Bacon in Practical Portfolio Performance Measurement and Attribution distinguishes between the two, specifying that "Maximum drawdown represents the maximum loss an investor can suffer in the fund buying at the highest point and selling at the lowest" and Largest Individual Drawdown to be "the largest individual uninterrupted loss in a return series". This is clear, however, other sources seem to be unclear or outright disagree. For example, Investopedia defines maximum drawdown as a "maximum observed loss from a peak to a trough of a portfolio" and doesn't mention Largest Individual Drawdown. CFI seems to agree, though they seem to emphasize (based on the graph provided) that a new peak must be reached if a drawdown is to be considered a maximum drawdown. Some sources seem to use the largest individual drawdown as a maximum drawdown (here, PerformanceAnalytics R package).

Questions the I have are:

  1. What is the difference between Maximum Drawdown and Largest Individual Drawdown?
  2. Does the new peak need to be reached for a drawdown to be considered for maximum drawdown (even if the non-peaked drawdown is larger than the peaked drawdown) as Wikipedia would suggest?
  3. Can trough value be before the peak value (as per what Investopedia's formula seem to suggest)?
  4. What is the commonly used definition of maximum drawdown?

1 Answer 1


From your (or rather Bacon's) description, it seems to me that the "Largest Individual Drawdown" would more properly be called a losing streak, or a run of negative returns. Since you seem to be using R, you could use rle to compute such streaks.

R <- c(rep(-0.02, 2), rep(0.03, 2), rep(-0.001, 5))
## [1] -0.020 -0.020  0.030  0.030 -0.001 -0.001 -0.001 -0.001 -0.001
## ==> two negative, two positive, five negative returns
rle(R < 0)  ## runs of returns < 0 get a TRUE
## Run Length Encoding
##   lengths: int [1:3] 2 2 5
##   values : logi [1:3] TRUE FALSE TRUE

Since drawdowns are computed from prices, not returns, I compute an artificial price series:

P <- cumprod(c(0, R) + 1)


All implementations I have ever seen would consider the first two negative returns as the maximum drawdown, whereas the later drop (time 5 to 10) is the longest losing streak.

See the function streaks in package PMwR (which I maintain).

streaks(P, up = 0)
##   start end state   return
## 1     1   3  down -0.03960
## 2     3   5    up  0.06090
## 3     5  10  down -0.00499

##   peak trough recover     max
## 1    1      3       5 0.03960
## 2    5     10      NA 0.00499

Question 2: No, it doesn't. (One usually speaks of "recovery" of a drawdown.)

Question 3: No: Imagine a series that goes up 1%, every day. Such a series has no drawdown. If the trough could be before the peak, it would have a drawdown.

Question 4: There is no universally accepted version, though I prefer this one (taken, with slight adjustments, from the PMwR manual):

Let the symbol $v$ be a time series of portfolio values, with observations indexed at $t=0, 1, 2 \ldots T$. The drawdown $D$ of this series at time $t$ is defined as \begin{equation} D_t = v^{\max}_t - v_t \end{equation} in which $v^{\max}_t$ is the running maximum, i.e. $v^{\max}_t=\max\{v_{t'}\,|\,{t'} \in [0,t]\}$. Note that $D$ is a vector of length $T+1$. [...] To compute relative drawdown, divide $D_t$ by $v^{\max}_t$. The maximum drawdown is the maximum of the vector.

In my experience, descriptions differ mainly in i) the sign (most people prefer positive; after all, you call it maximum drawdown): ii) whether absolute or relative (in equties, relative is much more common); and iii) whether the computation should be limited to a fixed historic window.

  • $\begingroup$ What if we had R <- c(rep(-0.02, 2), rep(0.03, 2), rep(-0.02, 5)). would the maximum drawdown then be at t=3 or t=10? $\endgroup$
    – cc88
    Jun 15, 2023 at 14:27
  • 1
    $\begingroup$ @cc88 Definitely at t=10: peak is at time 5 (when computing the cumprod as above), and at time 10 the series is 9.6% "underwater", which is the largest drawdown. $\endgroup$ Jun 16, 2023 at 18:04

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