From the tree to Black-Scholes

Chapter 5's error column was converging on about ₹450.57. This chapter writes down what is sitting there.

The limit, in words before symbols

The binomial price was a weighted average of payoffs, with binomial weights on the number of up-moves. Make the steps smaller and two things happen together.

The number of reachable prices grows. With $n$ steps there are $n+1$ terminal nodes; at a thousand steps the lattice is nearly a continuum.

The binomial weights become normal weights. A binomial distribution with many trials approaches a normal one — the ordinary central limit result from the Quantitative methods subject. Since the up-moves and down-moves add and subtract $\sigma\sqrt{\Delta t}$ in logs, what becomes normal is $\ln S_T$.

$\ln S_T \text{ normal} \quad\Longleftrightarrow\quad S_T \text{ lognormal}$

That is the limiting model, and it is what the source describes: the lognormal state-price density "can be recovered in principle by normal approximation of the binomial distribution." Derman and Kani state the assumption from the other direction — Black-Scholes "assumes that the index level executes a random walk with a constant volatility," and "if the Black-Scholes model is correct, then the index distribution at any options expiration is lognormal."

So the formula is not a separate theory from chapters 3 to 5. It is the same discounted risk-neutral average, taken in a limit where the average can be written out instead of summed.

The formula

For a European call on a non-dividend-paying underlying:

$C = S\,N(d_1) - Ke^{-rT} N(d_2)$

$d_1 = \frac{\ln(S/K) + \left(r + \tfrac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T}$

where $N(\cdot)$ is the cumulative standard normal distribution — the probability that a standard normal draw comes in below its argument.

And the put, which needs no separate derivation because chapter 2 already supplies it:

$P = C - S + Ke^{-rT} = Ke^{-rT}N(-d_2) - S\,N(-d_1)$

Five inputs: $S$, $K$, $T$, $r$, $\sigma$. Four of them are observable facts. The fifth is not, and the whole of chapters 10 and 11 is about that.

Reading the two terms

The formula is often presented as something to be computed rather than understood. It can be read.

$Ke^{-rT}N(d_2)$ — the present value of what you pay. You pay the strike $K$, but only if you exercise. $N(d_2)$ is the risk-neutral probability that the option finishes in the money, so this term is the strike, discounted, weighted by the chance of having to pay it. Chapter 3's warning applies in full: that probability is a pricing weight, not a forecast.

$S\,N(d_1)$ — the present value of what you receive. You receive the asset, but again only if you exercise. This term is the asset's current price weighted by $N(d_1)$, which is larger than $N(d_2)$ because the cases where you exercise are precisely the cases where the asset is worth more than average. $N(d_1)$ carries both the chance of exercising and the fact that the asset is dear when you do.

The difference is the option. Receive the asset, pay the strike, both contingent on the same event.

And $N(d_1)$ has a second meaning that chapter 8 takes up: it is the option's delta, the replicating holding of the underlying. The thing standing in front of $S$ in the price is the quantity of $S$ you must hold to hedge. That is not a coincidence; it falls out of differentiating the formula.

Where $\sigma^2/2$ comes from

The term $+\tfrac{1}{2}\sigma^2$ in $d_1$ looks arbitrary, and students reasonably ask why it is there.

It is the gap between an average return and an average price. If a price is lognormal, then the expected price grows at $\mu$ while the expected log return is $\mu - \tfrac{1}{2}\sigma^2$. The two differ because averaging multiplicative changes is not averaging their logs — the same arithmetic that makes the Measuring your return subject's CAGR fall short of an average annual return when returns are volatile.

So that half-variance is not an option-pricing artefact at all. It is volatility drag, appearing in a pricing formula because the formula is built on a multiplicative price process.

Checking the formula against the tree

If the derivation is right, the formula must sit at the end of chapter 5's column. Same contract: $S = K = 24{,}000$, $T = 30/365$, $r = 6.5\%$, $\sigma = 14\%$.

Steps in the tree	Value
64	₹449.07
256	₹450.20
1,024	₹450.48
4,096	₹450.55
Black-Scholes	₹450.57

The tree walks into the formula, which is the only evidence worth having that the two are the same model. If they disagreed, one of them would be wrong.

What the formula does and does not give you

It does give you a price that is consistent with the hedge. Every chapter so far has been one argument: an option is worth the cost of manufacturing it. The formula is the cost of manufacturing it when you can trade continuously.

It does give you sensitivities for free. Because it is a closed-form expression, its derivatives can be written down — and those derivatives are the Greeks of chapters 8 to 10. A tree gives you numbers; a formula gives you the shape of the dependence.

It does not give you a forecast. Chapter 3's result survives the limit: the underlying's expected return never enters. The source analyses the point directly and calls the irrelevance of the trend parameter a paradox.

It does not tell you volatility. Four inputs are read off a screen; $\sigma$ is the future standard deviation of returns over the life of the option, and nobody has it. Everything that can go wrong with Black-Scholes in practice enters through that one number — plus the assumptions behind the lognormal walk, which is chapter 7.

And it does not stop being useful when it is wrong. Derman and Kani describe how the market actually uses it: implied volatility is "a means of quoting prices." A formula can be a shared language for quoting without being a true description — a point worth holding onto before chapter 11 dismantles the constant-volatility assumption.

Working the problem

$S = 24{,}000$, $K = 24{,}000$, $T = 30/365 = 0.082192$, $r = 6.5\%$, $\sigma = 14\%$.

Step 1 — the volatility over the life of the option.

$\sigma\sqrt{T} = 0.14 \times \sqrt{0.082192} = 0.040137$

That number is worth pausing on. It says the index's standard deviation over the next thirty days is about 4.0% — roughly ±960 points. Everything else in the formula is arithmetic around this one quantity.

Step 2 — the drift term in the numerator.

$\left(r + \tfrac{1}{2}\sigma^2\right)T = \left(0.065 + 0.0098\right) \times 0.082192 = 0.0061480$

Step 3 — $d_1$ and $d_2$. With $S = K$, the $\ln(S/K)$ term is zero:

$d_1 = \frac{0 + 0.0061480}{0.040137} = 0.153175, \qquad d_2 = 0.153175 - 0.040137 = 0.113038$

Step 4 — the normal probabilities.

$N(d_1) = 0.560870, \qquad N(d_2) = 0.545000$

Step 5 — the two terms.

$S\,N(d_1) = 24{,}000 \times 0.560870 = ₹13{,}460.88$

$Ke^{-rT}N(d_2) = 23{,}872.12 \times 0.545000 = ₹13{,}010.30$

Step 6 — the price.

$C = 13{,}460.88 - 13{,}010.30 = ₹450.57$

What the two terms represent. The second is the discounted strike, weighted by the 54.5% risk-neutral chance of paying it. The first is the asset, weighted by 56.1% — the same event, but weighted to account for the asset being above average whenever that event occurs. They are subtracted because the option is an exchange: in the states where you exercise, you hand over the strike and receive the asset, and the price is the present value of that swap.

Two sanity checks worth running on any such calculation.

Against chapter 1's bound. $S - Ke^{-rT} = ₹127.88$, and ₹450.57 exceeds it. The price clears its arbitrage floor.

Against chapter 2's parity. The put comes to ₹322.70, and $C - P = 450.57 - 322.70 = ₹127.88$, matching $S - Ke^{-rT}$ exactly. Parity holds inside the formula, which it must, since the formula is a price.

And one observation about the at-the-money case. $N(d_2) = 54.5\%$, not 50%, even though the strike equals the current price. The index is expected to grow at the interest rate in the risk-neutral world, which tilts the exercise probability above a half. An at-the-money option is slightly more than a coin flip, and the tilt is interest, not optimism.

The point

Shrink the binomial steps to nothing and the weights become normal, so $\ln S_T$ is normal and $S_T$ lognormal, giving $C = S\,N(d_1) - Ke^{-rT}N(d_2)$ — the same discounted risk-neutral average chapters 3 to 5 built, written out instead of summed. The second term is the strike you pay weighted by the risk-neutral chance of paying it; the first is the asset you receive, weighted higher because it is worth more whenever you do receive it; and $N(d_1)$ doubles as the hedge ratio. The formula verifies against the tree at ₹450.57, respects chapter 1's bound and chapter 2's parity, and contains no forecast of the underlying — but it needs one number nobody can observe.