# Binomial tree with time dependent volatility

In the Cox approach for binomial trees, the up move $$u$$ and down move $$d$$ are given by: $$u = e^{\sigma \sqrt{dt}}$$ and $$d = e^{-\sigma \sqrt{dt}}$$. In this approach the volatility $$\sigma$$ is assumed constant. I am trying to build a tree with time dependent volatility.

Let volatility $$\sigma(t)$$ be time dependent. To have a recombined tree it requires that the variance $$\sigma_i \sqrt{dt_i}$$ is independent of $$i$$. This means that there are 2 degrees of freedom to choose because varying either $$dt_i$$ or $$\sigma_i$$ will result in a non-recombining tree. It is not clear to me how we can choose these?

As I understand it, $$\sigma_i$$ is the "forward" volatility on the time interval $$[t_i, t_{i+1}]$$. If I consider the volatility smile today and denote $$\sigma(K,t)$$ the implied volatility of an option with strike $$K$$ and expiry $$t$$, then we have the relationship: $$\sigma_{i}^2dt_i = \sigma(K,t_{i+1})^2t_{i+1} - \sigma(K,t_{i})^2t_{i}$$

The computation of these forward volatilities is then straightforward given that I know $$t_i$$ since I can compute the implied volatility $$\sigma(K,t)$$ for any $$t$$. However, given that $$t_i$$ are unknown I can't determine $$\sigma(K,t_i)$$ before first knowing $$t_i$$ ... I am dealing with an equation with too many unknowns.

• In any professional context, you would be way better off just switching to a trinomial tree Commented Sep 14, 2023 at 18:39

There's likely something far less crude than this but:

Fixing some initial time interval $$[t_0, t_1]$$ we have the volatility over that time interval as being given by:

Let the volatility over a time interval be denoted by: $$vol(t_i, dt_{i}) = \sigma_i \sqrt{d t_i} = \sigma_i \sqrt{(t_i + d t_i) - t_i} = \sigma_i \sqrt{t_{i+1} - t_i}$$

To be recombinant we require that, for any $$i$$: $$vol(t_i, dt_{i}) = vol(t_0, d t_0)$$

where we need $$vol$$ to satisfy:

$$vol(t_i, dt_{i}) = \sqrt{\sigma(K, t_{i} + d t_i)^2 (t_i+dt_{i}) - \sigma(K, t_i)^2t_i}$$

Since $$t_i$$ will be fixed we can only vary $$dt_i$$.

The time interval volatility will be equal to $$0$$ when $$dt = 0$$ and by increase to $$\infty$$ as $$t \to \infty$$ (by assumption). Assuming smoothness of $$\sigma(K, t)$$ and some other assumptions, $$vol(t_i, dt_i)$$ should be increasing in $$dt_i$$.

Hence just slowly increase $$dt_i$$ from $$0$$ until you satisfy $$vol(t_i, dt_{i}) = vol(t_0, d t_0)$$, do this at each step.