risk-neutral prop, three term model, Tuckman Chap. 7

[michael4129]

Hi David and FRM allies!

Could you explain to me how you come to the probability of 0,6489 in the comprehensive market risk notes page 36 or Excel File P2.Market-Tuckman--Chapter-7, Sheet 29.7. Three Steps?

This probability is the important factor to determine the risk neutral prices for node 1 and the final risk-neutral price.

As there is no formula embedded in cell F17, I thought that you might have derived the probability from solving it with Excel solver?

Thanks!

Michael

 

[David Harper CFA FRM]

Hi Michael,

I didn't use solver, I cheated and just used the input that Tuckman gives in the assignment: q = 0.6489.

It's just an extension of the one-step p, for which i did use a formula (on the first tab of the workbook). But in this case, we have (Tuckman 7.14 to 7.16):

• The unknown Price[1,1], which is $946.51 after we solve for q but is unknown when solving for q, = [$970.87*q + 975.61*(1-q)]/(1+0.055/2)
• The unknown Price[1,0], which is $955.78 after we solve for q but is unknown when solving for q, = [$975.61*q + 980.39*(1-q)]/(1+0.045/2)
• $925.21 = (0.8024*P[1,1] + 0.1976*P[1,0])/(1+0.05/2); i.e., bootstrapping the previously derived 80.24%. So, there is one equation with one unknown (q), that I was too lazy to solve, because Tuckman gives it as 0.6489:
• $925.21 = (0.8024*[$970.87*q + 975.61*(1-q)]/(1+0.055/2) + 0.1976* [$975.61*q + 980.39*(1-q)]/(1+0.045/2))/(1+0.05/2). I hope that helps, thanks!

 

[michael4129]

I see. You simply took the probabilites for the first node (0.8024; 0.1976) as given. Then it is easier to compute the q = 0.6489 for node 2. Thanks for clarification.

By the way, if you would use Solver for the formula (solving comprehensively for node 1 and 2):
=(((D18*($970.87*F17+975.61*(1-F17))/(1+0.055/2)))+((1-D18)*($975.61*F17+980.39*(1-F17))/(1+0.045/2)))/(1+0.05/2)

changing cells = probabilities
target value in cell B 27 = 925.21

then you would come to rather different probabilities:
Prob[1,1] = 0.90036 [comparing it to: 0.8024]
Prob[1,0] = 0.09964 [comparing it to: 0.1976]
Prob[2,1] = 0.45172 [comparing it to: 0.6489]
Prob[2,2] = 0.54828 [comparing it to: 0.3511]

So I guess there will be more than one solution...