Calibrating Binomial Interest Trees

CFA Level II Fixed Income
#1

Hey everyone,

I am struggling a bit with the correct calibration of the binomial interest tree. My understanding of the procedure is this (given one does not use Excel Solver but the calculator instead):

  1. First node is obviously equal to the 1 year spot rate = 1 year par rate

  2. 2nd year nodes are:

a) upper one: F(1,1) rate x e^sigma

b) lower one: F(1,1) rate x e^(-sigma)

  1. 3rd year nodes are:

a) upper one: F(2,1) rate x e^(2sigma)

b) middle one: F(2,1) rate

c) lower one: F(3,1) rate x e^(-2sigma)

This would satisfy the rule that the nodes in one year have to be 2 Sigmas apart. But with this approach I cannot reproduce the tree in the CFA Reading 36, 3.4. Am I doing something wrong?

Granted the numbers in the curriculum are obtained with Excel Solver, but my numbers with the calculator are very far off.

Thanks,

Tartaglia

#2

For the sake of completenes, these are the forward rates from the example in Reading 36, and the respective tree (sigma=0.15):

Current one-year rate1.000%

One-year rate, one year forward 1.400%

One-year rate, two years forward 1.350%

One-year rate, three years forward 1.860%

One-year rate, four years forward 3.500%

Time 1 1.0000%

Time 2 1.6121% 1.7862%

Time 3 2.8338% 1.1943% 1.3233%

Time 4 2.0994% 0.9803% 1.5552% 1.1521%

With my method I get for the first year:

1.2049 and 1.6265

Is this just a rounding error or a bigger problem?

For the third year, I would use in the middle section the actual forward rate, but that seems very far off.

#3

It’s a bigger problem.

The rates in the tree are not centered on the forward curve. They’re merely close to being centered on the forward curve.

#4

Does this mean, that the rates are set arbitrary around the forward curve? If the only condition is that the nodes in one time period are e^2sigma apart from each other, there could be an infinite amount of possible solutions, right?

If so, what does this imply for the exam (apart from the discussion of how likely it is that this is tested, I just want to understand this)? Is my approach above an acceptable way to calibrate the tree? How is everyone else calibrating the tree?

#5

The tree has to price par bonds correctly, so it’s not arbitrary.

You won’t have to calibrate a tree on the exam. They’ll give you the tree (or at least one rate at each time, and the volatility).

#6

But just so I understand the procedure correctly:

If I were given the forward rates, say F(2,1)=1.4% and interest rate volatility sigma=0.15, as in the example above how do I come up with the respective nodes in period 2? Is the method I suggested above correct?

#7

I just re-read chapter 3.4 in Reading 36 again and I do see your point now about pricing the par bond correctly.

There, they also begin their construction of the tree with two values for the nodes in period 2 as I suggested in my previous post above. However, then they double check if those node values corresponded with a specific term structure, i.e. price the benchmark par bond correctly.

If you use the values I suggested above, you do not get the correct value for the benchmark par bond, so you would have to try new values, until you finally get the value for the benchmark bond. I guess this is where Excel Solver comes into play and this is where this method is irrelevant for the exam, correct?

#8

hallelujah

#9

That’s pretty much it.

In fact, for the two rates at time 1 you get a quadratic equation in (say) the lower rate r1,L, which you solve using the quadratic formula, then get r1,H from that (r1,H = r1,L × e2σ). Similarly, because there are formulae for the roots of general cubic (3rd degree) and quartic (4th degree) polynomials, you could solve (analytically) for the rates at times 2 and 3 as well. Beyond that, you have to use numerical methods. And, to be fair, everyone uses numerical methods for times 1, 2, and 3 as well.

#10

Yeah, that makes sense.

Just to make sure I understand this properly (and because I could not find it in the curriculum- and I am struggling to figure this out on my own), could you provide the inputs for the quadratic formula to compute r1,L or r1,H?

As far as I understand, you know the forward rate, F(1,1) and you know the condition that needs to hold (r1,H = r1,L × e2σ). How would the quadratic equation look to find the analytic solution for t=1?

#11

You know the 2-period par rate, c2, and the par/spot/forward rate for T = 0: r0.

To value a 2-period par bond, you have:

1,000 = 0.5{1,000(1 + c2) / [(1 + r1,L × e2σ)(1 + r0)] + 1,000(c2) / (1 + r0)}

  • 0.5{1,000(1 + c2) / [(1 + r1,L)(1 + r0)] + 1,000(c2) / (1 + r0)}

If you multiply both sides of this equation by (1 + r1,L × e2σ)(1 + r1,L)(1 + r0), you get:

1,000(1 + r1,L × e2σ)(1 + r1,L)(1 + r0) = 0.5[1,000(1 + c2)(1 + r1,L) + 1,000(1 + c2)(1 + r1,L × e2σ)]

This is quadratic in r1,L.

Let’s simplify the notation:

  • a = e2σ
  • b = 1 + r0
  • c = 1 + c2
  • x = r1,L

Then we have:

1,000(1 + ax)(1 + x)b = 0.5[1,000_c_(1 + x) + 1,000_c_(1 + ax)]

1,000_b_[1 + (a + 1)x + ax2] = 500c + 500_cx_ + 500c + 500_acx_

2abx2 + [2_b_(a + 1) − ac – c]x − 2_c_ = 0

2abx2 + (a + 1)(2_b_ − c)x − 2_c_ = 0

#12

I have a lot to learn!

#13

Thank you so much S2000, this makes much more sense than the vague version in the curriculum.

Then at the end we can just solve the quadratic equation and we’ll have two solutions of which one we can choose one (or try out both).

Thank you for taking the time to elaborate on this topic!!

#14

So do I, my friend.

So do I.