Counting Distinct Nodes in Recombining Binomial Trees
Summary
The note explains how the relationship between the up and down factors determines whether a binomial price tree recombines. When the factors are reciprocals, an up move followed by a down move returns to the same value as the reverse sequence. After seven steps, this produces eight distinct values, one for each possible count of up moves.
When the factors do not multiply to one, move order affects the result, so the tree has a distinct node for every path: 128 after seven steps. The comparison illustrates why recombining trees require far less computation as the number of steps grows. The note also says the same recombining distinction applies to trinomial trees. Its node counts assume the stated conditions on the factors; other parameter relationships can also create recombination.
Key ideas
- Reciprocal up and down factors make an up move and a down move cancel regardless of their order.
- A recombining binomial tree has one distinct value for each possible number of up moves.
- With seven steps, the example has eight recombining values and 128 non-recombining values.
- Non-recombining tree size grows exponentially with the number of steps, while the example's recombining tree grows linearly.
- The same recombination concept applies to trinomial trees.
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Full text
# Finding distinct possible values in binomial tree
# Finding distinct possible values in binomial tree
I wonder how to solve this problem. Lets say we have a binomial tree with the following parameters:
$u=1.25,\ d = 1/u,\ T=15$.
How many distinct possible values are there for $X_{7}$?
## Answer by Kevin (score 2)
https://quant.stackexchange.com/a/48689
Noob2 has given the answer, a recombining tree has $N+1$ nodes after $N$ time points. In our case, you have $8$ distinct possible values after $7$ time steps. I will next illustrate the difference between a recombining and non-recombining tree.
For a binomial tree, if $u=\frac{1}{d}$, then an up-move followed by an down-move gives the same result as a down-move followed by an up-move (since $ud=1$): $S_0ud=S_0du=S_0$. This is a huge numerical advantange since the amount of potential nodes grows only linearly with the number of time steps.
If $ud\neq1$, then the order of up- and down-moves occurrring does matter and the amount of nodes grows exponentially. In particular, $S_0ud\neq S_0du\neq S_0$. After $N$ time steps, there are $2^N$ distinct nodes. So the difference between recombining and non-recombining is (for your example of $N=7$) the difference between $7+1=8$ and $2^7=128$. For numerical implementation, the former is of course much easier and less costy: just set $N=100$. Then, a recombining tree has $101$ nodes but a non-recombining tree has roughly $10^{30}$ nodes.
The bottom line ist that a recombining tree ``re-uses'' nodes it has already created and comes back to them. A non-recombining tree however creates for every existing node two new ones which leads to exponential grow. The same issue of (non-)recombining trees equally applies to the trinomial trees where you allow for an up/middle/down-move.
This image illustrates the amount of nodes of a recombining tree (the case you have since $ud=1$)
This image illustrates the amount of nodes for a non-recombining treeShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.