Usually you'd go in the other direction: given IV of KO and PEP and a forecast correlation of KO and PEP, you'd compare the calculated IV of KO/PEP vs its, say realized-kernel of overlapping 5 min bar RV. If there is a significant difference you have a stat-arb. High correlations are more predictable than vol, that is the motivation:
NEVER trust ChatGPT. I have a subscription to it, but use it only for code translation. It gets even that function wrong, but it at least outlines the solution. It tends to invent R/Python/Matlab packages and functions, and also to use functions that actually exist in ways that don't produce the desired returns. ChatGPT is no substitute for actual knowledge.
As always kevin, thank you. Although, why is it that KO vol is lower than PEP (.23)? Why could it not be 7 vols above -> .37? Im actually using this for Crypto since uniswap does have BTC/ETH. I can use current BTC ivols and uniswap to calc ETH vol. Side note, it does seem we can use BTC/ETH, BTC/USD, ETH/USD from uniswap (instantaneous vols) to get a good estimate of implied corr between the pairs. Would you also be able to help me solve for the instantaneous implied corr between BTC/USD and ETH/USD given the 3 three uniswap pools?
You are right, there are two solutions, I mis-read your question. The two are actually .21 and 39, depending on the angle (correlation) between KO and USD. Here is the calc: angleA = asin(sideA * sin(angleC) / sideC) angleB = pi - angleA - angleB sideB = sideC * sin(angleB) / sine(angleC) = .38997 angleC is the angle opposite side C (ie angle between KO and PEP or acos(.8)) the other solution is .3 - (.38997 - .3) or .21003 Sorry about that first reply, I should really read the question more carefully before firing off a response.
I mention this not to enlighten whatsoever and it is purely coincidental I'm sure, but .23 (original calc) and .39 are close to fibs .236 and 0.382
Kevin, if it is not a right angle triangle, dont we need 3 pieces of info? I understand that we have Side A (.3) and angle C (acos(.8)) but wont we need one other side or angle to solve? Unsure how you came up with side C in first equation I tried setting: SideA = .3 SideB = .09 Angle C = acos(.8) = 64.3 And here is the triangle i am getting back
acos returns angle in radians. > pracma::rad2deg(acos(.8)) [1] 36.8699 You are plugging in radians as degrees. Also, acos(.8) rounds to 64.4 radians
I am now getting your original answer of .234 for Side C. However when I use the formula you mentioned above I am getting .62. angleA = asin(.3 * sin(.368699) / .234) 2angleB = pi - .483 angleB = 1.32 sideB = .234 * sin(1.32) / sin(.368699) sideB = .625 Could you help me figure out where im going wrong in order to get the .389 that you mentioned?
It is quite possible I am wrong again. I have guests over and have been drinking all afternoon. I'll take a look at it when I have had less wine and give you a definitive answer.