I can declaim on math when drunk, but it might not make any sense! Unlike Aquarians, whose IQ jumps 20 points with every half bottle of Romanian vodka, mine goes in the opposite direction at roughly the same rate or higher. One rule of thumb: refrain from posting after alcohol ingestion.
OK, here is my final, and hopefully correct, answer. There is no KO vol that that satisfies the above criteria. The triangle relationship has zero, one, or two solutions for two sides and an angle. In this case there are zero solutions. If you hold the .3 (side a) and .09 (side c) fixed, then the minimum PEP/KO correlation (cos(angle C)) is around .945. If you hold .3 (side a) and .8 (cos(angle C)) fixed, then the minimum side c value is around .19. So either the implied cor is too low, or the implied KO/PEP is too low. Either way, you likely have a relative value/stat arb trade. This graph below illustrates: vector b (KO/USD IV) coming away from its common origin with vector a (PEP/USD IV) at an angle of 36.9 degrees (cos/corr = .8) has to cross (two solutions) or be tangent to (one solution) the circle inscribed by vector c (KO/PEP IV) with its origin at the other end of vector a. At an angle of 36.9 degrees or 64.5 radians, it (vector/side b) can't. Thus zero solutions.
Further to my last post on this thread: The min corr when holding a and c fixed comes from noting that the min corr (widest possible angle) comes when b is just tangent to the c circle (side b and side c are perpendicular). You can then use the Pythagorean Theorem to solve for side b: > sqrt(.3**2 - .09**2) [1] 0.2861818 Then plug that into the Law of Cosines, rewritten to put rho on the left hand side: > (.3**2 + .286118**2 - .09**2) / (2 * .3 * .2861818) [1] 0.9537266 Yielding ~.954 corr The min side c when holding a and angle C fixed comes from noting that the min length of c comes when a = b = .3 and solving for the sine of angles A and B (using the fact that the three angles add up to pi radians): > sin((pi - acos(.8)) / 2) [1] 0.9486833 Then plug that figure into the Law of Sines, rewritten to put c on the left hand side (sin(C) is obviously .6): > .6 / (0.9486833 / .3) [1] 0.1897367 Or about .19 Generally, if you are given two sides and and the angle between them, use the Law of Cosines to find the third side. If you are given two sides and an adjacent angle, use the Law of Sines.
Graph two posts above it incorrect. Labels "A" and "B" were switched. Too late to edit. Here is corrected version:
Elon is missing a few rocket scientists. Does intelligence make you a better trader? If so I need to hire some of you people!
Kev, i got to work with @sle on this problem and I wanted to post a real world example because the output im getting is iffy in a real world example. Im still going to use KO and PEP. Current Ivols PEP/USD 1 month = 16% KO/USD 1 month = 12% Forecasted 1 month Cor KO/USD, PEP/USD = 70% formula: c^2 = a^2 + b^2 - 2ab * cos(arccos(cor(a,b)) KO/PEPsigma^2 = KO/USDsigma^2 + PEP/USDsigma^2 - 2*KO/USDsigma *PEP/USDsigma*cos(arccos(.70)) In R: .12^2 + .16^2 - 2*.12*.16 * cos(acos(.70)) [1] 0.01312 Now when i look at the 1 month realized vol of KO/PEP (plotted below) the min value i get is .0235 over the past 20 years. I tried changing my cor value but even at 0 cor i get a value only .03. What i was expecting was a higher value that matches closer to what realized has been doing. Say .10. Ive even changed the vol numbers to .25 and .20 for KO and PEP but im not getting a value that would make sense in the real world To get the KO/PEP vol graph below in R: library(quantmod) getSymbols(c("KO", "PEP")) prices = na.omit(merge(Cl(KO), Cl(PEP))) ko.pep = prices[,1]/prices[,2] ko.pepVol = volatility(ko.pep, n = 20, N = 252, mean0 = T)