Trinomial Tree Pricing

A binomial tree gives the stock exactly two choices at every step: up or down. That's enough to converge to the right answer eventually, but "eventually" can mean thousands of steps, and some contracts — anything with a barrier level sitting between the tree's nodes, for instance — converge unevenly or even non-monotonically as you add steps. Give the stock a third choice at each step, staying roughly where it is, and both problems improve: the tree needs fewer steps for the same accuracy, and it can place a node exactly on a barrier.

Rain, cloudy, or sun

A weather forecast with only two categories — rain or shine — misses a real, common outcome: a day that's neither, one that stays about the same. Adding "cloudy, unchanged" as a third category doesn't just add resolution; it makes the middle-ground outcome, which is often the most likely one over a short step, explicitly representable instead of forcing it to average out of two extremes. A trinomial tree does exactly this to a stock's short-term path: up, down, or stay.

$$u = e^{\sigma\sqrt{2\Delta t}}, \quad d = \frac{1}{u}, \quad m = 1$$ $$p_u = \left(\frac{e^{r\Delta t/2} - e^{-\sigma\sqrt{\Delta t/2}}}{e^{\sigma\sqrt{\Delta t/2}} - e^{-\sigma\sqrt{\Delta t/2}}}\right)^{\!2}, \quad p_d = \left(\frac{e^{\sigma\sqrt{\Delta t/2}} - e^{r\Delta t/2}}{e^{\sigma\sqrt{\Delta t/2}} - e^{-\sigma\sqrt{\Delta t/2}}}\right)^{\!2}, \quad p_m = 1 - p_u - p_d$$

In plain English: the three branch sizes are fixed first — up by factor $u$, down by factor $d=1/u$, or stay at $m=1$ — and then the three probabilities are chosen so the tree's average move and average spread-of-outcomes exactly match what geometric Brownian motion predicts over that step. All three probabilities must come out positive, which is the real constraint on how big a single step can be.

Worked example 1: building one step. Take $\sigma=20\%$, $\Delta t=1$ month $=1/12$, $r=0$. Then $\sqrt{\Delta t/2}=\sqrt{1/24}=0.2041$, so $\sigma\sqrt{\Delta t/2}=0.0408$, and $e^{0.0408}=1.0417$, $e^{-0.0408}=0.9600$, a denominator of $0.0817$. With $r=0$, $e^{r\Delta t/2}=1$, so $p_u=\left(\frac{1-0.9600}{0.0817}\right)^2=(0.4896)^2=0.2397$ and $p_d=\left(\frac{1.0417-1}{0.0817}\right)^2=(0.5104)^2=0.2605$, leaving $p_m=1-0.2397-0.2605=0.4998$. Separately, $u=e^{0.20\sqrt{2/12}}=e^{0.0816}=1.0851$ and $d=1/1.0851=0.9216$. Starting at $S=100$ ($100), the three nodes after one step sit at $108.51 (prob 0.2397), $100 (prob 0.4998), and $92.16 (prob 0.2605) — about half the probability mass stays essentially where it started, a much more realistic picture of a single month than a binomial tree's forced either/or.

Worked example 2: why it converges faster. Price a 1-year at-the-money call, $K=100$, $r=0$, $\sigma=20\%$, whose true Black-Scholes value is about $7.97. A binomial tree with 50 steps typically lands within a few cents but oscillates step-count to step-count; a trinomial tree with the same 50 steps typically lands within a cent or two and oscillates less, because it has one more degree of freedom per step to match both the mean and the variance of the underlying process simultaneously, rather than matching variance only approximately as a two-branch tree must.

Both the binomial and trinomial lattices are discrete stand-ins for the continuous path above; the trinomial version simply has one more lever per step to keep its discrete steps honest about the true path's variance.

What this means in practice

The extra branch is what lets desks place a barrier level exactly on a row of nodes, which fixes the jagged, slowly-converging pricing errors that plague binomial trees on knock-in/knock-out options. Trinomial trees are also a natural stepping stone toward Finite Difference Option Pricing, since the middle branch's role is essentially a discretised diffusion term, the same object a finite-difference grid solves directly.