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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.

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  • $\begingroup$ In any professional context, you would be way better off just switching to a trinomial tree $\endgroup$
    – Brian B
    Sep 14, 2023 at 18:39

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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.

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