Recently I've been curious about volatility and leverage. Here's the problem: let's say you make or lose 10% of your total wealth based on the outcome of a coin flip. You make this bet a few times in a row. If you win then lose, you have 100*(1+10%) = $110 then 110*(1-10%) = $99. Likewise if you lose then win, 100*(1-10%) = $90 then 90*(1+10%) = $99. The more you play this game the more money you will lose, assuming the coin is fair. Basically you lose more when you're up and win less when you're down- both are bad.
This is one way of seeing why daily stock market returns must follow the log-normal distribution, which is right-skewed instead of a normal distribution. (in contrast to this Dutch book approach, you can also understand it as continuous returns are normally distributed, but daily/periodic are log-normal). If returns were symmetric, like the coin flipping game above, stocks would be pulled toward zero over time.
And this would increase with volatility. For example, lever the coin flipping game 2x, so it's 20% depending on if you win or lose (even without financing costs). 100*(1.2)*(.8) = $96 = 100*(.8)*(1.2) so now you end up with even less. If the game is positive expectation (such as +15%, -5%), i.e. you have an alpha generating strategy in market terminology, then leverage is a tradeoff between this downward pull of volatility and the upward pull of positive expected value.
It's obvious that 0% leverage give $0 downward pull, we showed 10$ is $1 down, 20% is $4 down, a quick calc shows 100*1.3*.7 = $91 i.e. $9 down, etc. This simple plot illustrates the exponential trend of losses with respect to leverage/volatility:
Fortunately the effect of levering up positive expectation is usually higher than the downward pull of volatility. The losses above are in terms of $ but since the starting amount is 100, in terms of percents it would be the same. The game starts out at 10% volatility so 20% corresponds to 2:1 leverage, 30% 3:1, 10% 1:1 (no leverage), and 0% volatility means all cash/out of the market. If you have theoretical alpha of say 10%, 1:1 leverage will pull it down 1%. 2:1 leverage will pull it down 4%, but also double it to 20% for a net of +7% [=(20-10)-(4-1)]. 3:1 increases the volatility to 30% as above decreasing the return 9% but it raised it to 30% theoretical for a net of +5% [=(30-20)-(9-4)].
You can see in the example above that leverage helps but the benefit decreases (+7%, then +5%, then ...). The alpha rises linearly with respect to leverage but returns are eaten away polynomially (degree 2) with respect to leverage on the bottom by volatility. Eventually the polynomial curve must cross the linear one and increasing leverage will then harm returns. The Kelly formula essentially finds this optimal crossover point. This is a toy example but perfectly applicable to trading.
In the next note I will analyze the ultrashort ETF, SKF. It's very interesting because it has to replicate the returns of the financial sector, leveraged 2x but inversely. So if the financial sector, XLF, goes up 3%, SKF has to go down 6% etc. The really intruiging thing to remember is that XLF's returns must be right skewed, log-normal, or else it would steadily fall by te effect of volatility. However SKF is -1*XLF meaning that it is left skewed and should therefore be negative expectation, and furthermore suffer extra volatility problems because it is double leveraged. My question is how can an creation like SKF be possible in a no-arbitrage environment if it always loses money. And if it is possible, where does the money go that it's losing... {can I have it?}
Topic:
Automated Trading,
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Money Management,
Stocks
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6 comments:
"If returns were symmetric, like the coin flipping game above, stocks would be pulled toward zero over time."
Stock returns are not only driven by random movements as dictated by the lognormal distribution. That is only the stochastic portion of returns (sigma*dX). There is usually some underlying "drift" or return (Mu) to the returns.
Base equation of most option pricing models:
dS/S = Mu*dT + sigma*dX
Anon,
Yes, thanks for making sure that's explicitly mentioned.
Regards,
Max
Hi Max,
Nice blog! As for your post, I think you are somewhat confused if I understood you correctly.
In your coin flipping example, you said "The more you play this game the more money you will lose, assuming the coin is fair." This is not true. Your expectation is always an even $100 no matter how many times you flip the coin. What is true is that the most likely outcome is $99 after two flips (so you lose $1).
You said "Basically you lose more when you're up and win less when you're down- both are bad." This is true. However you also win more when you are up are lose less when you are down -- both are good. When you flip two times, the possible outcomes and there probabilities are:
$100*(1.1)^2 = $121 --- 0.25
$100*(1.1)(0.9) = $99 --- 0.5
$100*(0.9)(0.9) = $81 --- 0.25
The expectation is exactly $100. In summary, high volatility does not imply losing. It does imply that the you will lose much more than 50% of the time. However this is compensated by the higher winnings when you win, so your expected winning/loss is always zero.
Note that coin flipping is in a discrete setting. The 10% return is really "accumulated return" in a continuous setting. The mean log-return should be
( log(1.1) + log(0.9) )/2 = -0.005
In general, using Taylor series approximation it is easy to get
( log(1+v) + log(1-v) )/2 = -v^2/2
where v stands for volatility. So in the coin flipping example, after 1000 flips, the most likely outcome is
$100*exp(-0.005*1000) = $0.67
This is also the median outcome, so there is about 50% chance you will lose at least $99.33! However, the mean outcome is still $100, so you will win/lose $0 on average.
Xun
Xun,
Thanks, I always hope to learn things I've missed by asking for comments.
I would like to correct the note. Could you tell me where this equation comes from,
"the most likely outcome is
$100*exp(-0.005*1000) = $0.67"?
How would you fix the note in general? I realize from what you've mentioned that I was wrong to conclude that higher volatility implies lower expectation because I ignored 1.1*1.1>1.2 and .9*.9>.8.
My goal is to understand why SKF and the other ultrashort ETFs get depressed over periods of more than a few days and to learn about probability. If you propose changes, I will be able to understand without much/any explanation.
Thanks again
Regards,
Max
Hi Max,
I am new to trading and trying to learn. I am not familiar with things like ultrashort ETF, so I am afraid that I don't have a better understanding than you do. I will try some explanations in my comment to your newer post.
" Could you tell me where this equation comes from, "the most likely outcome is $100*exp(-0.005*1000)=$0.67"? "
That equation is equivalent to: $100*(0.9^500*1.1^500)=$0.67. (Remember how we got the number -0.005?) Why did I use that equation then? Well, it emphasizes that the outcome after 1000 flips follows (approximately) a log-normal distribution.
For simplicity let's say we start with $1. After n flips we'll have
(1+X_1)(1+X_2)...(1+X_n)
where X_1, ..., X_n are independent random variables with identical distributions:
Prob(X_i = 0.1) = Prob(X_i = -0.1) = 1/2
Taking log of the outcome, we get a more familiar distribution:
Y_n := log(1+X_1) + ... + log(1+X_n)
This is a binomial distribution with mean (-0.005)*n. (Remember the mean of log(1+X_i) is -0.005.) The process {Y_n} is a random walk with a downward drift.
Y_n can be approximated by a normal distribution with the same mean. Normal distribution is all nice and symmetric: mean = median = mode = (-0.005)*n. (Mode is the most likely outcome.)
Now the outcome after n flips has distribution exp(Y_n) which can be approximated by a log-normal distribution LN. The graph of the probability density function (p.d.f.) of LN is left skewed as seen in your picture. For the log-normal distribution mean > median > mode, and they are exp(0), exp(-0.005*n) and exp(-0.015*n) respectively.
How useful is this approximation? The mean and median of exp(Y_n) are the same as those of LN respectively. However the mode of exp(Y_n) is different from that of LN. The reason is that exp(Y_n) is a discrete distribution whose outcomes are NOT evenly spaced, while its approximation LN is a continuous distribution.
It should be obvious that for any discrete distribution Y we should have median(exp(Y))=exp(median(Y)), and mode(exp(Y))=exp(mode(Y)). So both of the median and the mode of exp(Y_n) are equal to exp(-0.005*n).
However for LN, the mode is where the derivative of its p.d.f. equals 0, and this turns out to be exp(-0.015*n).
I hope this helps.
Xun
Thanks for the very detailed explanation Xun.
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