Matching Binomial and Black–Scholes Volatility for Option Pricing
Summary
The document examines why a one-step binomial call value differs from a Black–Scholes value when both are applied to a stock that can move up or down over one day. Its main method is to match the first two moments of the binomial log return to the Black–Scholes log-return distribution. Matching the log-return variance gives an annualized volatility; matching its mean also constrains the drift. The response then reports that using this volatility in Black–Scholes produces a price closer to the binomial expectation than the original calculation.
A second explanation frames Black–Scholes as the continuous-time limit of a binomial model and gives a volatility scaling rule based on step size, step count, and horizon. The discussion highlights that a one-period tree and a continuous-time model need not produce the same option price. The moment-matching approach aligns only selected distributional properties, so it does not make the models fully equivalent or guarantee identical option values.
Key ideas
- Annualized volatility depends on both the size of return moves and the time horizon.
- Matching log-return variance gives a different volatility estimate from treating a one-day return standard deviation as annual volatility.
- The binomial and Black–Scholes models can be aligned by matching the mean and variance of log returns.
- A one-step binomial option price can differ from a continuous-time Black–Scholes price.
- Black–Scholes can be interpreted as a limit of binomial models with increasingly many steps.
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Full text
# Calculate volatility under the binomial model for option pricing
# Calculate volatility under the binomial model for option pricing
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 by Quantuple (score 3, accepted)
https://quant.stackexchange.com/a/30759
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 by M. Jeunesse (score 1)
https://quant.stackexchange.com/a/30756
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 $\sigma=\sqrt{365}$.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.