# Non-recombining lattice in non-markovian models

Brigo&Mercurio Interest Rate Models - Theory and Practice, 2nd edition, when treating not markovian HJM models, says the following "the approximating lattice will not be recombining and the number of nodes in the tree will grow exponentially with the number of steps".

I don't see the relation between the Markov property $$\mathbb{E}[{f(W(t))}|\mathcal{F(s)}]=g(W(s))$$ and the recombination of the approximating lattice. Does anyone know where I could find a proof?

• I think that “recombining” is not a property of Markovness. It is driven by the assumption on the jump sizes, no? – Kermittfrog Nov 27 '20 at 20:26
• @Kermittfrog Also Rendleman in the 11th Chapter of his book link says that "in HJM there is no requirement that the binomial tree of 1-period interest rates recombines" – Matteo Campagnoli Nov 28 '20 at 20:39

## 1 Answer

I think I might have found the solution to my own question. The Markov property as stated above has no direct relation with the recombination of the approximating lattice. However, if we consider the "traditional" meaning of Markovness, that is being memoryless, things become clearer.

Consider a binomial tree, where the random variable $$X$$ can either go up to $$Xu$$ with probability $$p$$, or go down to $$Xd$$ with probability $$1-p$$. Needless to say that for free-arbitrage reasons $$0 and $$u\ge 1$$.

On the second node, the random variable can be in only three configurations: $$Xu^2$$, $$Xud$$ or $$Xd^2$$. This happens only if $$X$$ is Markov, because the evolution of $$X$$ does not depend on the previous values of $$X$$. Hence, the lattice is recombining.

I know this is far from a rigorous proof, but at least the relation between Markovness and recombining lattices is evident and understable.