Tuckman's three step binomial

[[email protected]]

Hi David,
In the study notes and spreadsheet pack, I am struggling to understand how prices were calculated for the 3 step binomial (tab 29,7) for 946,51 and 955,78.
I get how to calculate (q) and (q-1), but this requires us to know 946,51 and 955,78 values from the earlier period.
But using the risk-averse values of (p) and (p-1) i.e. 80%/20%, I am struggling to work out the 946,51 and 955,78 and how they can be derived from the 925,21 T0 value.
Can you help?
Thanks

 

[David Harper CFA FRM]

Hi @[email protected] Yes, great observation: my XLS does not actually perform the ultimate calculation (it is a feature I'd love to add at some point, but time has not permitted). Tuckman explains this in Chapter 7 but basically, if you look at the tree below, and we use the node[state, row] convention where (eg) P[1,0] = P[1st step state to the right, zero row starting from bottom] = $955.78, then in terms of the final 3-state tree, all of the node values are obvious immediately EXCEPT for P(1,1) and P(1,1) if the goal is to solve for risk-neutral probabilities (reminder: risk neutral probabilities are the probabilities that return a discounted expected value that is equal to the observed market price). My red circle is trying exactly to illustrate the idea of risk-neutral probabilities: we are "forcing" the P[0,0] node to equal 925.21 because presumably 925.21 is the correct market price as given by discounting at the 1.5 year spot rate (it's easy to miss this: the point of the 1.5 spot rate is to inform a "market price" which is then plugged into the tree). As Tuckman explains, then, we have two unknowns here, P(1,1) and P(0,1) but we have three relationships that involve them such that we can solve for q = 0.6489:

• [.8024*P(1,1) + 0.1976*P(1,0)]/(1+5%/2) = $925.21 (equation 7.14)
• [P(1,1) = $970.87*q + $975.61*(1-q)]/(1+5.5%/2) (equation 7.15)
• [P(1,0) = $975.61*q + $980.39*(1-q)]/(1+4.5%/2). (equation 7.16)

As Tuckman writes, "While equations (7.14) through (7.16) may appear complicated, substituting (7.15) and (7.16) into (7.14) results in a linear equation in the one unknown, q. Solving this resulting equation reveals that . Therefore, the risk-neutral interest rate process may be summarized by the following tree ...." I hope that clarifies! Thank you,

 

[QuantMan2318]

Hi @[email protected]

1. I am going under the assumption that you are not able to find out the value of the Bond in the second node (t = 1), you can arrive at the value, not from $925.21 but from the values at the right

2. You have to go from right to left to arrive at the value of the Bond at time 0

3. Therefore, take the two values: 970.87 and 975.61, the former is multiplied by the risk neutral probability of an up move at t=2 (0.6489), while the latter by risk neutral probability of a down move at t=2(0.3511), (you are assuming here that the risk neutral probability is given), this is discounted at the spot rate at t=1 for up move at 5.50% for one year

4. As the risk neutral probability is calculated from the Rf rate and the probability of up and down move, it is taken as a given here, or in other words, we assume that it has been implied from the market prices

we can either derive the price from the probability or the probability from the price but not both

Hope this helps

Thanks

 

[QuantMan2318]

P.S Please take my explain in addition to the great note that @David Harper CFA FRM has written

 

[[email protected]]

Thanks both. I am not sure if my algebra skills will be fit for purpose in the exam (or out of the exam!) to create and solve for 3 equations in the exam!

 

[silver7]

Hi @David Harper CFA FRM
For the second step: replicating portfolio in Page 18. How did you get all those values at around 600 in (blue/purple highlight)? I could not follow from here.


Thanks.

 

[David Harper CFA FRM]

Hi @silver7 It's a solution in two variables with two functions. We need the current portfolio of F(0.5) + F(1.0) that will have a payoff in six months of either zero or $3.00 (to replicate the option). So we need:

F(0.5) + 0.97324*F(1) = $0;
F(0.5) + 0.97800*F(1) = $3.0

-0.97324*F(1) + 0.97800*F(1) = $3.0
F(1)*[-0.97324 + 0.97800] = $3.0
F(1) = 3.0/[-0.97324 + 0.97800] = $629.34
And since F(0.5) + 0.97324*F(1) = $0 --> F(0.5) = -0.97324*F(1) = -0.97324 * $629.34 = $612.50

... although my XLS is dynamic to changing assumptions., so it uses the formula not the numbers. I hope that's helpful,

Here is Tuckman explaining this:

"To price the option by arbitrage, construct a portfolio on date 0 of underlying securities, namely six-month and one-year zero coupon bonds, that will be worth $0 in the up state on date 1 and $3 in the down state. To solve this problem, let and F1 be the face values of six-month and one-year zeros in the replicating portfolio, respectively. Then, these values must satisfy the following two equations:

F(0.5) + 0.97324*F(1) = $0 (7.3)
F(0.5) + 0.97800*F(1) = $3.0 (7.4)

Equation (7.3) may be interpreted as follows. In the up state, the value of the replicating portfolio’s now maturing six-month zero is its face value. The value of the once one-year zeros, now six-month zeros, is .97324 per dollar face value. Hence, the left-hand side of equation (7.3) denotes the value of the replicating portfolio in the up state. This value must equal $0, the value of the option in the up state. Similarly, equation (7.4) requires that the value of the replicating portfolio in the down state equal the value of the option in the down state." -- Tuckman, Bruce; Serrat, Angel. Fixed Income Securities: Tools for Today's Markets (Wiley Finance) (pp. 209-210). Wiley. Kindle Edition.

 

[[email protected]]

Hi David,
Just to double check, I get how the 3 equations can be used to solve for (q), but this requires us knowing (p) i.e. .8024 and .1976.
How are these (p) values derived before we then get to the (q) values?
Thanks

Hi @[email protected] Yes, great observation: my XLS does not actually perform the ultimate calculation (it is a feature I'd love to add at some point, but time has not permitted). Tuckman explains this in Chapter 7 but basically, if you look at the tree below, and we use the node[state, row] convention where (eg) P[1,0] = P[1st step state to the right, zero row starting from bottom] = $955.78, then in terms of the final 3-state tree, all of the node values are obvious immediately EXCEPT for P(1,1) and P(1,1) if the goal is to solve for risk-neutral probabilities (reminder: risk neutral probabilities are the probabilities that return a discounted expected value that is equal to the observed market price). My red circle is trying exactly to illustrate the idea of risk-neutral probabilities: we are "forcing" the P[0,0] node to equal 925.21 because presumably 925.21 is the correct market price as given by discounting at the 1.5 year spot rate (it's easy to miss this: the point of the 1.5 spot rate is to inform a "market price" which is then plugged into the tree). As Tuckman explains, then, we have two unknowns here, P(1,1) and P(0,1) but we have three relationships that involve them such that we can solve for q = 0.6489:

• [.8024*P(1,1) + 0.1976*P(1,0)]/(1+5%/2) = $925.32 (equation 7.14)
• [P(1,1) = $970.87*q + $975.61*(1-q)]/(1+5.5%/2) (equation 7.15)
• [P(1,0) = $975.61*q + $980.39*(1-q)]/(1+4.5%/2). (equation 7.16)

As Tuckman writes, "While equations (7.14) through (7.16) may appear complicated, substituting (7.15) and (7.16) into (7.14) results in a linear equation in the one unknown, q. Solving this resulting equation reveals that . Therefore, the risk-neutral interest rate process may be summarized by the following tree ...." I hope that clarifies! Thank you,

 

[David Harper CFA FRM]

Hi @[email protected] Tuckman retrieves p = 0.8024 in the prior step which depends on utilizing the observed one-year bond (market) price of s(1.0) = $950.42, see top row. It's not shown directly on the exhibit above, but in six months, the current one-year bond with have a price of either $973.24 = 1,000/(1 + 5.5%/2) or $978.00 = 1,000/(1+4.5%/2), such that [$973.24*p + $978.00*(1-p)]/(1+5.0%/2) = $950.42, solves for p = 0.8024. This is all in Tuckman .