Tuckman Chapter 7, risk neutral probability

FedHan

New Member
Hi guys,
pardon the eventual lapse, but under Learning Statement "Explain how the principles of arbitrage pricing of derivatives on fixed income extend over multiple periods", deriving the risk-neutral probability for the first time step, p = 0.8024, is quite straightforward ( $925.21 * (1 + spot_rate_6months/2) = p * 946.51 + (1-p) * 955.78.

Im having an issue with the second time step, q.
Doing $925.21 * (1+ 5.15%/2) = q * (970.87) + (1-q) * 975.61....in this case however q = 45%ish..not the 64.89% from David lecture.

Tx /pls advise
fh
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi @FedHan

My example merely follows Tuckman's "Arbitrage in a multiperiod setting," so we've got two one-year prices:
  • Price[2,2] = 970.87 = $1000/(1+6.0%/2); i.e., face of $1,000 at 1.5 years discounted back six months to t = 1.0 year
  • Price[2,1] = 975.61 = $1000/(1+5.0%/2); i.e., face of $1,000 at 1.5 years discounted back six months to t = 1.0 year
q = 0.6489 is a solution that informs Price[1,0] and Price[1,1] as in:
  • Price[1,1] $946.51 = ($970.87*0.6489 + $975.61*0.3511)/(1+5.5%/2); i.e., 5.5% is the six month rate at T=0.5 under the "up" path. I hope that helps!
 
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BerndSE

New Member
Subscriber
Hi there,

i would like to add a question regarding the values in the Binomial Tree. I understand how we derive Price[0,0], Price [2,1], Price[2,2], Price[2,3], but I do not understand how to derive Price[1,1] =946.51 and Price[2,1] = 955.78.

In the bullet points above this step is not described neither, but it seems that this values are given when the probabilities are calculated.

I would appreciate a short response on how to derive the two prices. Thanks!
 

ShaktiRathore

Well-Known Member
Subscriber
At node(1,1) the tree branches out to nodes (2,2) and (2,1),6 mnth rate at (1,1) is 5.5% that is up to 6% at (2,2) and 5% at (2,1) so we discount 1000 at 6% at node (2,2) ,P(2,2)=1000/(1+.06/2)=1000/1.03=970.87 and discount 1000 at 5% at node (2,1),P(2,1)=1000/1+.05/2=975.61.P(2,0)=1000/1+.04/2=980.30 so P(1,0)=.6489*975.61+.3511*980.3/1+.045/2=955.78,P(1,1)=.6489*P(2,2)+.3511*P(2,1)/1+.055/2=.6489*970.87+.3511*975.61/1+.055/2=946.51 as at node (1,1) 6 mnth rate is 5.5% and 4.5% at node (1,0), as both are emerding rates from node (0,0) where 6 mnth rate 5% goes to 5.5% at node (1,1) and 4.5% at node (1,0).
Thanks
 
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BerndSE

New Member
Subscriber
At node(1,1) the tree branches out to nodes (2,2) and (2,1),6 mnth rate at (1,1) is 5.5% that is up to 6% at (2,2) and 5% at (2,1) so we discount 1000 at 6% at node (2,2) ,P(2,2)=1000/(1+.06/2)=1000/1.03=970.87 and discount 1000 at 5% at node (2,1),P(2,1)=1000/1+.05/2=975.61.P(2,0)=1000/1+.04/2=980.30 so P(1,0)=.6489*975.61+.3511*980.3/1+.045/2=955.78,P(1,1)=.6489*P(2,2)+.3511*P(2,1)/1+.055/2=.6489*970.87+.3511*975.61/1+.055/2=946.51 as at node (1,1) 6 mnth rate is 5.5% and 4.5% at node (1,0), as both are emerding rates from node (0,0) where 6 mnth rate 5% goes to 5.5% at node (1,1) and 4.5% at node (1,0).
Thanks

Hi Shakti -
Thanks for your answer and I aggree with your calculation given risk-neutral probabilities are known before calculating the nodes.
But then I would like to add a question: How do we calculate the risk-neutral probabilities in this example?
Kind regards, Bernd
 

ShaktiRathore

Well-Known Member
Subscriber
Yes
As David cited above these risk neutral probabilities informs price P(1,1) and P(1,0),
P(1,1)=q(970.87)+(1-q)975.61/1.0275=975.61-4.74q/1.0275=949.498-4.613q
P(1,0)=q(975.61)+(1-q)980.3/1.0225=980.3-4.69q/1.0225=958.728-4.588q
P(0,0)*1.025=925.21*(1.025)=q*P(1,1)+(1-q)P(1,0)=q(949.5-4.61q)+(1-q)(958.73-4.59q)=-.02q^2-13.82q+958.73=>.02q^2+13.82q-9.389=0
I am getting .q~.67875 off mark of actual .6489 because i used approximations here. David would have used excel to arrive at exact answer otherwise method to arrive at answet would be same,may it helps
Thanks
 
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BerndSE

New Member
Subscriber
Hi Shakti,

thanks again for your response.
I would agree with your calculations, if the same probabilities were used for the two paths/time steps (from T=0 to T=0.5 and from T=0.5 to T=1.0). Anyhow, in the tree showing the risk-neutral probabilities, the "probabilities" from T=0 to T=0.5 differ from those probabilities from T=0.5 to T=1.0). If you would include the probabilities (labeled p here from T=0 to T=0.5) in your last equation, you would have one equation with two unknowns.

=> P(0,0)*1.025 = q*p* (P2,0) + p*(1-q)*(P2,1) + (1-p)*q*P(2,1) + (1-p)*(1-q)*P(2,2)

I believe your number (q = .711) differs, because you are assuming same probabilities for the two time steps. q=.711 is also pretty much between .8025 (=p) and .6489 (=q).

Kind regards :)
 

ShaktiRathore

Well-Known Member
Subscriber
Hi Shakti,

thanks again for your response.
I would agree with your calculations, if the same probabilities were used for the two paths/time steps (from T=0 to T=0.5 and from T=0.5 to T=1.0). Anyhow, in the tree showing the risk-neutral probabilities, the "probabilities" from T=0 to T=0.5 differ from those probabilities from T=0.5 to T=1.0). If you would include the probabilities (labeled p here from T=0 to T=0.5) in your last equation, you would have one equation with two unknowns.

=> P(0,0)*1.025 = q*p* (P2,0) + p*(1-q)*(P2,1) + (1-p)*q*P(2,1) + (1-p)*(1-q)*P(2,2)

I believe your number (q = .711) differs, because you are assuming same probabilities for the two time steps. q=.711 is also pretty much between .8025 (=p) and .6489 (=q).

Kind regards :)
Hi please check my post above again,
I have rechecked calculations and my answer q=.6587 comes pretty close to actual answer .6489. Actually i solved the quadratic eqn incorrectly now i solved again to get q=.6587. Yes i assumed onle one unknown the price P(0,0) is given . I assumed P(1,1) and P(1,0) are not given ,just i gave u method how to solve for risk neutral probability may be two prices are given in the question i dont know as i dont get the details of the question above and yes two equations for two unknowns can be formed to solve for two different risk neutral probabilities.


Thanks
 
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