## Determining Stock Prices

To agree on accurate pricing for any tradable asset is challenging—that’s why stock prices constantly change. In reality, companies hardly change their valuations on a day-to-day basis, but their stock prices and valuations change nearly every second. This difficulty in reaching a consensus about correct pricing for any tradable asset leads to short-lived arbitrage opportunities.

But a lot of successful investing boils down to a simple question of present-day valuation– what is the right current price today for an expected future payoff?

### Key Takeaways

- The binomial option pricing model values options using an iterative approach utilizing multiple periods to value American options.
- With the model, there are two possible outcomes with each iteration—a move up or a move down that follow a binomial tree.
- The model is intuitive and is used more frequently in practice than the well-known Black-Scholes model.

## Binomial Options Valuation

In a competitive market, to avoid arbitrage opportunities, assets with identical payoff structures must have the same price. Valuation of options has been a challenging task and pricing variations lead to arbitrage opportunities. Black-Scholes remains one of the most popular models used for pricing options but has limitations.

The binomial option pricing model is another popular method used for pricing options.

## Examples

Assume there is a call option on a particular stock with a current market price of $100. The at-the-money (ATM) option has a strike price of $100 with time to expiry for one year. There are two traders, Peter and Paula, who both agree that the stock price will either rise to $110 or fall to $90 in one year.

They agree on expected price levels in a given time frame of one year but disagree on the probability of the up or down move. Peter believes that the probability of the stock's price going to $110 is 60%, while Paula believes it is 40%.

Based on that, who would be willing to pay more price for the call option? Possibly Peter, as he expects a high probability of the up move.

## Binomial Options Calculations

The two assets, which the valuation depends upon, are the call option and the underlying stock. There is an agreement among participants that the underlying stock price can move from the current $100 to either $110 or $90 in one year and there are no other price moves possible.

In an arbitrage-free world, if you have to create a portfolio comprised of these two assets, call option and underlying stock, such that regardless of where the underlying price goes – $110 or $90 – the net return on the portfolio always remains the same. Suppose you buy "d" shares of underlying and short one call options to create this portfolio.

If the price goes to $110, your shares will be worth $110*d, and you'll lose $10 on the short call payoff. The net value of your portfolio will be (110d - 10).

If the price goes down to $90, your shares will be worth $90*d, and the option will expire worthlessly. The net value of your portfolio will be (90d).

If you want your portfolio's value to remain the same regardless of where the underlying stock price goes, then your portfolio value should remain the same in either case:

$\begin{aligned} &h(d) - m = l ( d ) \\ &\textbf{where:} \\ &h = \text{Highest potential underlying price} \\ &d = \text{Number of underlying shares} \\ &m = \text{Money lost on short call payoff} \\ &l = \text{Lowest potential underlying price} \\ \end{aligned}$

So if you buy half a share, assuming fractional purchases are possible, you will manage to create a portfolio so that its value remains the same in both possible states within the given time frame of one year.

$\begin{aligned} &110d - 10 = 90d \\ &d = \frac{ 1 }{ 2 } \\ \end{aligned}$

This portfolio value, indicated by (90d) or (110d - 10) = 45, is one year down the line. To calculate its present value, it can be discounted by the risk-free rate of return (assuming 5%).

$\begin{aligned} \text{Present Value} &= 90d \times e^ { (-5\% \times 1 \text{ Year}) } \\ &= 45 \times 0.9523 \\ &= 42.85 \\ \end{aligned}$

Since at present, the portfolio is comprised of ½ share of underlying stock (with a market price of $100) and one short call, it should be equal to the present value.

$\begin{aligned} &\frac { 1 }{ 2} \times 100 - 1 \times \text{Call Price} = \$42.85 \\ &\text{Call Price} = \$7.14 \text{, i.e. the call price of today} \\ \end{aligned}$

Since this is based on the assumption that the portfolio value remains the same regardless of which way the underlying price goes, the probability of an up move or down move does not play any role. The portfolio remains risk-free regardless of the underlying price moves.

In both cases (assumed to up move to $110 and down move to $90), your portfolio is neutral to the risk and earns the risk-free rate of return.

Hence both the traders, Peter and Paula, would be willing to pay the same $7.14 for this call option, despite their differing perceptions of the probabilities of up moves (60% and 40%). Their individually perceived probabilities don’t matter in option valuation.

Supposing instead that the individual probabilities matter, arbitrage opportunities may have presented themselves. In the real world, such arbitrage opportunities exist with minor price differentials and vanish in the short term.

But where is the much-hyped volatility in all these calculations, an important and sensitive factor that affects options pricing?

The volatility is already included by the nature of the problem's definition. Assuming two (and only two—hence the name “binomial”) states of price levels ($110 and $90), volatility is implicit in this assumption and included automatically (10% either way in this example).

## Black-Scholes

But is this approach correct and coherent with the commonly used Black-Scholes pricing? Options calculator results (courtesy of OIC) closely match with the computed value:

Unfortunately, the real world is not as simple as “only two states.” The stock can reach several price levels before the time to expiry.

Is it possible to include all these multiple levels in a binomial pricing model that is restricted to only two levels? Yes, it is very much possible, but to understand it takes some simple mathematics.

## Simple Math

To generalize this problem and solution:

"X" is the current market price of a stock and "X*u" and "X*d" are the future prices for up and down moves "t" years later. Factor "u" will be greater than one as it indicates an up move and "d" will lie between zero and one. For the above example, u = 1.1 and d = 0.9.

The call option payoffs are "P_{up}" and "P_{dn}" for up and down moves at the time of expiry.

If you build a portfolio of "s" shares purchased today and short one call option, then after time "t":

$\begin{aligned} &\text{VUM} = s \times X \times u - P_\text{up} \\ &\textbf{where:} \\ &\text{VUM} = \text{Value of portfolio in case of an up move} \\ \end{aligned}$

$\begin{aligned} &\text{VDM} = s \times X \times d - P_\text{down} \\ &\textbf{where:} \\ &\text{VDM} = \text{Value of portfolio in case of a down move} \\ \end{aligned}$

For similar valuation in either case of price move:

$s \times X \times u - P_\text{up} = s \times X \times d - P_\text{down}$

$\begin{aligned} s &= \frac{ P_\text{up} - P_\text{down} }{ X \times ( u - d) } \\ &= \text{The number of shares to purchase for} \\ &\phantom{=} \text{a risk-free portfolio} \\ \end{aligned}$

The future value of the portfolio at the end of "t" years will be:

$\begin{aligned} \text{In Case of Up Move} &= s \times X \times u - P_\text{up} \\ &=\frac { P_\text{up} - P_\text{down} }{ u - d} \times u - P_\text{up} \\ \end{aligned}$

$\begin{aligned} \text{In Case of Down Move} &= s \times X \times d - P_\text{down} \\ &=\frac { P_\text{up} - P_\text{down} }{ u - d} \times d - P_\text{down} \\ \end{aligned}$

The present-day value can be obtained by discounting it with the risk-free rate of return:

$\begin{aligned} &\text{PV} = e(-rt) \times \left [ \frac { P_\text{up} - P_\text{down} }{ u - d} \times u - P_\text{up} \right ] \\ &\textbf{where:} \\ &\text{PV} = \text{Present-Day Value} \\ &r = \text{Rate of return} \\ &t = \text{Time, in years} \\ \end{aligned}$

This should match the portfolio holding of "s" shares at X price, and short call value "c" (present-day holding of (s* X - c) should equate to this calculation.) Solving for "c" finally gives it as:

Note: If the call premium is shorted, it should be an addition to the portfolio, not a subtraction.

$c = \frac { e(-rt) }{ u - d} \times [ ( e ( -rt ) - d ) \times P_\text{up} + ( u - e ( -rt ) ) \times P_\text{down} ]$

Another way to write the equation is by rearranging it:

Taking "q" as:

$q = \frac { e (-rt) - d }{ u - d }$

Then the equation becomes:

$c = e ( -rt ) \times ( q \times P_\text{up} + (1 - q) \times P_\text{down} )$

Rearranging the equation in terms of “q” has offered a new perspective.

Now you can interpret “q” as the probability of the up move of the underlying (as “q” is associated with P_{up} and “1-q” is associated with P_{dn}). Overall, the equation represents the present-day option price, the discounted value of its payoff at expiry.

## This "Q" is Different

How is this probability “q” different from the probability of an up move or a down move of the underlying?

$\begin{aligned} &\text{VSP} = q \times X \times u + ( 1 - q ) \times X \times d \\ &\textbf{where:} \\ &\text{VSP} = \text{Value of Stock Price at Time } t \\ \end{aligned}$

Substituting the value of "q" and rearranging, the stock price at time "t" comes to:

$\begin{aligned} &\text{Stock Price} = e ( rt ) \times X \\ \end{aligned}$

In this assumed world of two-states, the stock price simply rises by the risk-free rate of return, exactly like a risk-free asset, and hence it remains independent of any risk. Investors are indifferent to risk under this model, so this constitutes the risk-neutral model.

Probability “q” and "(1-q)" are known as risk-neutral probabilities and the valuation method is known as the risk-neutral valuation model.

The example scenario has one important requirement – the future payoff structure is required with precision (level $110 and $90). In real life, such clarity about step-based price levels is not possible; rather the price moves randomly and may settle at multiple levels.

To expand the example further, assume that two-step price levels are possible. We know the second step final payoffs and we need to value the option today (at the initial step):

Working backward, the intermediate first step valuation (at t = 1) can be made using final payoffs at step two (t = 2), then using these calculated first step valuation (t = 1), the present-day valuation (t = 0) can be reached with these calculations.

To get option pricing at number two, payoffs at four and five are used. To get pricing for number three, payoffs at five and six are used. Finally, calculated payoffs at two and three are used to get pricing at number one.

Please note that this example assumes the same factor for up (and down) moves at both steps – u and d are applied in a compounded fashion.

## A Working Example

Assume a put option with a strike price of $110 is currently trading at $100 and expiring in one year. The annual risk-free rate is 5%. Price is expected to increase by 20% and decrease by 15% every six months.

Here, u = 1.2 and d = 0.85, x = 100, t = 0.5

using the above derived formula of

$q = \frac { e (-rt) - d }{ u - d }$

we get q = 0.35802832

value of put option at point 2,

$\begin{aligned} &p_2 = e (-rt) \times (p \times P_\text{upup} + ( 1 - q) P_\text{updn} ) \\ &\textbf{where:} \\ &p = \text{Price of the put option} \\ \end{aligned}$

At P_{upup} condition, underlying will be = 100*1.2*1.2 = $144 leading to P_{upup }= zero

At P_{updn} condition, underlying will be = 100*1.2*0.85 = $102 leading to P_{updn }= $8

At P_{dndn} condition, underlying will be = 100*0.85*0.85 = $72.25 leading to P_{dndn }= $37.75

p_{2} = 0.975309912*(0.35802832*0+(1-0.35802832)*8) = 5.008970741

Similarly, p_{3} = 0.975309912*(0.35802832*8+(1-0.35802832)*37.75) = 26.42958924

$p_1 = e ( -rt ) \times ( q \times p_2 + ( 1 - q ) p_3 )$

And hence value of put option, p_{1} = 0.975309912*(0.35802832*5.008970741+(1-0.35802832)* 26.42958924) = **$18.29.**

Similarly, binomial models allow you to break the entire option duration to further refined multiple steps and levels. Using computer programs or spreadsheets, you can work backward one step at a time to get the present value of the desired option.

## Another Example

Assume a European-type put option with nine months to expiry, a strike price of $12 and a current underlying price at $10. Assume a risk-free rate of 5% for all periods. Assume every three months, the underlying price can move 20% up or down, giving us u = 1.2, d = 0.8, t = 0.25 and a three-step binomial tree.

Red indicates underlying prices, while blue indicates the payoff of put options.

Risk-neutral probability "q" computes to 0.531446.

Using the above value of "q" and payoff values at t = nine months, the corresponding values at t = six months are computed as:

Further, using these computed values at t = 6, values at t = 3 then at t = 0 are:

That gives the present-day value of a put option as $2.18, pretty close to what you'd find doing the computations using the Black-Scholes model ($2.30).

## The Bottom Line

Although using computer programs can make these intensive calculations easy, the prediction of future prices remains a major limitation of binomial models for option pricing. The finer the time intervals, the more difficult it gets to predict the payoffs at the end of each period with high-level precision.

However, the flexibility to incorporate the changes expected at different periods is a plus, which makes it suitable for pricing American options, including early-exercise valuations.

The values computed using the binomial model closely match those computed from other commonly used models like Black-Scholes, which indicates the utility and accuracy of binomial models for option pricing. Binomial pricing models can be developed according to a trader's preferences and can work as an alternative to Black-Scholes.