8.3 One-period binomial model and the risk-neutral probability

Motivation

Parity pinned down the difference of call and put prices without any model. To price a call alone we need a model of how $S$ moves. The simplest model, two states in one period, already contains the central idea of derivatives pricing: an option can be replicated by trading the underlying and a bond, and its price is the cost of the replicating portfolio. No probabilities of up or down moves enter.

Setup

One period of length $\Delta t$ . Today the stock is $S_0$ . At $\Delta t$ it is either $uS_0$ or $dS_0$ with $0<d<u$ . Both states have positive real-world probability. A bond grows at $e^{r\Delta t}$ . A claim pays $V_u$ in the up state and $V_d$ in the down state. Ignore dividends for now.

No arbitrage and replication#

Lemma 8.3.1no arbitrage in the one-period model

The model is arbitrage-free if and only if $d<e^{r\Delta t}<u$ .

Proof

If $e^{r\Delta t}\le d$ , borrow $S_0$ and buy one share: zero cost, terminal value $S_0(d-e^{r\Delta t})\ge0$ in the down state and $S_0(u-e^{r\Delta t})>0$ in the up state. That is an arbitrage. If $e^{r\Delta t}\ge u$ , short the share and lend: symmetric. Conversely, if $d<e^{r\Delta t}<u$ , any zero-cost portfolio $(\Delta \text{ shares}, -\Delta S_0 \text{ in bonds})$ has terminal values $\Delta S_0(u-e^{r\Delta t})$ and $\Delta S_0(d-e^{r\Delta t})$ , which have opposite signs unless $\Delta=0$ . So no zero-cost portfolio is non-negative in both states and positive in one.∎

Proposition 8.3.2replication

If $d<e^{r\Delta t}<u$ , the portfolio of $\Delta$ shares and $B$ rupees in bonds with

\Delta = \frac{V_u-V_d}{S_0(u-d)},\qquad B = e^{-r\Delta t}\,\frac{uV_d-dV_u}{u-d} \tag{8.8}

pays exactly $V_u$ and $V_d$ . The unique arbitrage-free price of the claim is

V_0=\Delta S_0+B = e^{-r\Delta t}\big[pV_u+(1-p)V_d\big],\qquad p=\frac{e^{r\Delta t}-d}{u-d}\in(0,1). \tag{8.9}

Proof

We need $\Delta uS_0+Be^{r\Delta t}=V_u$ and $\Delta dS_0+Be^{r\Delta t}=V_d$ . Subtracting gives $\Delta S_0(u-d)=V_u-V_d$ , which is the formula for $\Delta$ . Substituting back, $Be^{r\Delta t}=V_u-u\frac{V_u-V_d}{u-d}=\frac{uV_d-dV_u}{u-d}$ . Then

V_0=\frac{V_u-V_d}{u-d}+e^{-r\Delta t}\frac{uV_d-dV_u}{u-d} =e^{-r\Delta t}\Big[\frac{e^{r\Delta t}-d}{u-d}V_u+\frac{u-e^{r\Delta t}}{u-d}V_d\Big],

which is (8.9). By Lemma 8.3.1, $0<p<1$ . If the claim traded at a price other than $V_0$ , buy the cheaper of {claim, replicating portfolio} and sell the dearer: the payoffs cancel in both states and the price difference is riskless profit.∎

The risk-neutral probability#

Three remarks carry through the rest of the module.

The real probability of an up move does not appear. Two traders who disagree about the chance of an up move, but agree on $S_0,u,d,r$ , must agree on the option price. What matters is the set of possible moves (volatility), not their likelihood (drift).
$p$ is a probability measure under which the stock earns the risk-free rate. Indeed $pu S_0+(1-p)dS_0=S_0e^{r\Delta t}$ , which is just the definition of $p$ rearranged. Call it the risk-neutral probability $\mathbb Q$ . Pricing is "expected payoff under $\mathbb Q$ , discounted at $r$ ". This is a computational device. It does not say investors are risk-neutral; it says that, given the stock price, the option's price is determined, and the formula that delivers it looks like a risk-neutral expectation.
With a dividend yield $q$ , a share held over $\Delta t$ becomes $e^{q\Delta t}$ shares, and the same algebra gives

p = \frac{e^{(r-q)\Delta t}-d}{u-d}. \tag{8.10}

This is a one-period version of the fundamental theorem of asset pricing (Harrison and Kreps 1979; Harrison and Pliska 1981): a model has no arbitrage if and only if there is an equivalent probability measure under which discounted asset prices (with dividends reinvested) are martingales. The market is complete (every claim is replicable) if and only if that measure is unique. Two states and two assets give uniqueness; add a third state and you lose it (Problem 8.7).

One-period pricing examples#

Worked example 8.3Hull's textbook case

$S_0=20$ , $u=1.1$ , $d=0.9$ , $r=12\%$ , $\Delta t=0.25$ , call with $K=21$ . Then $V_u=1$ , $V_d=0$ , $\Delta=1/(20\times0.2)=0.25$ , $B=-e^{-0.03}\times0.9\times1/0.2=-4.367$ , $V_0=5-4.367=0.633$ . Check: $p=(e^{0.03}-0.9)/0.2=0.6523$ and $e^{-0.03}\times0.6523=0.633$ . This matches Hull's printed 0.633 (Hull, Ch. 13) and is a unit test in test_m08_binomial.py.

Worked example 8.4a Nifty-sized version

Let $S_0=24{,}500$ , one month ( $\Delta t=1/12$ ), $u=1.04$ , $d=0.96$ , $r=5.5\%$ , and an ATM call $K=24{,}500$ . Then $V_u=980$ , $V_d=0$ , so $\Delta=980/(24{,}500\times0.08)=0.5$ and $B=e^{-0.004583}(0-0.96\times980)/0.08=-11{,}706.22$ . The call is worth $0.5\times24{,}500-11{,}706.22=543.78$ points. The risk-neutral probability is $p=(e^{0.004583}-0.96)/0.08=0.5574$ ; it exceeds one half because the stock must earn 5.5% a year under $\mathbb Q$ with symmetric moves. To replicate one lot you would hold $0.5\times65=32.5$ units of Nifty (in practice half a lot of futures) and borrow ₹7.61 lakh.

Practitioner notes

The hedge ratio $\Delta$ in (8.8) is a finite difference, $\partial V/\partial S$ in the limit. It is the delta of 8.9 and the number of futures a market maker holds against an option.
Replication fails in practice when the underlying can move to more than two places (gaps), when you cannot trade continuously, and when costs eat the rebalancing. The rest of the module refines the model; M09 measures what happens when it fails.