Finance-Theory · Unit 13 · Video 1 · Interactive Practice

The Binomial Option Pricing Model: A One-Period Coin-Flip World

IKey Formulas

Formula	Name	What it says
$S_1 = uS_0$ with prob. $p$ $S_1 = dS_0$ with prob. $1-p$ $u > d$	One-period stock move	Tomorrow's price is a Bernoulli trial; $u$ and $d$ are gross factors, so a 10% rise is $u = 1.1$ and a 10% fall is $d = 0.9$ . Calling the larger factor $u$ is only a labelling convention.
$C_u = \text{Max}[uS_0 - K,\ 0]$ $C_d = \text{Max}[dS_0 - K,\ 0]$	Call payoffs tomorrow	The call expires tomorrow, so its value then is its payoff; each payoff arrives with the probability of the stock move that produces it, $p$ or $1-p$ .
$\dfrac{E[S_1]}{S_0} = pu + (1-p)d$	Expected gross return	For a growth stock with $u = 1.1$ and $d = 0.99$ this is $0.99 + 0.11p$ , above 1 whenever $p > \tfrac{1}{11}$ .
$C_0 = f(S_0,\ K,\ u,\ d,\ p,\ r)$	Today's call price	Six candidate inputs; $r$ is the gross riskless rate from today to tomorrow, $r = 1 + \text{net rate}$ . One of the six turns out to be redundant.

Key Insight: If the stock can take only two values tomorrow, so can the option. The probabilities $p$ and $1-p$ must sum to one, but nothing forces $u$ and $d$ to sum to anything, or even to sit on either side of 1.

IITwo Stock Prices, Two Call Payoffs

Tomorrow's two stock prices, $uS_0$ and $dS_0$ , fix the call's two payoffs, $C_u$ and $C_d$ .

IIIExpected Gross Return of the Coin Flip

Weighting $u$ and $d$ by $p$ and $1-p$ gives the stock's expected gross return, and so its drift.

Stock

IVQuiz Questions

Problem 1 · The Call's Two Payoffs

Given: A stock trades at $30 today and tomorrow is worth $uS_0$ or $dS_0$ with $u = 1.2$ and $d = 0.8$ . A call on it with strike $27 expires tomorrow. Find the call's payoffs $C_u$ and $C_d$ .

✅ Correct! The call is in the money only in the up state: it pays $9 there and nothing in the down state.

❌ Those are the stock's two prices. The call pays the stock price minus the strike, and only when that is positive.

❌ Close, but a call payoff is never negative. In the down state the stock is worth $24, below the $27 strike, so the holder lets the call expire:

C_d = \text{Max}[dS_0 - K,\ 0] = 0

❌ Not quite. $3 is today's price minus the strike; the payoff uses tomorrow's price

uS_0

, not

S_0

❌ Not quite. Compute

uS_0

and

dS_0

first, then apply

\text{Max}[S_1 - K,\ 0]

to each.

Show solution

Tomorrow's two stock prices: $uS_0 = 1.2 \times 30 = 36$ and $dS_0 = 0.8 \times 30 = 24$ dollars.

C_u = \text{Max}[36 - 27,\ 0] = 9, \qquad C_d = \text{Max}[24 - 27,\ 0] = \text{Max}[-3,\ 0] = 0

So the call pays $9 if the stock rises and $0 if it falls.

Problem 2 · Gross Factors, Not Net Rates

Given: A stock at $120 today will either rise 35% or fall 15% by tomorrow. A call on it with strike $90 expires tomorrow. Find the factors $u$ , $d$ and the down-state payoff $C_d$ .

What are u and d?

What is the down-state payoff?

✅ Correct! A 15% fall multiplies the price by

d = 0.85

, leaving $102, still $12 above the strike.

❌ Those are net rates of change. The model multiplies today's price by a gross factor:

u = 1 + 0.35

and

d = 1 - 0.15

❌ Close, but $d$ is a gross factor too. A 15% fall keeps 85% of the price, so

d = 1 - 0.15 = 0.85

, not

0.15

❌ The labels are reversed. By convention

u > d

: the larger factor, 1.35, is

u

❌ Check $d$ . With

d = 0.15

the stock would drop to $18 and the call would pay nothing, but a 15% fall leaves

dS_0 = 0.85 \times 120 = 102

dollars.

❌ Not quite. $30 is today's price minus the strike; the down-state payoff uses

dS_0

❌ That is $C_u$ . In the up state the stock is worth $162 and the call pays $72; the down state has the stock at $102.

❌ Check the factors. A rise of

x

percent gives

u = 1 + x/100

; a fall of

y

percent gives

d = 1 - y/100

❌ Check the payoff. Apply

\text{Max}[dS_0 - K,\ 0]

with the gross factor

d

Show solution

Factors: $u = 1 + 0.35 = 1.35$ and $d = 1 - 0.15 = 0.85$ ; $u > d$ as the convention requires.

Stock tomorrow: $uS_0 = 1.35 \times 120 = 162$ and $dS_0 = 0.85 \times 120 = 102$ dollars.

C_d = \text{Max}[102 - 90,\ 0] = 12, \qquad C_u = \text{Max}[162 - 90,\ 0] = 72

The call is in the money in both states: it pays $12 after a fall and $72 after a rise.

Problem 3 · Expected Return and Drift

Given: A stock's gross factors are $u = 1.15$ and $d = 0.95$ , and the probability of a rise is $p = 0.4$ . Find the expected gross return $E[S_1]/S_0$ , and the threshold that $p$ must exceed for the stock to drift upward.

What is the expected gross return?

The stock drifts upward whenever p exceeds

✅ Correct!

E[S_1]/S_0 = 0.95 + 0.20p

is 1.03 at

p = 0.4

, an expected rise of 3%, and it exceeds 1 exactly when

p > \tfrac{1}{4}

❌ That is the plain average of $u$ and $d$ . It would be right only for a fair coin; here the rise has probability 0.4, not 0.5.

❌ The probabilities are swapped.

p = 0.4

belongs to the rise

u

, and

1 - p = 0.6

to the fall

d

❌ That is only the up-state term.

pu = 0.46

leaves out the fall; add

(1-p)d = 0.6 \times 0.95 = 0.57

❌ Not quite. The factors need not be symmetric around 1: the fall here is only 5% against a 15% rise, so a smaller

p

already tips the balance.

❌ Close, but that is $(u-1)/(u-d)$ . Solve

0.95 + 0.20p > 1

for

p

: the threshold is

(1-d)/(u-d)

❌ Not quite.

1 - d = 0.05

still has to be divided by

u - d = 0.20

❌ Check the expectation. Weight each gross factor by its probability:

pu + (1-p)d

❌ Check the threshold. Set

d + (u-d)p = 1

and solve for

p

Show solution

Expected gross return:

\frac{E[S_1]}{S_0} = pu + (1-p)d = 0.4 \times 1.15 + 0.6 \times 0.95 = 0.46 + 0.57 = 1.03

an expected rise of 3%.

Drift threshold: rewrite the expectation as $d + (u - d)p = 0.95 + 0.20p$ . It exceeds 1 when

0.20p > 0.05 \quad\Longleftrightarrow\quad p > \frac{1-d}{u-d} = \frac{0.05}{0.20} = \frac{1}{4}

With $p = 0.4 > \tfrac{1}{4}$ the stock drifts upward, consistent with the 3% found above.

Problem 4 · Same Tree, Different Payoff

Given: A stock at $60 today moves to $uS_0$ or $dS_0$ tomorrow with $u = 1.15$ and $d = 0.75$ . A put with strike $63 expires tomorrow and pays $\text{Max}[K - S_1,\ 0]$ . Find its payoffs $P_u$ and $P_d$ in the up and down states.

✅ Correct! The same two stock prices, $69 and $45, feed a different payoff rule: the put pays only in the down state.

❌ Those are the call's payoffs. A call with this strike pays $6 when the stock rises to $69; the put pays when the stock ends below the strike.

❌ Close, but a put payoff is never negative either. With the stock at $69, above the $63 strike, the put expires unused:

P_u = 0

❌ Not quite. $3 is the strike minus today's price; the down-state payoff uses

dS_0 = 45

dollars.

❌ Not quite. Compute

uS_0

and

dS_0

, then apply

\text{Max}[K - S_1,\ 0]

to each.

Show solution

Tomorrow's stock prices: $uS_0 = 1.15 \times 60 = 69$ and $dS_0 = 0.75 \times 60 = 45$ dollars.

P_u = \text{Max}[63 - 69,\ 0] = 0, \qquad P_d = \text{Max}[63 - 45,\ 0] = 18

Two stock values tomorrow again mean two option values tomorrow, each arriving with the probability of the stock move behind it: $0 with probability $p$ , $18 with probability $1 - p$ .

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