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ScottyfromAussie
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Joined: April 7th, 2011, 3:20 am

Probably really basic

February 3rd, 2012, 3:34 pm

Hey guys,I came across an author who used a bit of a simple formula to estimate the probability of losing (PL) on two positions if you trade them together.Basically he said If PL of your strategy is 0.6 and you trade two different assets your overall risk is0.6*0.6=0.36The big problem being of course, he's assuming independence and thats not reality.So I thought, wellif the correlation between the two assets is 0 (independency) then the PL combined would be 0.36 (probability of both positions losing).if the correlation between the two assets was 1, then the PL combined would be 0.6, since its like buying two positions in the same asset.if the correlation between the two assets was -1, then the PL combined would be 0, since if one position made a profit, the other would have to make a loss.I then graphed the combined PL versus correlation and it formed a quadratic curve. From there its pretty simple to calculate the quadratic equation and then find the real probability of both positions losing.e.g. if you have a strategy that had a PL of 30%, with a correlation between the two assets of 70%. The probability of both positions losing is 0.2244.My question is,Does anyone know how to develop an equation that would incorporate 5 different assets so I can calculate their actual probability of losing? (as a function of the strategy PL and their correlations to each other).Its probably some intermediate calculus but I wasn't able to find anything when I googled it.p.s. Please note I am aware that correlations between assets continuously change. I don't expect the equation to reveal Gods plans for future events
Last edited by ScottyfromAussie on February 2nd, 2012, 11:00 pm, edited 1 time in total.
 
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secret2
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Joined: July 28th, 2010, 10:29 pm

Probably really basic

February 3rd, 2012, 5:35 pm

What do you think the equation is for the 2 asset case?
 
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acastaldo
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Joined: October 11th, 2002, 11:24 pm

Probably really basic

February 3rd, 2012, 6:10 pm

Quoteas a function of the strategy PL_i and their correlations to each otherA note of caution: Correlation is very helpful in understanding the inter-relationship between normally-distributed variables, but it is somewhat misleading and incomplete in describing relationships among 0/1 variables, especially when there are more than 2 involved.
 
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ScottyfromAussie
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Probably really basic

February 4th, 2012, 3:22 am

QuoteOriginally posted by: secret2What do you think the equation is for the 2 asset case?It varies depending on what values the correlation and PL are. even a slightly change will change the equation.But to solve for it is easy.If we think of my example below and graph it on a xy plane, we'll get three known points.(say our PL of our strategy is 0.6)PL wil be our y axis and correlation will be our x axis.when correlation between assets is -1, if one position makes a profit, the other position HAS to be make a loss, so the probability of both assets losing is 0)(-1,0) is our first pointwhen the correlation between asset is 0, the two assets are now independent, so solving for their combined PL is easy0.6*0.6 = 0.36(0,0.36) is our second pointfinally when the correlation between our assets is 1, the two assets now behave like they are the same, so taking a position in both is really like taking 2 positions in oneOur PL thus must be 0.6(1,0.6) is our third point.if you graph these points, you'll notice they form a curve, thus the equation is likely to be quadratic. And its very easy to solve for the quadratic once you know 3 points like we have above.In this case our equation is y = (-0.06x^2) + 0.3x +0.36So if we had a strategy that lost 60% of the time and a correlation between assets of 40%, our probability of both positions making aloss is 47.04%. This is a lot more realistic than just assuming the assets are independent and calculating the probability of both positions making a loss to be 36%As you can see, its not perfect (correlations change over time) but its better than just assuming independence.If you were to use this, I'd recommend you add a buffer region around the combined PL to allow for error, constantly update the correlation value and add a strategy for combating times when the market crashes and the correlation explodes.But my original question was, does anyone know how to make an equation that would incorporate 5 assets? 2 is easy but I don't know where to start with 5.
Last edited by ScottyfromAussie on February 3rd, 2012, 11:00 pm, edited 1 time in total.
 
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KajetanSikorski
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Probably really basic

February 6th, 2012, 2:00 am

EDIT Sorry my indicator variables are positive when there is a lose and negative when there is a profit. I don't know why I did it that way but I don't really wanna go through and change everything.Let me answer your question by first showing you the rigorous way to do the problem you know how to do. Basically the issue I have with your solution is that there isn't really an a priori reason to fit a quadratic. (I mean it makes sense to suspect that you would need a quadratic since second order moments are involved but there isn't really a rigorous justification.)In your 2 asset example there are fundamentally 4 (actually 3 due to symmetry) quantities involved. Probability that only asset 1 loses, probability and 1 and 2 lose, etc.p_1, p_12, p_2, p_0The last one being the probability neither loses. In principle p_1, and p_2 could be different but the way you stated your problem it seems like you're assuming they're the same.The random variables we're working with are X_1 and X_2: kind of indicator variables that can be 1 or -1 the former representing loss and the latter profit.What is the correlation given by? Just compute the Assuming q is the probability a particular asset loses, irrespective of what the other does, and a symmetry between X_1 and X_2 we have (after a bit of algebra)Since we're dealing with probability we also knowp_12 + p_1 + p_2 + p_0 = p_12 + 2*p_1 +p_0 = 1We also have the law of total probability.p_12 +p_1 = qSo we have a system of 3 equations and 3 unknowns.Solving this for p_12 we getp_12 = q (c + q - c q)You'll notice that this can be negative, for some choices of q and c. It's possible that's because I made algebra mistake, but more likely I think it just indicates that some choices of c are incompatible with some choices of q. For example if c = -1, then actually q HAS to be 1/2 just due to symmetry. Every time there is one asset that is up and one down and by observing that labeling them 1 and 2 is arbitrary you notice that only 1/2 makes sense for q (p_12 is negative if q<1/2, and I'm guessing p_0 is negative if q>1/2).So.....if you have three assets, you would do the same thing as I did except the variables would be p_0, p_1,p_2,p_3, p_12, p_23,p_13,p_{123} so if you assume symmetries again then variables would become p_{non at a time}, p_{one at a time}, p_{two at a time}, p_{three at a time}. So now you'd get a 4x4 system. So for the case you asked about you'd need to solve a 6x6 system of equations.
Last edited by KajetanSikorski on February 5th, 2012, 11:00 pm, edited 1 time in total.
 
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KajetanSikorski
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Probably really basic

February 6th, 2012, 2:05 am

By the way in the three asset case you would need up to third order moments, not just a correlation coefficient. If you had 4 assets you would need up to fourth order moments, etc. If you don't have/want to include those moments then you have a couple options as to what to do:- Use the principle of maximum entropy to choose a distribution among those compatible with the second order moments (or minimum cross entropy if for some reason you wanted to use a different prior than uniform)- Come up with bounds for the probability of loss instead of a strict value. I'm sure you could do this using one of the famous probability inequalities, perhaps the multidimensional chebyshev inequality? I would be interested to hear from others if they have any thoughts on this.
Last edited by KajetanSikorski on February 5th, 2012, 11:00 pm, edited 1 time in total.
 
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ScottyfromAussie
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Probably really basic

February 6th, 2012, 12:20 pm

Hi Kaje,Thanks for your response, it looks solid. I'll compute through when I've had a decent sleep Thanks again, Scotty