0
$\begingroup$

I am trying to simulate the distribution of Geometric Brownian Motion drawdowns from samples of multivariate Normal and Laplace distributions under the same covariance structure. Drawdowns are defined to be the largest peak to trough decline of a cumulative return series where the trough comes after the peak. A cumulative return series is simply the compounded growth of sampled returns (forming something like a geometric brownian motion).

As expected, samples drawn from multivariate Laplace distributions show larger extreme values.

But when I accumulate the returns and take the average of the worst, say 1%, 0.1%, .01% of drawdowns, the normal and laplacian values are quite similar. In fact, the magnitude of the normal drawdowns are not consistently less than those of the laplacian drawdowns. I have been simulating for anywhere between 4 and 30 stocks over 252 days and up to 50,000 drawdowns per simulation (over the 252 days).

There is no portfolio rebalancing. In other words, I allow the weights of the portfolio holdings to drift over the 252 days.

This is not what I expected. Any theoretical insights regarding this finding are appreciated.

$\endgroup$
3
  • $\begingroup$ Hi and welcome, could you please add your code to the question so that we may analyse it? $\endgroup$ Aug 9 '21 at 13:41
  • $\begingroup$ I will add code. Unfortunately it is part of a very large application that I created in R that depends on external data and i will have to adjust it so it can be a good MRE. $\endgroup$ Aug 10 '21 at 14:13
  • $\begingroup$ If my answer helps you, there's no need to add your code, but you might want to reference it or copy parts of my formulas over or such. $\endgroup$ Aug 10 '21 at 14:28
1
$\begingroup$

In the following I am arguing that the result is due to the law of large numbers and the convergence of the sum of Laplace distributed random variables towards a normal distribution.

Ansatz

Let's keep it simple and assume - without loss of generality - a univariate setting. As in your question text, we define the the draw down $DD$ as the largest peak-to-trough movement along an observed time series $S_i$ of length $N$, $S_1, S_2,\ldots S_N$, i.e.

$$ \begin{align} DD_N &\equiv \max_{i,j}S_i-S_j \\ \mathrm{s. t.} &\quad 1\leq i<j\leq N \end{align} $$

As in your question, given two points $i<j$ in our series, we introduce the cumulative (log) return $R(i,j)\equiv \sum_{k=i+1}^jr_k$ and identify $$ S_j\equiv S_ie^{R(i,j)} $$ with $r_k\equiv\ln(S_k)-\ln(S_{k-1})$ the $k$th log return observation.

Statement 1: Longer (observation) time periods will result in larger price variation, and hence in larger drawdown (statistically):

Given the setup above we find that a move from $i$ to $j$ (which could be the drawdown) is

$$ \begin{align} S_j-S_i&=S_ie^{R(i,j)}-S_i\\ &=S_i\left(e^{R(i,j)}-1\right)\\ &=S_i\left(e^{\sum_{k=i+1}^jr_k}-1\right) \end{align} $$

Hence the drawdown $DD_N$ is (of course) directly linked to the cumulative sum of individual return contributions. Somewhat informally: the larger the potential period our measure covers ($j-i+1$ will increase in probability when $N$ increases), the larger the potential for a large drawdown is. Again, informally

$$ DD_N > DD_M \quad \forall N>M $$

This should be no news to you: The longer the time horizon $j-i+1$ during which we allow for (bad) individual returns to accumulate by letting $N$ grow, the larger the potential drawdown can be.

Statement 2: Longer cumulative return series will converge to a normal distribution

Yet, as $N$ increases, and under proper technical conditions, the distribution of the sum of $N$ (sufficiently) independent random variables will converge to that of a normal distribution by the central limit theorem. Thus, for a sufficiently large number $N$ in your example, the maximum draw down sampled from Laplace distributed returns will converge to that of a normal distribution with appropriately chosen parameters.

We can see that this is indeed the case if we have a look at the kurtosis of a sum of $N$ Laplace distributed random variables centered around zero. The moment generating function of the Laplace distribution is

$$ M_X(t)\equiv \mathrm{E}\left(e^{tX}\right)=\frac{1}{1-b^2t^2} $$

for some dispersion parameter $b$. Likewise, the MGF of the sum of $N$ independent Laplace trials is $$ M_{Z_N=\sum_{i=1}^NX_i}(t)\equiv \mathrm{E}\left(e^{t\sum_{i=1}^NX_i}\right)=\frac{1}{\left(1-b^2t^2\right)^N} $$

From the same wiki page, we know how to calculate the first four moments of the distribution using the moment generating function and find

$$ \mathrm{E}(X)=0\quad\mathrm{E}(X^2)=2b^2\quad\mathrm{E}(X^3)=0\quad\mathrm{E}(X^4)=24b^4\quad $$

and

$$ \mathrm{E}(Z_N)=0\quad\mathrm{E}(Z_N^2)=2b^2N\quad\mathrm{E}(X^3)=0\quad\mathrm{E}(X^4)=12b^4N(N+1)\quad $$

And we can now show that the kurtosis of $Z_N$ will converge to that of a normally distributed random variable as $N\to \infty$

$$ \begin{align} \mathrm{Kurt}(Z_N)&\equiv \frac{\mathrm{E}\left(\left(Z-\mathrm{E}(Z)\right)^4\right)}{\mathrm{E}\left(\left(Z-\mathrm{E}(Z)\right)^2\right)^2}\\ &=\frac{\mathrm{E}\left(Z^4\right)}{\mathrm{E}\left(Z^2\right)^2}\\ &=\frac{12b^4N(N+1)}{4b^4N^2}\\ &=3\left(1+\frac{1}{N}\right) \end{align} $$

Supporting simulation study

Please find below the results of a corresponding simulation study conducted using R. I have based the simulation on n trading days worth of simulated daily returns (using the normal and the Laplace distribution). For convenience, I am assuming a volatility of 20%, a mean return of nil, return series of length $N=2, 20, 50, 252$ observations. I have stuck to your setup of nSim = 50000 simulated paths.

NB: I have chosen the parameters of the Laplace distribution such that the simulated return volatility is indeed 20% and I have adjusted the simulated series such that the expected value is $E(S_j|S_i)=S_i$ under both setups.

nSim  <- 50000
sigma <- 0.2
dt    <- 1 / 252

maxDD <- function(z){
    dd <- 0.0
    for (i in 1:(length(z)-1)){
        cand <- max(z[i]-z[-(1:i)])
        if (cand>dd){dd <- cand}
    }
    dd
}

n <- 2

normal <- normal <- sapply(1:nSim,function(i){
    maxDD(exp(c(0,cumsum(
        -0.5 * sigma^2 * dt + sigma*sqrt(dt)*rnorm(n=n)
                         ))))})

laplace <- sapply(1:nSim,function(i){
    maxDD(exp(c(0,cumsum(
        -log(2) * sigma^2 * dt + sigma*sqrt(dt)*ExtDist::rLaplace(n=n,mu = 0,b=sqrt(2)/2)
                         ))))})

qqplot(normal,laplace)
abline(a=0,b=1)
title(main=paste("N = ",n))

Finally, I have created QQ-plots of the resulting Drawdown distributions for each hypothetical series length below. As we can see from the diagrams, the simulated drawdown distributions converge as per your observation. For small $N$, the tails of the Laplace have an effect over the normal distribution, but as soon as we simply 'allow' for more returns to enter the drawdown, the influence of the central limit theorem outweighs the tails fo the Laplace distribution.

N = 2 N = 20 N = 50 N = 252

$\endgroup$
8
  • $\begingroup$ Kermittfrog. This is incredibly insightful. Thank you so much. I would have expected this without compounding, but not with drawdowns under drift. I will have to study your write-up and try to visualize when the law of large numbers wouldn't hold. $\endgroup$ Aug 10 '21 at 14:09
  • $\begingroup$ Kermittfrog. Almost by definition, compounding introduces serial correlation. Why does this not affect the assumptions of the law of large numbers that relies on a lack of such correlation? $\endgroup$ Aug 10 '21 at 18:23
  • $\begingroup$ You need to introduce serial correlation in your simulated returns as well, then. $\endgroup$ Aug 10 '21 at 19:02
  • $\begingroup$ Kermittfrog. So a single drawdown is a serially correlated wealth curve that points down, yet you invoke the law of large numbers (which relies on independence). Are you saying for the law of large numbers to not hold, the drawdowns themselves must be dependent on one another? $\endgroup$ Aug 10 '21 at 19:33
  • $\begingroup$ I understand the drawdown to be the maximum loss from peak to trough in a time series of some length as you‘ve introduced it. This will include one or more consecutive days of (cumulative) down moves. The longer this series becomes (in trading days), the more the CLT will weigh in I’d say $\endgroup$ Aug 10 '21 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.