Calculate volatility under the binomial model for option pricing

Question

The original question is quoted below.

The underlying stock price is now \$100, and tomorrow it will be either \$101 (with probability $p$) or \$99 (with probability $1-p$). A call option with value $c$ which expire tomorrow has exercise price \$100. Find the the value of $c$ under the Black-Scholes model. Ignore interest rate.

When I tackle this question, I first derive $p=\frac{1}{2}$. To use call option price formula, we need $S, E, r, T-t, \sigma$. From the question, it is clear that $$S=100, E=100, r=0, T-t=\frac{1}{365}$$ So we only need $\sigma$. Since $\sigma$ is measured by the standard deviation of the return $\frac{dS}{S}$, I proceed as follow: $$E(return)=(1/100)(0.5)+(-1/100)(0.5)=0$$ $$Var(return)=(1/100-0)^2(0.5)+(-1/100-0)^2(0.5)=0.0001$$ $$sd(return)=\sqrt{0.0001}=0.01$$ Apply the explicit price formula for call option under the Black-Scholes model, I found $$d_1=0.00026171, d_2=-0.00026171, N(d_1) = 0.500104407965456, N(d_2)=0.499895592034544$$ Thus, the desired price is $$SN(d_1)-Ee^{-r(T-t)}N(d_2)=0.0209$$ The procedure seems logical to me. However,since the profit from the call option is $$(1)(0.5)+(0)(0.5)=0.5$$ I expected the price to be close or equal to \$0.5. How come they differ so much?

Answer 1

From your answer to my comment, here is what I would do.

Over the horizon $[0,\Delta t]$, the BS model tells you that the expected log-return is $$ \Bbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \left(\mu-\frac{1}{2}\sigma^2\right)\Delta t$$ with a variance $$ \Bbb{V}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \sigma^2 \Delta t$$

Over the same horizon, your binomial model tells you that: $$ \Bbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = 0.5 \ln(101/100)+0.5\ln(99/100) \approx -5e^{-5}$$ \begin{align} \Bbb{V}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] &=(0.5\ln^2(101/100) + 0.5\ln^2(99/100)) - (-5e^{-5})^2 \approx 1e^{-4} \end{align}

If you want your models to be mutually consistent, they should at least agree on the two first moments of the log-return's distribution. This constrains you to choose: $$ (\mu-\frac{1}{2}\sigma^2)\Delta t = -5e^{-5},\quad \sigma^2 \Delta t = 1e^{-4} $$ which gives, with $\Delta t=1/365$ $$\sigma = \sqrt{1e^{-4}\dot\,365} = 1e^{-2}\sqrt{365} = 0.1911$$ $$\mu = -5e^{-5}\dot\,365+\frac{1}{2}(0.1911)^2 \approx 1e^{-5}$$ Now you can use BS formula as you propose to find a price around $0.4$.

Answer 2

Black Scholes can be seen as the continuous limit of a binomial model when the number of steps go to infinity.

(It can be seen as a result of Donsker's theorem)

Thus it is normal that your call price in the one-period model is different than the one in the BS model.

If you have $n$ steps in your binomial to describe the period $[0,T]$ and if your increment on one step in $\pm h$, then the equivalent volatility is $h\sqrt{\frac{n}{T}}$.

So here $n=1$, $T=\frac{1}{365}$ and $h=1$ so $\sigma=\sqrt{365}$.