# Basic binomial option pricing example

A security is currently trading at 100, and with 99% probability it will be at 110 tomorrow, and with 1% probability at 90. What is the value of an ATM call option today expiring tomorrow? Assume nil interest rates.

If you sell the option at $C$ and immediately hedge with $n$ shares of the underlying, you'll be left with $C - 100n$. Tomorrow your portfolio will be worth $C - 100n + 110n = C + 10n$ with 99% chance and $C - 100n + 90n = C-10n$ with 1% chance. In order for you to not lose or make money on selling the option, in the first case your portfolio must be worth 10, and in the second case 0. In other words, $C$ and $n$ are uniquely determined by

\begin{cases} C + 10n = 10 \\ C - 10n = 0. \end{cases} Solving yields $C = 5$ and $n = 1/2$.

On the other hand, one sees that there is a 99% chance the option has value 10 dollars tomorrow and a 1% chance it has no value, yielding $C = (10)(.99) = 9.90.$

Which is the correct valuation? Is the stock mispriced if we know there is a 99% chance of an upward movement?

• When thinking about the distribution, it makes more sense to think about the forward, which will be the expected value of the distribution, in your case, the stock should be worth (with no dividends or rates) approx 0.99*110+0.01*90=109.8. Ie buying and holding the stock you would expect to net you 9.8, while the option you expect to net you 9.9. – will Aug 22 '17 at 9:11

The correct valuation is C=5, assuming the stock is trading at 100. Any other option price allows an immediate arbitrage opportunity between the option and the stock.

Whether the stock should be trading at 100 is a separate question. It depends on the risk tolerance of investors, and the correlation of this stock with other investments, for example.

Your example is a bit odd;

tomorrow, you expect a stock to be worth either $110$, with $99\%$ probability, or $90$ with $1\%$ probability. So your expectation of the value tomorrow is $0.99\cdot110 + 0.01\cdot 90 = 109.8$ - this means that the intrinsic value of your option is $\max(0, 109.8-100) = 9.8$. If the option is worth less than this, then there is an arbitrage, as it has negative time value.

The value of the option is

$$\begin{eqnarray} \mathrm{Call} &=& \int_0^\infty \phi(S_T) \cdot \max(S_t-K,0)\ \mathrm{d}S_T\\ &=&\phi(110) \cdot (110-100)^+ + \phi(90)\cdot(90-100)^+ \\ &=& \phi(110) \cdot 10 = 99\% \cdot 10\\ &=& 9.9 \end{eqnarray}$$

This is an odd / contrived example though, since the spot price is so far from the forward, and we have said there are no dividends and interest rates are zero. So you're only going to be able to explain this with some fairly contrived scenarios (i.e. maybe the company is selling loads of stock to raise money and driving the price down, such that it's not at an equillibrium value, and the market knows this will stop tomorrow? Who knows.)