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Give the reasoning for using recombining trees when pricing options.

FRM® Part II · Market Risk Measurement and Management · Arbitrage Pricing with Term Structure Models · FRM2.MR.12.g

The idea

In a non-recombining binomial tree an up move followed by a down move leads to a different node than a down move followed by an up move. So the tree has 2ⁿ nodes at step n. In a recombining tree the two orders reach the same node, so step n has only n + 1 nodes. The total through n steps, counting the start, is written Total. Total = (n + 1)(n + 2) ÷ 2. That growth is polynomial instead of exponential, which is what makes small time steps feasible. Recombining requires the rate after up-down to equal the rate after down-up. This is true for additive or multiplicative moves with constant sizes. The limitation is path dependence. A payoff depending on the route taken, such as a prepayment-dependent mortgage cash flow, cannot be valued on a plain recombining tree.

Formulas

Total nodes in a recombining tree through n steps

Total = (n + 1) × (n + 2) ÷ 2

n
the number of steps, with the starting node counted

Step n alone has n + 1 nodes, against 2ⁿ in a non-recombining tree. Recombining needs up-down and down-up to give the same rate.

A practice question

From the question bank, with the reason each wrong option is wrong.

A recombining binomial tree of six-month rates covers five years in ten steps. The number of distinct nodes at the final date (step 10) is:

  1. 10Why not: Counts the steps rather than the nodes; a recombining tree has n + 1 nodes at step n.
  2. 20Why not: Doubles the number of steps; node count grows by one per step.
  3. 1024Why not: That is 2^10, the number of terminal nodes of a non-recombining tree.
  4. 11 - correct

Step n of a recombining tree has n + 1 nodes (0 to n up moves), so step 10 has 11 nodes.

A harder one

A four-step binomial tree, counting the starting node and the nodes at each of the four later dates, is built in two ways: non-recombining and recombining. The number of nodes the recombining tree saves is closest to:

  1. 16 - correct
  2. 15Why not: Reports the number of nodes in the recombining tree (1 + 2 + 3 + 4 + 5) rather than the saving.
  3. 31Why not: Reports the number of nodes in the non-recombining tree rather than the difference.
  4. 11Why not: Compares only the final date: 16 non-recombining terminal nodes minus 5 recombining ones, ignoring the earlier dates.

Non-recombining: 1+2+4+8+16 = 31. Recombining: 1+2+3+4+5 = 15. Saving: 31 - 15 = 16.

The hardest tier

In a recombining binomial tree with five steps, consider the node reached after exactly 2 up moves and 3 down moves. In a non-recombining tree, how many separate nodes would represent this same state?

  1. 6Why not: Reports the number of nodes at step 5 of the recombining tree (n + 1), not the paths to one node.
  2. 32Why not: Reports 2^5, the total number of paths through the tree.
  3. 10 - correct
  4. 20Why not: Counts ordered selections (5 x 4) instead of combinations, double-counting each pair of up moves.

The number of paths with 2 ups out of 5 steps is C(5,2) = 10; a non-recombining tree values this state 10 times.

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