Monte Carlo Drawdown

So, the other day I was messing around with the returns of something and I wanted to see if, with the basic statistics of the returns, what are the risks implicit (at least the ones I can reasonably see).

Since I’m a naive fellow, I wrote a very naive program in Python that uses Monte Carlo to estimate the mean drawdown I’d suffer and the maximum drawdown.

First, from the empirical results I extract some statistics: the win rate winrate, which is the probability of a positive return and the return to risk rr, which measures how much I win for every loss, on average.

With these I generate (pseudo)random numbers (on a normal distribution; I know, naive). If the number is up to winrate, I win; otherwise I lose. If I do this a many number of times (say, a thousand) I’ll have a (thousand) series of outcomes. From these I extract the mean drawdown and the maximum drawdown.

The jupyter notebook can be found here. The code is below:

First I load the libraries I need and read the file with the empirical results; I do some visualisation and so on.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import math
import random

file_data = trades2.csv
trades = pd.read_csv(file_data)

cumulativo = trades.results.cumsum()


Then I find out the basic assumptions of this strategy:

loss = pd.DataFrame(list(filter(lambda x: x <= 0, trades.results)))
gain = pd.DataFrame(list(filter(lambda x: x > 0, trades.results)))

winrate = round((len(gain)/(len(gain)+len(loss))),2)
lossrate = 1 - winrate

rr = round(-gain.mean()/loss.mean(),2)[0]

And here is where I generate the possible results (notice I normalize the possible outcome: so the loss is -1 and the gain is rr)

montecarlo = [[rr if np.random.uniform(0,1,len(trades))[j] <= winrate else -1 for j in range(1,len(trades)-1)] for i in range(1,1000)]

result = pd.DataFrame(montecarlo).transpose()

plt.axhline(y=0,linewidth=1, color=#000000’)
plt.plot(result.cumsum(), alpha = 0.5)

The calculation of the drawdown is pretty simple:

drawdown = result.cumsum() - result.cumsum().cummax()
maxdd = -drawdown.min()

# mean drawdown

# maximum drawdown I take to be the one that occurs 5% of the cases

Mean DD is about 11 and maximum DD is about 19. One might say this is not a good strategy, but one has to consider other factors as well.