Finance-Theory · Unit 13 · Video 5 · Interactive Practice

Calibrating the Binomial Model and the One Principle Behind All Pricing

IKey Formulas

Formula	Name	What it says
$E[S_1] = p\,uS_0 + (1-p)\,dS_0$	First moment	Expected price tomorrow under the actual probability $p$
$\mathrm{Var}[S_1] = p(1-p)(u-d)^2 S_0^2$ , $\mathrm{SD}[S_1] = \sqrt{p(1-p)}\,(u-d)\,S_0$	Second moment	$u$ and $d$ enter the volatility only through the spread $u - d$
$p = \tfrac{1}{2}:\ \ \tfrac{1}{2}(u+d) = 1 + \mu,\ \ \tfrac{1}{2}(u-d) = \sigma$	Calibration	Match the historical mean $\mu$ and standard deviation $\sigma$ of the per-period change (as fractions of $S_0$ ); two targets fix $u$ and $d$
$C_0 = \dfrac{1}{r}\big[p^* C_u + (1-p^)\,C_d\big]$ , $p^ = \dfrac{r-d}{u-d}$	One-period price	$r$ is the gross riskless return; $C_0$ contains no $p$

Key Insight: One principle prices nearly every security: identify its cash flows, build a portfolio of instruments you can already price that produces the same cash flows, and the security must cost what that portfolio costs, or there is an arbitrage. The binomial call is one case: $C_0 = V_0 = S_0\Delta^* + B^*$ .

IIThe Grid of Possible Prices

With $n$ periods of up factor $u$ and down factor $d$ , the recombining tree ends in exactly $n + 1$ possible prices.

Periods n 4 Step per period

IIICalibrating u and d to History

Two historical targets, a mean and a standard deviation, pin down $u$ and $d$ once $p$ is fixed by convention.

History p u d

💡 Challenge: fit the same history with $p = \tfrac{1}{5}$ and then $p = \tfrac{4}{5}$ . Each $p$ fits the same two moments with a different pair $(u, d)$ ; the price formula then uses only $u$ , $d$ and $r$ , never $p$ , so what matters is getting $u$ and $d$ right.

IVWider Spread, Higher Call Value

Holding $S_0 = K = 100$ , $r = 1.03$ and $p = \tfrac{1}{2}$ fixed, widening $u - d$ raises volatility and the call's value together.

u − 1 = 1 − d 0.05

VQuiz Questions

Problem 1 · Volatility of Tomorrow's Price

Given: A stock trades at $50 today ( $S_0 = 50$ ). Tomorrow it moves up by $u = 1.10$ with actual probability $p = 0.6$ , or down by $d = 0.95$ with probability $0.4$ . Find the standard deviation of tomorrow's price $S_1$ .

✅ Correct!

\mathrm{SD}[S_1] = \sqrt{0.24}\,(0.15)(50) \approx 3.67

: the spread

u - d

scaled by

\sqrt{p(1-p)}

❌ Close, but that uses $p = \tfrac{1}{2}$ .

\tfrac{1}{2}(u-d)S_0

is the standard deviation only when

p = \tfrac{1}{2}

; here

\sqrt{p(1-p)} = \sqrt{0.24} \approx 0.49

, not

0.5

❌ That is the variance.

p(1-p)(u-d)^2S_0^2 = 13.5

is in squared dollars; take its square root.

❌ Not quite.

(u - d)S_0 = 7.5

is the gap between the two outcomes; the standard deviation multiplies it by

\sqrt{p(1-p)}

❌ Not quite. Use

\mathrm{Var}[S_1] = p(1-p)(u-d)^2S_0^2

and take the square root.

Show solution

The two outcomes are $uS_0 = 55$ and $dS_0 = 47.5$ , so the expected price is

E[S_1] = 0.6(55) + 0.4(47.5) = 33 + 19 = 52.

The variance simplifies to

\mathrm{Var}[S_1] = p(1-p)(u-d)^2S_0^2 = (0.6)(0.4)(0.15)^2(50)^2 = 0.24 \times 0.0225 \times 2500 = 13.5.

Check directly: $0.6(55-52)^2 + 0.4(47.5-52)^2 = 0.6(9) + 0.4(20.25) = 5.4 + 8.1 = 13.5$ . Then

\mathrm{SD}[S_1] = \sqrt{13.5} \approx 3.67,

so tomorrow's price has a standard deviation of about $3.67.

Problem 2 · Fit u and d

Given: Over past periods of the same length, the stock's price change averaged $+1\%$ per period with a standard deviation of $4\%$ per period. Fix $p = \tfrac{1}{2}$ by convention. Find the $u$ and $d$ that reproduce both moments.

✅ Correct! The midpoint

\tfrac{1}{2}(1.05 + 0.97) = 1.01

carries the drift and the half-spread

\tfrac{1}{2}(0.08) = 0.04

carries the volatility.

❌ The drift is missing. These give

E[S_1] = S_0

and a

4\%

standard deviation; history says the mean change is

+1\%

, so both factors must shift up by

0.01

❌ Close, but the spread is too narrow. With

p = \tfrac{1}{2}

the standard deviation is HALF the spread:

\tfrac{1}{2}(1.03 - 0.99) = 0.02

, only

2\%

. The spread must be

2\sigma = 0.08

❌ Not quite. Adding the drift to

u

and subtracting it from

d

widens the spread instead of shifting it: the mean is

\tfrac{1}{2}(1.05 + 0.95) = 1

, with no drift, and the standard deviation is

5\%

❌ Not quite. Solve

\tfrac{1}{2}(u + d) = 1.01

and

\tfrac{1}{2}(u - d) = 0.04

Show solution

With $p = \tfrac{1}{2}$ the moments per unit of $S_0$ are

\frac{E[S_1]}{S_0} = \tfrac{1}{2}(u + d), \qquad \frac{\mathrm{SD}[S_1]}{S_0} = \sqrt{\tfrac{1}{2}\cdot\tfrac{1}{2}}\,(u - d) = \tfrac{1}{2}(u - d).

Match them to history:

\tfrac{1}{2}(u + d) = 1.01, \qquad \tfrac{1}{2}(u - d) = 0.04.

Adding and subtracting: $u = 1.01 + 0.04 = 1.05$ and $d = 1.01 - 0.04 = 0.97$ .

Check: $\tfrac{1}{2}(1.05 + 0.97) = 1.01$ and $\tfrac{1}{2}(1.05 - 0.97) = 0.04$ . Two equations, two unknowns, once $p$ is fixed.

Problem 3 · Widen the Spread

Given: A stock at $100 and a one-period call struck at $100 ( $S_0 = K = 100$ ); gross riskless return $r = 1.02$ (discount by dividing by $r$ ); $p = \tfrac{1}{2}$ . In the narrow case $u = 1.04,\ d = 0.96$ ; in the wide case $u = 1.12,\ d = 0.88$ . The narrow-case call is worth about $2.94.

What is the wide-case call value $C_0$ ?

By what factor did the call value rise?

✅ Correct! The up-state payoff triples (

4 \to 12

) while

p^*

falls from

\tfrac{3}{4}

\tfrac{7}{12}

3 \times \tfrac{7}{9} = \tfrac{7}{3}

. More volatility, more call value.

❌ That uses the actual probability.

\tfrac{1}{2}(12)/1.02 \approx 5.88

weights by

p

; the price weights by

p^* = (1.02 - 0.88)/(1.12 - 0.88) = \tfrac{7}{12}

❌ Close, but undiscounted.

p^* C_u = \tfrac{7}{12}(12) = 7

is paid tomorrow; divide by

r = 1.02

❌ Not quite.

12/1.02

prices the up-state payoff as if it were certain; weight it by

p^* = \tfrac{7}{12}

❌ Only half the story.

C_u

triples, but the risk-neutral weight on the up state falls from

\tfrac{3}{4}

\tfrac{7}{12}

, a factor of

\tfrac{7}{9}

❌ Not quite.

\tfrac{7}{9}

is only the change in

p^*

; the up-state payoff also triples.

❌ Not quite.

E[S_1] = 100

in both cases, but the call's payoff is one-sided: the extra upside counts and the extra downside is floored at zero.

❌ Check the wide-case price.

C_u = 12

C_d = 0

p^* = (r - d)/(u - d)

, then divide by

r

❌ Check the factor. Divide the wide-case

C_0

by the narrow-case

C_0

Show solution

Narrow case: $uS_0 = 104$ , $dS_0 = 96$ , so $C_u = 4$ , $C_d = 0$ and

p^* = \frac{1.02 - 0.96}{1.04 - 0.96} = \frac{0.06}{0.08} = \frac{3}{4}, \qquad C_0 = \frac{\tfrac{3}{4}(4)}{1.02} = \frac{3}{1.02} \approx 2.94.

Wide case: $uS_0 = 112$ , $dS_0 = 88$ , so $C_u = 12$ , $C_d = 0$ and

p^* = \frac{1.02 - 0.88}{1.12 - 0.88} = \frac{0.14}{0.24} = \frac{7}{12}, \qquad C_0 = \frac{\tfrac{7}{12}(12)}{1.02} = \frac{7}{1.02} \approx 6.86.

Factor: $\dfrac{7/1.02}{3/1.02} = \dfrac{7}{3} \approx 2.33$ , the product of the payoff effect $12/4 = 3$ and the probability effect $\tfrac{7/12}{3/4} = \tfrac{7}{9}$ .

The standard deviation of $S_1$ went from $\tfrac{1}{2}(0.08)(100) = 4$ to $\tfrac{1}{2}(0.24)(100) = 12$ while the expected price stayed at 100.

Problem 4 · Same Principle, Different Payoff

Given (illustrative figures): A one-period European put struck at $100 pays $\max[K - S_1, 0]$ . The stock is at $100 ( $S_0 = K = 100$ ), $u = 1.25$ , $d = 0.85$ , gross riskless return $r = 1.01$ (discount by dividing by $r$ ). Price it by replication: find $\Delta$ shares and $B$ dollars of bonds whose payoff matches the put in both states.

How many shares does the replicating portfolio hold?

What is the put worth today?

✅ Correct! Short

0.375

of a share and lend about $46.41: the portfolio pays 0 up and 15 down, exactly like the put, so the put costs about $8.91.

❌ Check the sign. The put pays more when the stock falls, so the replicating portfolio is SHORT the stock:

\Delta = (P_u - P_d)/((u - d)S_0) = (0 - 15)/40

❌ Not quite. Divide the payoff gap by the gap in stock prices,

(u - d)S_0 = 125 - 85 = 40

, not by

S_0

alone.

❌ Not quite.

-15/0.4

forgets

S_0

: the stock prices differ by

(u - d)S_0 = 40

dollars, not by

0.4

❌ Wrong state. The put pays in the DOWN state, whose risk-neutral weight is

1 - p^* = 0.6

, not

p^* = 0.4

❌ Close, but undiscounted.

(1 - p^*)P_d = 0.6 \times 15 = 9

arrives tomorrow; divide by

r = 1.01

❌ That weights by $p = \tfrac{1}{2}$ .

\tfrac{1}{2}(15)/1.01 \approx 7.43

uses a guessed actual probability. No

p

was given because none is needed; the replicating portfolio (or

p^*

) sets the price.

❌ Check $\Delta$ .

\Delta = (P_u - P_d)/((u - d)S_0)

with

P_u = 0

P_d = 15

❌ Check the price.

V_0 = S_0\Delta + B

with

B = (uP_d - dP_u)/((u - d)r)

Show solution

Cash flows. $uS_0 = 125$ gives $P_u = \max[100 - 125, 0] = 0$ ; $dS_0 = 85$ gives $P_d = 15$ .

Replicate. Solve $125\Delta + 1.01B = 0$ and $85\Delta + 1.01B = 15$ :

\Delta = \frac{P_u - P_d}{(u - d)S_0} = \frac{0 - 15}{40} = -0.375, \qquad B = \frac{uP_d - dP_u}{(u - d)\,r} = \frac{1.25(15) - 0.85(0)}{0.4(1.01)} = \frac{18.75}{0.404} \approx 46.41.

Check: $1.01B = 18.75/0.4 = 46.875$ exactly, so up: $125(-0.375) + 46.875 = 0$ ; down: $85(-0.375) + 46.875 = 15$ .

Price = cost of the portfolio.

P_0 = S_0\Delta + B \approx -37.5 + 46.41 = 8.91.

Same answer with $p^* = (1.01 - 0.85)/(1.25 - 0.85) = 0.4$ : $P_0 = \tfrac{1}{1.01}\big[0.4(0) + 0.6(15)\big] = \tfrac{9}{1.01} \approx 8.91$ . Any other price is an arbitrage against the portfolio.

Solved: 0 / 4