One-Step Binomial Option Pricing

Worked Solution

How to Think About It: The one-step binomial model is the simplest non-trivial derivatives pricing framework, and it contains every idea that matters: replication, no-arbitrage, and risk-neutral valuation. Before writing formulas, think about what you are doing economically. You have two states of the world and two instruments (the stock and the bond). Two instruments, two equations -- you can match any single payoff exactly. That is replication. Once you can replicate the option, its price must equal the cost of the replicating portfolio, or there is free money. The "risk-neutral probability" is not a real-world probability -- it is the synthetic probability that makes the expected return on the stock equal the risk-free rate. It falls out of the no-arbitrage condition.

Quick Sanity Checks:

• If the call is deep ITM in both states ($K$ very small), the call is worth roughly $S_0 - Ke^{-rT}$, like a forward minus strike. The delta should be close to 1 and borrowing should be close to $Ke^{-rT}$.
• If the call is deep OTM in both states ($K$ very large), the call is worthless, delta is 0, and borrowing is 0.
• If the call pays off only in the up state (the typical interesting case), the price should be between 0 and $S_0 - Ke^{-rT}$, and $\Delta$ should be between 0 and 1.
• The risk-neutral probability $q$ must satisfy $0 < q < 1$, which is guaranteed by the no-arbitrage condition $d < e^{rT} < u$.

Derivation:

*Part 1 -- Risk-neutral probability and price:*

The option pays $C_u = \max(uS_0 - K,\, 0)$ in the up state and $C_d = \max(dS_0 - K,\, 0)$ in the down state. Define the risk-neutral probability $q$ as the value that makes the stock's expected gross return equal the risk-free return:

$q \cdot u + (1 - q) \cdot d = e^{rT}$

Solving:

$q = \frac{e^{rT} - d}{u - d}$

The call price is the discounted expected payoff under this measure:

$C_0 = e^{-rT}\bigl[q\, C_u + (1 - q)\, C_d\bigr]$

*Part 2 -- Replicating portfolio:*

Hold $\Delta$ shares and invest $B$ dollars in the risk-free bond ($B < 0$ means borrowing). The portfolio must match the option in both states:

$\Delta \cdot uS_0 + B\, e^{rT} = C_u$ $\Delta \cdot dS_0 + B\, e^{rT} = C_d$

Subtract the second from the first:

$\Delta = \frac{C_u - C_d}{(u - d)\, S_0}$

This is the option delta -- the sensitivity of the option price to the stock price. Substitute back to get the bond position:

$B = e^{-rT}\frac{u\, C_d - d\, C_u}{u - d}$

Note $B \leq 0$ when $C_d = 0$ (the typical case), confirming you are borrowing to fund the stock purchase.

*Part 3 -- Equivalence:*

The cost of the replicating portfolio is:

$\Delta\, S_0 + B = \frac{C_u - C_d}{u - d} + e^{-rT}\frac{u\, C_d - d\, C_u}{u - d}$

$= \frac{1}{u - d}\Bigl[(C_u - C_d) + e^{-rT}(u\, C_d - d\, C_u)\Bigr]$

$= \frac{e^{-rT}}{u - d}\Bigl[e^{rT}(C_u - C_d) + u\, C_d - d\, C_u\Bigr]$

$= \frac{e^{-rT}}{u - d}\Bigl[C_u(e^{rT} - d) - C_d(e^{rT} - u)\Bigr]$

$= e^{-rT}\Bigl[\frac{e^{rT} - d}{u - d}\, C_u + \frac{u - e^{rT}}{u - d}\, C_d\Bigr]$

$= e^{-rT}\bigl[q\, C_u + (1 - q)\, C_d\bigr] = C_0$

The replicating portfolio cost equals the risk-neutral price. This must hold; otherwise there is an arbitrage.

Practical Interpretation: In practice, $\Delta$ tells you the hedge ratio -- how many shares to hold per option sold. On a trading desk, you would sell the call at $C_0$, immediately buy $\Delta$ shares, and fund the excess with borrowing. At expiry, your portfolio matches the option payoff exactly regardless of which state occurs. The risk-neutral probability $q$ is not your view on whether the stock goes up -- it is the market-implied probability backed out from the requirement of no arbitrage. Every model in derivatives pricing (Black-Scholes, local vol, stochastic vol) is built on exactly this logic, just with more time steps and more states.

Answer: The risk-neutral probability is $q = (e^{rT} - d)/(u - d)$. The call price is $C_0 = e^{-rT}[q\, C_u + (1-q)\, C_d]$. The replicating portfolio holds $\Delta = (C_u - C_d)/((u-d)S_0)$ shares and borrows $B = e^{-rT}(uC_d - dC_u)/(u-d)$, with total cost $\Delta S_0 + B = C_0$.