Question
Given a time series, I want to calculate the maximum drawdown, and I also want to locate the beginning and end points of the maximum drawdown so I can calculate the duration. I want to mark the beginning and end of the drawdown on a plot of the timeseries like this:
So far I've got code to generate a random time series, and I've got code to calculate the max drawdown. If anyone knows how to identify the places where the drawdown begins and ends, I'd really appreciate it!
https://stackoverflow.com/questions/22607324
italiano
english
français
española
中国
日本の
العربية
Deutsch
한국어
Português
Russianimport pandas as pd
import matplotlib.pyplot as plt
import numpy as np
# create random walk which I want to calculate maximum drawdown for:
T = 50
mu = 0.05
sigma = 0.2
S0 = 20
dt = 0.01
N = round(T/dt)
t = np.linspace(0, T, N)
W = np.random.standard_normal(size = N)
W = np.cumsum(W)*np.sqrt(dt) ### standard brownian motion ###
X = (mu-0.5*sigma**2)*t + sigma*W
S = S0*np.exp(X) ### geometric brownian motion ###
plt.plot(S)
# Max drawdown function
def max_drawdown(X):
mdd = 0
peak = X[0]
for x in X:
if x > peak:
peak = x
dd = (peak - x) / peak
if dd > mdd:
mdd = dd
return mdd
drawSeries = max_drawdown(S)
MaxDD = abs(drawSeries.min()*100)
print MaxDD
plt.show()
Solution
Just find out where running maximum minus current value is largest:
n = 1000
xs = np.random.randn(n).cumsum()
i = np.argmax(np.maximum.accumulate(xs) - xs) # end of the period
j = np.argmax(xs[:i]) # start of period
plt.plot(xs)
plt.plot([i, j], [xs[i], xs[j]], 'o', color='Red', markersize=10)
OTHER TIPS
behzad.nouri solution is very clean, but it's not a maximum drawdow (couldn't comment as I just opened my account and I don't have enough reputation atm).
What you end up having is the maximum drop in the nominal value rather than a relative drop in value (percentage drop). For example, if you would apply this to time series that is ascending over the long run (for example stock market index S&P 500), the most recent drop in value (higher nominal value drops) will be prioritized over the older decrease in value as long as the drop in nominal value/points is higher.
For example S&P 500:
By applying this method to period after 2000, you'll see Corona Virus Crisis rather than 2007-08 Financial Crisis
Related code (from behzad.nouri) below:
n = 1000
xs = np.random.randn(n).cumsum()
i = np.argmax(np.maximum.accumulate(xs) - xs) # end of the period
j = np.argmax(xs[:i]) # start of period
plt.plot(xs)
plt.plot([i, j], [xs[i], xs[j]], 'o', color='Red', markersize=10)
You just need to divide this drop in nominal value by the maximum accumulated amount to get the relative ( % ) drawdown.
( np.maximum.accumulate(xs) - xs ) / np.maximum.accumulate(xs)
on the back of this I added unerwater analysis if that helps anyone...
def drawdowns(equity_curve):
i = np.argmax(np.maximum.accumulate(equity_curve.values) - equity_curve.values) # end of the period
j = np.argmax(equity_curve.values[:i]) # start of period
drawdown=abs(100.0*(equity_curve[i]-equity_curve[j]))
DT=equity_curve.index.values
start_dt=pd.to_datetime(str(DT[j]))
MDD_start=start_dt.strftime ("%Y-%m-%d")
end_dt=pd.to_datetime(str(DT[i]))
MDD_end=end_dt.strftime ("%Y-%m-%d")
NOW=pd.to_datetime(str(DT[-1]))
NOW=NOW.strftime ("%Y-%m-%d")
MDD_duration=np.busday_count(MDD_start, MDD_end)
try:
UW_dt=equity_curve[i:].loc[equity_curve[i:].values>=equity_curve[j]].index.values[0]
UW_dt=pd.to_datetime(str(UW_dt))
UW_dt=UW_dt.strftime ("%Y-%m-%d")
UW_duration=np.busday_count(MDD_end, UW_dt)
except:
UW_dt="0000-00-00"
UW_duration=np.busday_count(MDD_end, NOW)
return MDD_start, MDD_end, MDD_duration, drawdown, UW_dt, UW_duration
Your max_drawdown already keeps track of the peak location. Modify the if to also store the end location mdd_end when it stores mdd, and return mdd, peak, mdd_end.
I agree with k0rnik.
A short example for prooving that formula given by behzad.nouri can produce wrong result.
xs = [1, 50, 10, 180, 40, 200]
pos_min1 = np.argmax(np.maximum.accumulate(xs) - xs) # end of the period
pos_peak1 = np.argmax(xs[:pos_min1]) # start of period
pos_min2 = np.argmax((np.maximum.accumulate(xs) -
xs)/np.maximum.accumulate(xs)) # end of the period
pos_peak2 = np.argmax(xs[:pos_min2]) # start of period
plt.plot(xs)
plt.plot([pos_min1, pos_peak1], [xs[pos_min1], xs[pos_peak1]], 'o',
label="mdd 1", color='Red', markersize=10)
plt.plot([pos_min2, pos_peak2], [xs[pos_min2], xs[pos_peak2]], 'o',
label="mdd 2", color='Green', markersize=10)
plt.legend()
mdd1 = 100 * (xs[pos_min1] - xs[pos_peak1]) / xs[pos_peak1]
mdd2 = 100 * (xs[pos_min2] - xs[pos_peak2]) / xs[pos_peak2]
print(f"solution 1: peak {xs[pos_peak1]}, min {xs[pos_min1]}\n rate :
{mdd1}\n")
print(f"solution 2: peak {xs[pos_peak2]}, min {xs[pos_min2]}\n rate :
{mdd2}")
Further the price of an asset cannot be negative so
xs = np.random.randn(n).cumsum()
is not correct. It could be better to add:
xs -= (np.min(xs) - 10)
This solution is tested and works but here I'm computing the maximum duration drawdown and NOT the duration of the maximum drawdown. The solution can be easily adapted to find the duration of the maximum drawdown.
def max_dur_drawdown(dfw, threshold=0.05):
"""
Labels all drawdowns larger in absolute value than a threshold and returns the
drawdown of maximum duration (not the max drawdown necessarily but most often they
coincide).
Args:
dfw (pd.DataFrame): monthly data, the pre-computed drawdowns or underwater.
threshold (float): only look at drawdowns greater than this in absolute value e.g. 5%
Returns:
dictionary containing the start, end dates and duration in months for the maximum
duration drawdowns keyed by column name.
"""
max_dur_per_column = {}
columns = dfw.columns.copy()
mddd_start = {}
mddd_end = {}
mddd_duration = {}
for col in columns:
# run the drawdown labeling algorithm
dfw['sign'] = 0
dfw['sign'].loc[dfw[col] == 0] = +1
dfw['sign'].loc[dfw[col] < 0] = -1
# find the sign change data points
dfw['change'] = dfw['sign'] != dfw['sign'].shift(1)
# the first change doesn't count
dfw['change'].iloc[0] = False
# demarcate the lef and right of the drawdowns
left = dfw[(dfw['change'] == True) & (dfw['sign'] == -1)].index.values
right = dfw[(dfw['change'] == True) & (dfw['sign'] == 1)].index.values
min_len = min(len(left), len(right))
intervals = pd.IntervalIndex.from_arrays(left[0:min_len], right[0:min_len])
# find the minimum value per drawdown interval so we label all data points to the left of it.
min_per_int = list(map(lambda i: (i.left, i.right, dfw[col][(dfw.index >= i.left) & (dfw.index < i.right)].min()), intervals))
# filter out drawdowns lower in absolute value than a threshold
min_per_int = list(filter(None.__ne__, list(map(lambda x: None if x[2] >= -threshold else x, min_per_int))))
# label only the negative part of the underwater NDD stands for negative-side drawdown.
dfw['NDD'] = 0
mddd_start[col] = None
mddd_end[col] = None
mddd_duration[col] = 0
for i in min_per_int:
# find the index of the data point that is minimum this is an argmin
min_idx = dfw[(dfw.index >= i[0]) & (dfw.index < i[1]) & (abs(dfw[col] - i[2]) < 1e-15)].index[0]
# compute the duration and update the maximum duration if needed
tmp_dur = int(np.round((min_idx - i[0]) / np.timedelta64(1, 'M')))
if tmp_dur > mddd_duration[col]:
mddd_start[col] = i[0].date()
mddd_end[col] = min_idx.date()
mddd_duration[col] = tmp_dur
return mddd_start, mddd_end, mddd_duration
Example usage:
# compute cumulative returns
dfc = pd.DataFrame(dfr['S&P500'] / dfr['S&P500'][0])
# compute drawdowns
dfw = dfc / dfc.cummax() - 1
print(max_dur_drawdown(dfw))
