because any input is wrong but because two outcomes is not enough of a world for a payoff with a kink in it. The repair is more states, and it converges.