Tuckman Chapter 7, risk neutral probability

[FedHan]

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]

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!

 

[FedHan]

Thanks David...although there is a typo, Price[1,1] should be $946.51 instead of $956.51

 

[BerndSE]

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]

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

 

[BerndSE]

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]

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

 

[BerndSE]

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]

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