Fixed Income- Pathwise valuation

How to calculate the number of independent paths in a path-wise valuation method?

Schweser notes say 2(n-1), n being the number of periods.

However, at a number of places I don’t get the right answer using this formula? Help please!

It’s 2_n_ _ 2n __ −1 _.

With 1 period, there’s 21−1 =20 = 1 path (i.e., discount at today’s spot rate).

With 1 _ 2 _ period_ s _, there are _ 22−1 = _ 21 = 2 paths:

  • Up
  • Down

With 2 _ 3 _ periods, there are _ 23−1 = _ 22 = 4 paths:

  • Up, up
  • Up, down
  • Down, up
  • Down, down

With 3 _ 4 _ periods, there are _ 24−1 = _ 23 = 8 paths:

  • Up, up, up
  • Up, up, down
  • Up, down, up
  • Up, down, down
  • Down, up, up
  • Down, up, down
  • Down, down, up
  • Down, down, down

And so on.

A 30-year, monthly binomial tree (for valuing a 30-year, monthly-pay mortgage) will have _ 2360−1 = 2359 _ paths, or slightly more than 2,348,542,582,773,830,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

_ 1,174,271,291,386,915,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 _.

I’ll leave it to you to enumerate them all, but I’ll get you started:

  • Up, up, (356 _ 355 _ more ups), up, up

this is super helpful…thanks S2000magician…you are a saviour!

My pleasure.

i am confused now. in the Practise Problems in CFAI website. under Ahti Maalouf Case Scenario, the four-year bond requires 23 = 8 paths. Why is is not 24 = 16 paths

The interest rate in each node represents a one-period forward rate.

For a four-year bond, the final (4th year) cash flow (principal + coupon) will be discounted using the 1-year forward rate in Year 3, which means you only need the interest rate tree up to Year 3 only, hence there is only 2(4-1) = 8 paths.

As you should be, because I was mistaken. I was thinking of a tree for stock prices, not an interest rate tree.

Indeed, for a binomial interest rate tree, it should be 2n−1. I’ve fixed my earlier post.

Thanks for your confirmation.!

My pleasure.