Pricing by replication
Introduction to Financial Options / Lecture 04
What does it cost to manufacture an option payoff?
From parity to pricing
Introduction to Financial Options / Lecture 04
Lecture 03 showed that matching payoffs must have matching prices.
Last time
A call minus a put is a financed share.
Today
Can stock and cash reproduce a call by themselves?
If we can build it, we can price it.
One share. Two possible tomorrows.
Introduction to Financial Options / Lecture 04
Today the share costs $100. At expiry it will be either $80 or $120.
Down state
S_d=\$80
Up state
S_u=\$120
For now, interest is zero. The two outcomes are a model, not a forecast.
What must a call deliver?
Introduction to Financial Options / Lecture 04
Take a European call with strike $100.
Share ends at $80
Call payoff $0
Share ends at $120
Call payoff $20
C_T=(S_T-100)^+
Our building materials
Introduction to Financial Options / Lecture 04
Stock
Hold \Delta shares. Their expiry value is \Delta S_T.
Cash
Hold B dollars today. At zero interest it is still B at expiry.
\text{Stock + cash at expiry}=\Delta S_T+B
Choose one \Delta and one B that work in both states.
Two states give two equations
Introduction to Financial Options / Lecture 04
Down state
80\Delta+B=0
Up state
120\Delta+B=20
What stock position makes the difference between the two portfolio values equal $20?
Subtract to find the hedge
Introduction to Financial Options / Lecture 04
\Delta=\frac{C_u-C_d}{S_u-S_d}=\frac{20-0}{120-80}=\tfrac12
Half a share gains $20 when the whole share rises $40.
\Delta is the stock hedge ratio: option payoff change per dollar of stock change.
Then find the cash leg
Introduction to Financial Options / Lecture 04
80(\tfrac12)+B=0\quad\Longrightarrow\quad B=-40
Down
\tfrac12(80)-40=0
Up
\tfrac12(120)-40=20
The call is replicated by half a share and $40 of borrowing.
The option price follows
Introduction to Financial Options / Lecture 04
We can assemble the call’s payoff today for one known cost.
C_0=\Delta S_0+B=\tfrac12(100)-40=\boxed{\$10}
Any other call price would create two portfolios with identical future payoffs but different prices today.
Move the strike. Rebuild the hedge.
Introduction to Financial Options / Lecture 04
Keep S_0=\$100 and equally spaced up/down states. Change the strike or the size of the move.
Zero rates. The two states have equal pricing weights because their average is $100.
Spot the arbitrage
Introduction to Financial Options / Lecture 04
The call trades at $9. Its replicating portfolio costs $10. What trade pays you today and has no expiry risk?
Buy cheap
Buy the call for $9.
Sell expensive
Short the replicating portfolio for $10.
Collect $1 now. At expiry the two payoffs cancel in both states.
The probability formula drops out
Introduction to Financial Options / Lecture 04
C_0=\frac{S_0-S_d}{S_u-S_d}C_u+\frac{S_u-S_0}{S_u-S_d}C_d
Here the weights are \tfrac12 and \tfrac12, so C_0=\tfrac12(20)+\tfrac12(0)=\$10.
The weights come from replication. They are not an estimate of how often the stock really rises.
A pricing probability is not a forecast
Introduction to Financial Options / Lecture 04
Real-world probability
How often do we actually expect the up state?
Risk-neutral probability
Which weight makes the discounted stock price a fair average?
S_0=qS_u+(1-q)S_d\qquad(r=0)
The option price is fixed by trading, even if investors disagree about the first probability.
Two probabilities. One option price.
Introduction to Financial Options / Lecture 04
S_0=\$100, S_d=\$80, S_u=\$120, K=\$100, and r=0. The call pays $20 only in the up state.
P · real-world forecast
Suppose P(\text{up})=70\%.
\mathbb E^P[C_T]=0.70(20)=\$14
Expected payoff, not the replication price.
Q · pricing weights
Replication gives Q(\text{up})=50\%.
C_0=\mathbb E^Q[C_T]=0.50(20)=\$10
The no-arbitrage price at zero rates.
P asks what may happen. Q weights the same outcomes to price traded payoffs. A bullish forecast does not change the hedge cost.
Bring interest back
Introduction to Financial Options / Lecture 04
q=\frac{S_0e^{rT}-S_d}{S_u-S_d},\qquad C_0=e^{-rT}\big[qC_u+(1-q)C_d\big]
The cash leg grows at the risk-free rate. The pricing weight makes the expected stock value grow at that rate too.
We assume no dividends, frictionless stock trading, and one borrowing/lending rate.
When does the model avoid arbitrage?
Introduction to Financial Options / Lecture 04
S_d<S_0e^{rT}<S_u\quad\Longleftrightarrow\quad 0<q<1
If the risk-free growth of the share’s starting price lies outside the two outcomes, stock and cash already offer an arbitrage.
The familiar “probability” range is a consequence of a no-arbitrage condition.
Price the matching put
Introduction to Financial Options / Lecture 04
Return to S_d=\$80, S_u=\$120, K=\$100, and zero rates.
Down
Put payoff $20
Up
Put payoff $0
\Delta_P=\frac{0-20}{120-80}=-\tfrac12,\qquad P_0=\$10
Parity survives the tree
Introduction to Financial Options / Lecture 04
C_0-P_0=S_0-Ke^{-rT}
In our zero-rate, at-the-money tree: \$10-\$10=\$100-\$100.
Binomial pricing and put-call parity are two uses of the same rule: equal future payoffs, equal current prices.
One step is not enough for time
Introduction to Financial Options / Lecture 04
Now allow a move today to tomorrow, then another move tomorrow to expiry.
Each step multiplies the share by 1.2 or 0.8. The middle paths meet at 96.
Start at expiry and work backward
Introduction to Financial Options / Lecture 04
For a K=\$100 call, terminal payoffs at 144 / 96 / 64 are 44 / 0 / 0.
At the 120 node
Next values are 44 and 0. Call value: 22.
At the 80 node
Next values are 0 and 0. Call value: 0.
C_0=\tfrac12(22)+\tfrac12(0)=\boxed{\$11}
The hedge changes after a move
Introduction to Financial Options / Lecture 04
At 100, hedge the first step
\Delta_0=\frac{22-0}{120-80}=0.55
After an up move to 120
\Delta_u=\frac{44-0}{144-96}\approx0.917
There is no single buy-and-hold stock position for the whole life of this option.
More steps, smaller moves
Introduction to Financial Options / Lecture 04
In a standard binomial tree, a short step of length \Delta t uses
u=e^{\sigma\sqrt{\Delta t}},\qquad d=e^{-\sigma\sqrt{\Delta t}}
Shorter time step
Smaller typical price moves.
More hedge dates
The portfolio can be adjusted more often.
This is a new tree parameterization, not the fixed 1.2 / 0.8 example above.
Trees can handle early exercise
Introduction to Financial Options / Lecture 04
At each node of an American option, compare exercising now with continuing.
V_{\text{node}}=\max\left(\text{exercise value},\ \text{discounted continuation value}\right)
The same backward tree can price European options or add the American exercise choice.
The model’s clean edges
Introduction to Financial Options / Lecture 04
What makes replication work
Trade stock and cash at every node. Borrow and lend at one rate. Know the possible moves.
What real markets add
Bid/ask, stock borrow, jumps, uncertain volatility, dividends, and discrete hedge times.
The tree is a pricing argument under assumptions, not a promise of a riskless trading profit in practice.
Check the logic yourself
Introduction to Financial Options / Lecture 04
A put pays $30 if the share ends at $70, and $0 if it ends at $130. Today the share is $100. Rates are zero. Find the hedge and put price.
\Delta=(0-30)/(130-70)=-\tfrac12. The cash holding is $65, so the put costs $15.
Summary
Introduction to Financial Options / Lecture 04
- In a two-state model, stock and cash can exactly reproduce an option payoff.
- Matching payoffs force the option price to equal the replication cost.
- Risk-neutral probabilities are a shortcut derived from that hedge, not forecasts.
- In a multi-step tree, the stock hedge changes after each move.
- Smaller steps and more frequent rehedging point toward a continuous-time model.
Next: derive Black-Scholes by continuously adjusting the stock hedge.