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Deriving the Risk-Neutral Probabilities in a One-Period Binomial Model

Article Quant Q&A · Author: James

Summary

The document explains how a call option can be replicated in a one-period, two-state stock model. A portfolio holding a suitable number of shares and borrowing or lending cash is chosen to match the call’s payoff in both the up and down states. Its initial cost therefore determines the option price.

The key algebraic step is to verify that the two state weights in the pricing formula reproduce both the current stock price and the value of one unit of cash. Expanding each weighted sum makes the terms cancel and leaves the stock price in the first identity and one in the second. These weights support the risk-neutral valuation interpretation. The explanation addresses the algebra behind the identities rather than developing the full model; it assumes the replication setup and does not discuss how to apply it when markets have more states or replication is unavailable.

Key ideas

  • A call can be replicated in a two-state model with shares and a cash position.
  • The share holding is chosen so the portfolio has the same payoff in the up and down states.
  • The pricing weights reproduce both the current stock price and the value of cash.
  • Expanding the weighted sums shows directly why the two identities hold.

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Full text
# Single period risk-neutral probability derivation


# Single period risk-neutral probability derivation












Let $S_u$ be the price of stock in the up-state one period from now. Let $S_d$ be the price of the stock in the down state.

Let $C_u$ be the payoff of a call option at time $1$ in the up-state and similarly for $C_d$.

If we define $\delta = \frac{C_u-C_d}{S_u-S_d}$, then if we purchase $\delta$ shares at time zero and borrow $e^{-r}(\delta S_u - C_u) = e^{-r}(\delta S_d - C_d)$, this portfolio replicates the payoff of the call option.

So at initiation we must have $C(0) = \delta S - e^{-r}(\delta S_u - C_u)$.

substituting for $\delta$ can get us

$$ C = \frac{S-e^{-r}S_d}{S_u-S_d}C_u + \frac{e^{-r}S_u - S}{S_u-S_d}C_d$$

now the author claims in the book I am reading that a "little algebra" quickly implies

$$S = \frac{S-e^{-r}S_d}{S_u-S_d} S_u + \frac{e^{-r}S_u - S}{S_u-S_d} S_d$$ and

$$1 = \frac{S-e^{-r}S_d}{S_u-S_d} e^r + \frac{e^{-r}S_u - S}{S_u-S_d} e^r$$

Now ive been looking at these last two equations for over an hour and I cannot for the life of me see how you algebraically get to them. Can someone help me out?

The above comes from page 12 of "a course in derivative securities" by Kerry Back...to the comment below, I agree, it does seem like a contradiciton but perhaps I am missing something.

## Answer by Francis (score 1)

https://quant.stackexchange.com/a/60829

I think the author is just saying that

\begin{equation} \begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} S_u + \frac{e^{-r}S_u - S}{S_u - S_d} S_d &= \frac{S S_u - e^{-r}S_d S_u + e^{-r}S_uS_d - SS_d}{S_u - S_d} \\ &= \frac{S (S_u - S_d)}{S_u - S_d} \\ &= S \end{split} \end{equation}

and that

\begin{equation} \begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} e^r + \frac{e^{-r}S_u - S}{S_u - S_d} e^r &= \frac{S e^r - S_d + S_u - S e^r}{S_u - S_d} \\ &= \frac{S_u - S_d}{S_u - S_d} \\ &= 1 \end{split} \end{equation}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.