Saturday, June 19, 2010

Estimating Probability of Drawdown

I've shown several examples of how to use LSPM's probDrawdown function as a constraint when optimizing a leverage space portfolio.  Those posts implicitly assume the probDrawdown function produces an accurate estimate of actual drawdown.  This post will investigate the function's accuracy.

Calculation Notes:
To calculate the probability of drawdown, the function traverses all permutations of the events in your lsp object over the given horizon and sums the probability of each permutation that hits the drawdown constraint.  The probability of each permutation is the product of the probability of each event in the permutation.

In the example below, there are 20 events in each lsp object and the probability of drawdown is calculated over a horizon of 10 days, yielding 20^10 permutations to traverse - for each iteration.  So don't be surprised when the code takes more than a couple minutes to run.

For a more detailed discussion about the calculation, see:
pp. 89-138 of The Leverage Space Trading Model, and/or
pp. 377-414 of The Handbook of Portfolio Mathematics
both by Ralph Vince.

The results below were run on daily SPY from 2008-01-01 to 2009-01-01, using 20 days of data to estimate the probability of a 5% drawdown over the next 10 days.  Results on daily QQQQ over the same period, and monthly SPX from 1950-present produced similar results.

I chose a prediction horizon of 10 periods to provide a fairly smooth actual probability of drawdown curve without making the probDrawdown calculation time too long.  Using 10 (instead of 20) days of data in the lsp object only changed results slightly.

The chart below shows that probDrawdown nearly always over estimates actual drawdown over the next 10 periods and hardly ever under estimates it.  While it's comforting the function doesn't under estimate risk, I would prefer a less biased estimator.

Notice that the above calculation assumes each event is independently distributed.  Brian Peterson suggested a block bootstrap to attempt to preserve any dependence that may exist.  My next post will investigate if that materially improves estimates.


# Pull data and calculate differences
symbol <- "SPY"
getSymbols(symbol, from="2008-01-01", to="2009-01-01")
sym <- get(symbol)
symDiff <- diff(Cl(sym))
symDiff[1] <- 0

NP <- 20    # number of periods to use in lsp object
HR <- 10    # drawdown horizon
DD <- 0.05  # drawdown level

# Initialize projected / actual drawdown objects
NR <- NROW(symDiff)
prjDD <- xts(rep(0,NR),index(symDiff))
actDD <- xts(rep(0,NR),index(symDiff))

# Socket cluster with snow to speed up probDrawdown()
cl <- makeSOCKcluster(2)

# Start loop over data
for( i in (NP+1):(NR-HR) ) {
  # Objects to hold data for the last 20 days and next 10 days
  last20 <- symDiff[(i-NP):i]
  next10 <- symDiff[(i+1):(i+HR)]
  maxLoss <- -Cl(sym)[i]

  # Portfolios to estimate drawdown and calculate actual drawdown
  prjPort <- lsp(last20, f=1, maxLoss=maxLoss)
  actPort <- lsp(next10, f=1, maxLoss=maxLoss)
  # Estimate probability of drawdown
  prjDD[i] <- probDrawdown(prjPort, DD, HR, snow=cl)
  # Calculate actual drawdown probability
  actDD[i] <- sum(HPR(actPort)/cummax(HPR(actPort)) <= (1-DD)) / HR
# End loop over data

# Chart results


Ralph Vince said...

Thanks for another great post Josh. I've looked at this at length. Remember -- the way you are calculating probability of drawdown, RD, is on a sample and replacement basis where the probabilities associated at each T to the horizon are unchanged. Thus, if, say, a large loss is seen at time T = t, the probablity of seeing that same large loss, in succession (i.e. at T = t+1), is the same as it was at T = t. The probability of drawdown using this method will give you the exact probability for these kinds of cases.
However, in real life (again, depending on market, system trading time frame) your scenarios may be such that his is not the case -- and I suspect in the data you are using that it is not. In other words, the probability of any given scenario, (any given row in the joint probabilities table) arising is the same regardless of what the previous row was.
Ultimately, the code you post here is a sort of test for dependency of exactly that(!) and may be even more interesting in its own right along those lines. -Ralph Vince

Ralph Vince said...


Ah, I know why this is so -- and it has to do with confidence intervals. I will explain this in detail in Tampa and it will make perfect sense.

Joshua Ulrich said...

Hi Ralph,

Thanks for the thoughts in both your comments. I look forward to the detailed explanation.

Ralph Vince said...

Josh -- interesting paper that confirms perhaps what you are pointing out here, but from an altogether different standpoint: