Finance-Theory · Unit 13 · Video 4 · Interactive Practice

From Binomial Tree to Black-Scholes: Multiperiod Pricing and the Continuous-Time Limit

IKey Formulas

Formula	Name	What it says
$C_0 = \dfrac{1}{r^2}\Big[p^{2} C_{uu} + 2p^(1-p^)\,C_{ud} + (1-p^)^2 C_{dd}\Big]$	Two-period call	The one-period formula applied at every node; the 2 counts the two paths, up-down and down-up, that meet at $udS_0$ ; the true probability $p$ never appears
$C_0 = \dfrac{1}{r^n}\displaystyle\sum_{k=0}^{n}\binom{n}{k}p^{k}(1-p^)^{n-k}\,\mathrm{Max}\big[0,\ u^k d^{n-k} S_0 - K\big]$	$n$ -period binomial formula	$\binom{n}{k}$ paths with $k$ up moves, each with risk-neutral probability $p^{k}(1-p^)^{n-k}$ , all paying $\mathrm{Max}[0, u^k d^{n-k}S_0 - K]$ ; the sum is discounted over $n$ periods, with $p^* = \frac{r-d}{u-d}$ and $r$ the gross riskless return per period
$\tfrac{1}{2}\sigma^2 S^2 \dfrac{\partial^2 C}{\partial S^2} + rS\dfrac{\partial C}{\partial S} + \dfrac{\partial C}{\partial t} = rC$	Black-Scholes-Merton PDE (the $n \to \infty$ limit)	Boundary conditions $C(S,T) = \mathrm{Max}[S-K, 0]$ and $C(0,t) = 0$ ; here $r$ is the continuously compounded rate and $\sigma$ , the volatility, is what remains of $u$ and $d$

Key Insight: Nothing in the binomial formula fixes the length of a period, so the time to expiry can be cut as finely as needed; in the limit the replication condition $C_0 = V_0$ becomes the Black-Scholes PDE, and neither $p$ nor the stock's expected return ever enters. All of it rests on independent, identically distributed coin tosses.

IIRolling Back a Two-Period Tree

The one-period formula applied node by node, backward from expiry, gives the same $C_0$ as the weights $p^{*2}$ , $2p^*(1-p^*)$ , $(1-p^*)^2$ .

Step 1 Strike K 95

IIICounting Paths: the $n$ -Period Formula

$2^n$ paths finish at only $n+1$ prices; each price contributes paths × one path's risk-neutral probability × payoff.

Periods n 8

IVThe Continuous-Time Limit

Holding the horizon fixed while cutting it into ever more periods, the binomial price converges to the Black-Scholes price.

Periods n 16

💡 The dashed line, $\approx 10.45$ , is the value the Black-Scholes-Merton PDE gives for these inputs (illustrative: volatility $\sigma$ of 20% a year, a riskless rate of 5% a year continuously compounded). At every $n$ the tree's per-period $u$ , $d$ and gross $r$ are chosen to match that $\sigma$ and that rate over each, ever shorter, period, which keeps the spread of final prices from blowing up or collapsing. How $u$ and $d$ are chosen from data is the subject of the next video.

VQuiz Questions

Problem 1 · A Two-Period Call

Given: A stock trades at $50. Each period it moves up by $u = 1.2$ or down by $d = 0.9$ , and the gross riskless return per period is $r = 1.02$ (discount by dividing by $r$ ). Investors believe an up move has probability $p = 0.65$ . Find the value today of a European call with strike $50 expiring in two periods.

✅ Correct! With

p^* = 0.4

the weights are

0.16

0.48

0.36

; the belief

p = 0.65

plays no part in the price.

❌ Missing the coefficient 2. Two paths, up-down and down-up, reach

udS_0

, so the middle payoff carries weight

2p^*(1-p^*) = 0.48

, not

0.24

❌ That uses the true probability. Replication prices the call with

p^* = \frac{r-d}{u-d}

; investors can disagree about

p

and still agree on the price.

❌ Discounted only once. The payoff arrives two periods from now, so the expected payoff is divided by

r^2 = 1.0404

, not by

r

❌ Not quite. Find

p^*

, the three end prices and their payoffs, then weight them

p^{*2}

2p^*(1-p^*)

(1-p^*)^2

and divide by

r^2

Show solution

The risk-neutral probability is

p^* = \frac{r-d}{u-d} = \frac{1.02 - 0.9}{1.2 - 0.9} = \frac{0.12}{0.3} = 0.4, \qquad 1 - p^* = 0.6

After two periods the stock (in dollars) is at $u^2S_0 = 72$ , $udS_0 = 54$ or $d^2S_0 = 40.5$ , so the call pays $C_{uu} = 22$ , $C_{ud} = 4$ , $C_{dd} = 0$ . Then

C_0 = \frac{1}{1.02^2}\Big[(0.4)^2(22) + 2(0.4)(0.6)(4) + (0.6)^2(0)\Big] = \frac{3.52 + 1.92}{1.0404} = \frac{5.44}{1.0404} \approx 5.23

The call is worth about $5.23. The probability $p = 0.65$ was never used: it does not appear in the two-period formula.

Problem 2 · Paths, Periods and End Prices

Given: Traders agree that over the next 5 hours (300 minutes) a stock has $1{,}024$ possible price paths in a recombining binomial tree with equal periods. Find the number and length of the periods, the number of distinct prices at expiry, and the number of paths that end at the price reached with exactly 3 up moves.

Periods

Distinct prices at expiry

Paths ending at the price with exactly 3 up moves

✅ Correct!

2^{10} = 1{,}024

gives 10 periods of 30 minutes; the tree recombines onto

n + 1 = 11

end prices, and

\binom{10}{3} = 120

of the paths end at

u^3d^7S_0

❌ Each step doubles the number of paths. It does not add two:

n

steps give

2^n

paths, so solve

2^n = 1{,}024

❌ Not a square root.

32^2 = 1{,}024

, but the path count is

2^n

, not

n^2

❌ One period per path overcounts. Every period doubles the number of paths, so

n

periods already give

2^n

❌ That counts paths, not prices. An up move then a down move lands where a down then an up lands, so paths with the same number of ups share an end price.

❌ Off by one. The number of up moves

k

runs from

0

n

, which is

n + 1

values.

❌ Many orders reach the same node. Any arrangement of 3 ups among the 10 moves ends at

u^3d^7S_0

❌ Not a product. Choosing which 3 of the 10 moves are ups is

\binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}

❌ Paths are not spread evenly. The middle end prices collect far more paths than the extremes; each end price gets

\binom{10}{k}

of them.

❌ Check the periods. Solve

2^n = 1{,}024

, then divide 300 minutes by

n

❌ Check the end prices. In a recombining tree the end price depends only on the number of up moves

k

❌ Check the path count. Count the ways to place 3 up moves among 10 periods.

Show solution

Periods. Each period doubles the number of paths, so $n$ periods give $2^n$ paths and $2^n = 1{,}024 = 2^{10}$ means $n = 10$ . Each period lasts $300/10 = 30$ minutes.

End prices. A path with $k$ up moves ends at $u^k d^{10-k} S_0$ whatever the order of the moves, so the end prices are indexed by $k = 0, 1, \dots, 10$ : that is $n + 1 = 11$ distinct prices.

Paths to the 3-up price. The number of ways to choose which 3 of the 10 moves are up moves is

\binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120

Check: summing $\binom{10}{k}$ over all $k$ gives $2^{10} = 1{,}024$ paths.

Problem 3 · Three Periods, Two Routes

Given: A stock trades at $100; each period it moves up by $u = 1.2$ or down by $d = 0.9$ , and the gross riskless return per period is $r = 1.05$ (discount by dividing by $r$ ). A European call has strike $100 and expires in three periods. Find the call's value $C_u$ at the up node after one period, and its value $C_0$ today.

Value at the up node after one period, $C_u$

Value today, $C_0$

✅ Correct! Backward induction through

C_u \approx 29.93

and

C_d \approx 6.71

and the three-period formula with weights

1, 3, 3, 1

both give

C_0 \approx 17.45

❌ Discounted only once. From the up node the call still has two periods to run, so its expected payoff is divided by

r^2

❌ Missing the coefficient 2. From

uS_0

, two paths reach

u^2dS_0

, so that payoff is weighted

2p^*(1-p^*) = 0.5

❌ That is the payoff from exercising now. A European call cannot be exercised at the up node; its value there is the discounted risk-neutral expectation of the payoffs two periods later.

❌ The binomial coefficients are missing. Three paths end at

u^2dS_0

, so its payoff carries

\binom{3}{2}p^{*2}(1-p^*)

, not

p^{*2}(1-p^*)

❌ Discounted only once. The payoff arrives three periods from now; divide by

r^3 = 1.157625

❌ One in-the-money price is missing. Besides

u^3S_0

, the price

u^2dS_0

is above the strike too and adds its term.

❌ Check $C_u$ . From

uS_0

the call faces a two-period tree: weight its three end payoffs and divide by

r^2

❌ Check $C_0$ . Use the

n = 3

formula, or apply the one-period formula to

C_u

and

C_d

Show solution

First $p^* = \frac{1.05 - 0.9}{1.2 - 0.9} = \frac{0.15}{0.3} = 0.5$ . The end prices (in dollars) are $u^3S_0 = 172.8$ , $u^2dS_0 = 129.6$ , $ud^2S_0 = 97.2$ and $d^3S_0 = 72.9$ , so the payoffs are $72.8$ , $29.6$ , $0$ and $0$ .

$C_u$ . From $uS_0 = 120$ two periods remain, with end payoffs $72.8$ , $29.6$ , $0$ :

C_u = \frac{1}{1.05^2}\Big[(0.5)^2(72.8) + 2(0.5)(0.5)(29.6) + (0.5)^2(0)\Big] = \frac{18.2 + 14.8}{1.1025} = \frac{33}{1.1025} \approx 29.93

$C_0$ from the $n$ -period formula with $n = 3$ (only $k = 3$ and $k = 2$ are in the money):

C_0 = \frac{1}{1.05^3}\Big[\tbinom{3}{3}(0.5)^3(72.8) + \tbinom{3}{2}(0.5)^3(29.6)\Big] = \frac{9.1 + 11.1}{1.157625} = \frac{20.2}{1.157625} \approx 17.45

Check by rolling back. From $dS_0 = 90$ : $C_d = \frac{1}{1.1025}\big[(0.5)^2(29.6)\big] = \frac{7.4}{1.1025} \approx 6.71$ , and then $C_0 = \frac{1}{1.05}\big[0.5(29.93) + 0.5(6.71)\big] \approx 17.45$ . The call is worth about $17.45 today.

Problem 4 · What the Limit Means

Given: An option's time to expiry is held fixed while the number of binomial periods $n \to \infty$ , with $u$ and $d$ both closing in on 1, and the binomial price converges to the Black-Scholes price. Which statement about this limit is correct?

✅ Correct! The equation is the arbitrage condition written in continuous time: just as

p

never entered the tree's price, the expected return never enters the PDE.

❌ That $r$ is the riskless rate. The stock's expected return appears nowhere in the equation, just as the true probability

p

never entered the tree's price.

❌ The two $r$ 's differ. In the tree

r

is a gross return per period (one plus the one-period interest rate); in the PDE

r

is the continuously compounded rate.

❌ The other way round. With

u

and

d

fixed, more and more periods would make the spread of final prices blow up; they must shrink toward 1 at the rate that keeps the spread realistic.

Show solution

In one period the price is pinned by replication: $\Delta^*$ shares and $B^*$ dollars of bonds match the call in both states, and the call must cost what that portfolio costs, $C_0 = V_0 = S_0\Delta^* + B^*$ . Repeating that condition over infinitely many, infinitely short periods turns it into the PDE

\tfrac{1}{2}\sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} + rS\frac{\partial C}{\partial S} + \frac{\partial C}{\partial t} = rC

Its coefficients are the volatility $\sigma$ (what remains of $u$ and $d$ ) and the continuously compounded riskless rate $r$ ; no expected return appears, just as $p$ is absent from the binomial formula. Option (c) confuses the tree's gross per-period $r$ with this continuously compounded rate, and (d) has the limit backward: the moves must shrink toward 1, or the spread of final prices blows up as $n$ grows.

Solved: 0 / 4

From Binomial Tree to Black-Scholes: Multiperiod Pricing and the Continuous-Time Limit

IKey Formulas

IIRolling Back a Two-Period Tree

IIICounting Paths: the nnn-Period Formula

IVThe Continuous-Time Limit

VQuiz Questions

IIICounting Paths: the $n$ -Period Formula