NOTE: Don’t believe everything I said in this comment! I elaborate on some of the problems with it in the responses, but I’m leaving this original comment up because I think it’s instructive even though it’s not correct.
There is a theoretical account for why portfolios leveraged beyond a certain point would have poor returns even if prices follow a random process with (almost surely) continuous sample paths: leverage decay. If you could continuously rebalance a leveraged portfolio this would not be an issue, but if you can’t do that then leverage exhibits discontinuous behavior as the frequency of rebalancing goes to infinity.
A simple way to see this is that if the underlying follows Brownian motion dS/S=μdt+σdW and the risk-free return is zero, a portfolio of the underlying leveraged k-fold and rebalanced with a period of T (which has to be small enough for these approximations to be valid) will get a return
r=kS(T)S(0)−k=kexp(ΔlogS)−k
On the other hand, the ideal leveraged portfolio that’s continuously rebalanced would get
ri=(S(T)S(0))k−1=exp(kΔlogS)−1
If we assume the period T is small enough that a second order Taylor approximation is valid, the difference between these two is approximately
E[ri−r]≈k(k−1)2E[(Δlog(S))2]≈k(k−1)2σ2T
In particular, the difference in expected return scales linearly with the period in this regime, which means if we look at returns over the same time interval changing T has no effect on the amount of leverage decay. In particular, we can have a rule of thumb that to find the optimal (from the point of view of maximizing long-term expected return alone) leverage in a market we should maximize an expression of the form
kμ−k(k−1)2σ2
with respect to k, which would have us choose something like k=μ/σ2+1/2. Picking the leverage factor to be any larger than that is not optimal. You can see this effect in practice if you look at how well leveraged ETFs tracking the S&P 500 perform in times of high volatility.
I didn’t follow the math (calculus with stochastic processes is pretty confusing) but something seems obviously wrong here. I think probably your calculation of E[(Δlog(S)2)] is wrong?
Maybe I’m confused, but in addition to common sense and having done the calculation in other ways, the following argument seems pretty solid:
Regardless of k, if you consider a short enough period of time, then with overwhelming probability at all times your total assets will be between 0.999 and 1.001.
So no matter how I choose to rebalance, at all times my total exposure will be between 0.999k and 1.001k.
And if my exposure is between 0.999k and 1.001k, then my expected returns over any time period T are between 0.999kTμ and 1.001kTμ. (Where μ is the expected return of the underlying, maybe that’s different from your μ but it’s definitely just some number.)
So regardless of how I rebalance, doubling k approximately doubles my expected returns.
So clearly for short enough time periods your equation for the optimum can’t be right.
But actually maximizing EV over a long time period is equivalent to maximizing it over each short time period (since final wealth is just linear in your wealth at the end of the initial short period) so the optimum over arbitrary time periods is also to max leverage.
Thanks for the comment—I’m glad people don’t take what I said at face value, since it’s often not correct...
What I actually maximized is (something like, though not quite) the expected value of the logarithm of the return, i.e. what you’d do if you used the Kelly criterion. This is the correct way to maximize long-run expected returns, but it’s not the same thing as maximizing expected returns over any given time horizon.
My computation of E[(Δlog(S))2] is correct, but the problem comes in elsewhere. Obviously if your goal is to just maximize expected return then we have
and to maximize this we would just want to push k as high as possible as long as μ>0, regardless of the horizon at which we would be rebalancing. However, it turns out that this is perfectly consistent with
E[Ik(T)Vk(T)]1/T≈1+k(k−1)2σ2
where Ik is the ideal leveraged portfolio in my comment and Vk is the actual one, both with k-fold leverage. So the leverage decay term is actually correct, the problem is that we actually have
dIkIk=kdSS+k(k−1)2dS2S2=(kμ+k(k−1)2σ2)dt+kσdz
and the leverage decay term is just the second term in the sum multiplying dt. The actual leveraged portfolio we can achieve follows
dVkVk=kμdt+kσdz
which is still good enough for the expected return to be increasing in k. On the other hand, if we look at the logarithm of this, we get
dlog(Vk)=(kμ−k22σ2)dt+kσdz
so now it would be optimal to choose something like k=μ/σ2 if we were interested in maximizing the expected value of the logarithm of the return, i.e. in using Kelly.
The fundamental problem is that Ik is not the good definition of the ideally leveraged portfolio, so trying to minimize the gap between Vk and Ik is not the same thing as maximizing the expected return of Vk. I’m leaving the original comment up anyway because I think it’s instructive and the computation is still useful for other purposes.
NOTE: Don’t believe everything I said in this comment! I elaborate on some of the problems with it in the responses, but I’m leaving this original comment up because I think it’s instructive even though it’s not correct.
There is a theoretical account for why portfolios leveraged beyond a certain point would have poor returns even if prices follow a random process with (almost surely) continuous sample paths: leverage decay. If you could continuously rebalance a leveraged portfolio this would not be an issue, but if you can’t do that then leverage exhibits discontinuous behavior as the frequency of rebalancing goes to infinity.
A simple way to see this is that if the underlying follows Brownian motion dS/S=μdt+σdW and the risk-free return is zero, a portfolio of the underlying leveraged k-fold and rebalanced with a period of T (which has to be small enough for these approximations to be valid) will get a return
r=kS(T)S(0)−k=kexp(ΔlogS)−kOn the other hand, the ideal leveraged portfolio that’s continuously rebalanced would get
ri=(S(T)S(0))k−1=exp(kΔlogS)−1If we assume the period T is small enough that a second order Taylor approximation is valid, the difference between these two is approximately
E[ri−r]≈k(k−1)2E[(Δlog(S))2]≈k(k−1)2σ2TIn particular, the difference in expected return scales linearly with the period in this regime, which means if we look at returns over the same time interval changing T has no effect on the amount of leverage decay. In particular, we can have a rule of thumb that to find the optimal (from the point of view of maximizing long-term expected return alone) leverage in a market we should maximize an expression of the form
kμ−k(k−1)2σ2with respect to k, which would have us choose something like k=μ/σ2+1/2. Picking the leverage factor to be any larger than that is not optimal. You can see this effect in practice if you look at how well leveraged ETFs tracking the S&P 500 perform in times of high volatility.
I didn’t follow the math (calculus with stochastic processes is pretty confusing) but something seems obviously wrong here. I think probably your calculation of E[(Δlog(S)2)] is wrong?
Maybe I’m confused, but in addition to common sense and having done the calculation in other ways, the following argument seems pretty solid:
Regardless of k, if you consider a short enough period of time, then with overwhelming probability at all times your total assets will be between 0.999 and 1.001.
So no matter how I choose to rebalance, at all times my total exposure will be between 0.999k and 1.001k.
And if my exposure is between 0.999k and 1.001k, then my expected returns over any time period T are between 0.999kTμ and 1.001kTμ. (Where μ is the expected return of the underlying, maybe that’s different from your μ but it’s definitely just some number.)
So regardless of how I rebalance, doubling k approximately doubles my expected returns.
So clearly for short enough time periods your equation for the optimum can’t be right.
But actually maximizing EV over a long time period is equivalent to maximizing it over each short time period (since final wealth is just linear in your wealth at the end of the initial short period) so the optimum over arbitrary time periods is also to max leverage.
Thanks for the comment—I’m glad people don’t take what I said at face value, since it’s often not correct...
What I actually maximized is (something like, though not quite) the expected value of the logarithm of the return, i.e. what you’d do if you used the Kelly criterion. This is the correct way to maximize long-run expected returns, but it’s not the same thing as maximizing expected returns over any given time horizon.
My computation of E[(Δlog(S))2] is correct, but the problem comes in elsewhere. Obviously if your goal is to just maximize expected return then we have
E[R(T)]=E[V(T)]V(0)=T−1∏i=0E[V(i+1)V(i)|Fi]=T−1∏i=0E[kS(i+1)S(i)−k|Fi]=kT(exp(μ)−1)Tand to maximize this we would just want to push k as high as possible as long as μ>0, regardless of the horizon at which we would be rebalancing. However, it turns out that this is perfectly consistent with
E[Ik(T)Vk(T)]1/T≈1+k(k−1)2σ2where Ik is the ideal leveraged portfolio in my comment and Vk is the actual one, both with k-fold leverage. So the leverage decay term is actually correct, the problem is that we actually have
dIkIk=kdSS+k(k−1)2dS2S2=(kμ+k(k−1)2σ2)dt+kσdzand the leverage decay term is just the second term in the sum multiplying dt. The actual leveraged portfolio we can achieve follows
dVkVk=kμdt+kσdzwhich is still good enough for the expected return to be increasing in k. On the other hand, if we look at the logarithm of this, we get
dlog(Vk)=(kμ−k22σ2)dt+kσdzso now it would be optimal to choose something like k=μ/σ2 if we were interested in maximizing the expected value of the logarithm of the return, i.e. in using Kelly.
The fundamental problem is that Ik is not the good definition of the ideally leveraged portfolio, so trying to minimize the gap between Vk and Ik is not the same thing as maximizing the expected return of Vk. I’m leaving the original comment up anyway because I think it’s instructive and the computation is still useful for other purposes.