In everyone's binomial trees online I see constant U and D. Even when I read Option Volatility and Pricing by Natenburg, all his diagrams use a constant U and D (where U is the upwards magnitude from one step to the next. For example 1.05). The whole premise of a binomial model is that, as the amount of steps increase, the option price derived from the model should get closer and closer to Black Scholes' (see below).

Binomial Vs Black Scholes

enter image description here

The problem is that, with a constant U and D, it doesn't. The value of the option continues to grow indefinitely with the number of steps. If we fix the time to maturity, and increase the number of steps in between now and maturity, we increase the range of potential prices the underlying can reach. This increases volatility and hence increases the value of the option.

Holding all else constant, increasing the number of steps only increases the value of the option, which makes me believe that I'm missing some other adjustment.

My question in: What adjustments to U and D, or to the model, do I need to make such that I can approximate the BS Model (as per the image above).

A constant U and D does not seem correct. I've attached another picture from the book by Natenburg below for reference. If I increase the number of steps from 3 to 20, the value of the option goes from 5.22 to 9.24.

enter image description here


1 Answer 1


You can choose $U=e^{(r-\sigma^2/2)\delta t + \sigma \sqrt{\delta t}}$ and $D=e^{(r-\sigma^2/2)\delta t - \sigma \sqrt{\delta t}}$ to have the binomial model converge to the BS model when $\delta t=T/N \rightarrow 0$. The intuition is that in the binomial model as $\delta t\rightarrow 0$ you have $E[(S_{t+\delta t} - S_t)/S_t] \approx r \delta t$ and $\text{VAR}[(S_{t+\delta t} - S_t)/S_t] \approx \sigma^2 \delta t$

  • $\begingroup$ Generally we say that U and D vary with the time interval $\Delta t$ but as you can see from the equation $\Delta t=T/N$ above they also vary with the number of steps $N$ since the number of steps determines the basic time interval. (But I still prefer to say that U and D are a function of $\Delta t$. So I would write $U=f(r,\sigma;\Delta t)$). $\endgroup$
    – nbbo2
    Commented Jan 28, 2020 at 11:43
  • $\begingroup$ In the book I referenced above they do $e^{σ√δt}$ where $Δt=T/N$, but while T is 0.75 years, they use $75$, which I find strange. They they divide the result of $√δt$ by 10. I'll play around with the formula and report back. $\endgroup$
    – user43953
    Commented Jan 28, 2020 at 18:00
  • $\begingroup$ @Antoine Conze Rather than my stock prices flatting out on both sides (gyazo.com/8f841ccff30bf8fe8e81858acbd34403), they seem to be expanding: gyazo.com/91692683f83ef1fcf334d5efe798bc25. I'm not sure why, it's quite frustrating. $Δt=T/N$ is constant for each step $n$. What should be changing in terms of the $U$ and $D$ calculations at each step? $\endgroup$
    – user43953
    Commented Jan 28, 2020 at 18:57
  • $\begingroup$ They will "expand": after $N$ steps the max stock price is $S_0 U^N = e^{(r-\sigma^2/2)T + \sigma \sqrt{T} \sqrt{N}}$. But the probability of reaching the max is circa $1/2^N$. Think of the limit to the exponential brownian motion: its values are unbounded and the distribution is log normal. $\endgroup$ Commented Jan 29, 2020 at 8:22

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