使用python计算最大回撤

1. 单期简单收益率

R t = P t − P t − 1 P t − 1 R_{t}=\dfrac {P _{t}-P_{t-1}}{P_{t-1}} Rt=Pt−1Pt−Pt−1
说明：

R t R_t Rt 为单期简单收益率
P t P_t Pt 为t期的价格
P t − 1 P_{t-1} Pt−1 为t-1期的价格

import datetime
import pandas as pd
pd.core.common.is_list_like = pd.api.types.is_list_like

price = pd.Series([3.42,3.51,3.68,3.43,3.56,3.67], index=[datetime.date(2015,7,x) for x in range(3,9)])
price

2015-07-03    3.42
2015-07-04    3.51
2015-07-05    3.68
2015-07-06    3.43
2015-07-07    3.56
2015-07-08    3.67
dtype: float64

利用ffn库计算单期简单收益

import ffn

r = ffn.to_returns(price)
r

2015-07-03         NaN
2015-07-04    0.026316
2015-07-05    0.048433
2015-07-06   -0.067935
2015-07-07    0.037901
2015-07-08    0.030899
dtype: float64

2. 最大回撤

最大回撤(Maximum Drawdown, MDD) 用来衡量投资（特别是基金）的表现，

2.1 回撤：某资产在时刻T的回撤是指资产在(0,T)的最高峰值与现在价值 P T P_T PT之间的回落值，用数学公式表达为：

D ( T ) = max ⁡ { 0 , max ⁡ t ∈ ( 0 , T ) P t − P T } D\left( T\right) =\max \left\{ 0,\max _{t\in (0,T)}P_t-P_{T}\right\} D(T)=max{0,maxt∈(0,T)Pt−PT}

2.2 对应的回撤率为：

d ( T ) = D ( T ) max ⁡ t ∈ ( 0 , T ) p t d\left(T\right) = \dfrac {D\left( T\right) }{\max _{t\in \left( 0,T\right) }p_{t}} d(T)=maxt∈(0,T)ptD(T)

知道回撤的含义之后，最大回撤就比较容易理解了，资产在T时刻的最大回撤MDD(T),就是资产在时段(0,T)内回撤的最大值，对应的数学公式为：

M D D ( T ) = max ⁡ τ ∈ ( 0 , T ) D ( τ ) = max ⁡ τ ∈ ( 0 , T ) [ max ⁡ t ∈ ( 0 , τ ) P t − P τ ] MDD\left(T\right) = {\max _{\tau \in \left(0,T\right)}} D\left(\tau\right) = {\max _{\tau \in \left(0,T\right)}} \begin{aligned} \left[ \max _{t\in \left( 0,\tau \right) }P_{t}-P_{\tau}\right] \end{aligned} MDD(T)=maxτ∈(0,T)D(τ)=maxτ∈(0,T)[t∈(0,τ)maxPt−Pτ]

相应的最大回撤率为：
m d d ( T ) = max ⁡ τ ∈ ( 0 , T ) d ( τ ) M D D ( T ) max ⁡ t ∈ ( 0 , T ) P t mdd\left(T\right) = {\max _{\tau \in \left(0,T\right)}} d\left(\tau \right) \dfrac {MDD\left(T\right)}{\max _{t \in \left(0,T\right)}P_t} mdd(T)=maxτ∈(0,T)d(τ)maxt∈(0,T)PtMDD(T)

直观的讲，MDD(T)对应的是在(0,T)时段内资产价值从最高峰回落到最低谷的幅度。最大回撤常用来描述投资者在持有资产是可能面临的最大亏损。

2.3 利用收益率计算最大回撤

如果某资产的收益率序列为R₁,R₂,…,R_T,在初始时刻0时，我们投资1元在该资产上并一直持有到T时刻，则初始值为1元的资产价值就会随时间变化为：（1+R₁）,（1+R₁）（1+R₂）,（1+R₁）（1+R₂）（1+R₃）,…, ∏ k = 1 T ( 1 + R k ) \prod _{k=1}^{T}\left( 1+R_{k}\right) ∏k=1T(1+Rk)

时刻T对应的回撤值为：
D ( T ) = max ⁡ { 0 , max ⁡ t ∈ ( 0 , T ) ∏ k = 1 t ( 1 + R k ) − ∏ k = 1 T ( 1 + R k ) } D(T) = \max \left\{ 0, \max _{t \in (0,T)} \prod _{k=1}^{t}\left( 1+R_{k}\right) - \prod _{k=1}^{T}\left( 1+R_{k}\right) \right\} D(T)=max{0,maxt∈(0,T)∏k=1t(1+Rk)−∏k=1T(1+Rk)}

相应的回撤率为：
d ( T ) = D ( T ) max ⁡ t ∈ ( 0 , T ) ∏ k = 1 t ( 1 + R k ) d(T) = \dfrac {D\left( T\right) }{\max _{t\in \left( 0,T\right) }\prod _{k=1}^{t}\left( 1+R_{k}\right) } d(T)=maxt∈(0,T)∏k=1t(1+Rk)D(T)

最大回撤为：
M D D ( T ) = max ⁡ τ ∈ ( 0 , T ) D ( τ ) = max ⁡ τ ∈ ( 0 , T ) [ max ⁡ t ∈ ( 0 , τ ) ∏ k = 1 t ( 1 + R k ) − ∏ k = 1 τ ( 1 + R k ) ] MDD(T) = \max _{\tau \in \left( 0,T\right) }D\left( \tau \right) = \max _{\tau \in \left( 0,T\right) } \begin{aligned} \left[\max _{t\in \left( 0,\tau \right) }\prod _{k=1}^{t}\left( 1+R_{k}\right) - \prod _{k=1}^{\tau}\left( 1+R_{k}\right) \right]\end{aligned} MDD(T)=maxτ∈(0,T)D(τ)=maxτ∈(0,T)[t∈(0,τ)maxk=1∏t(1+Rk)−k=1∏τ(1+Rk)]

相应的最大回撤率为：
m d d ( T ) = max ⁡ τ ∈ ( 0 , T ) d ( τ ) = M D D ( T ) max ⁡ t ∈ ( 0 , T ) ∏ k = 1 t ( 1 + R k ) mdd(T) = \max _{\tau \in \left( 0,T\right) }d(\tau) = \dfrac {MDD(T)}{\max _{t \in \left( 0,T\right) }\prod _{k=1}^{t}\left( 1+R_{k}\right)} mdd(T)=maxτ∈(0,T)d(τ)=maxt∈(0,T)∏k=1t(1+Rk)MDD(T)

value = (1 + r).cumprod()
value

2015-07-03         NaN
2015-07-04    1.026316
2015-07-05    1.076023
2015-07-06    1.002924
2015-07-07    1.040936
2015-07-08    1.073099
dtype: float64

D = value.cummax() - value
D

2015-07-03         NaN
2015-07-04    0.000000
2015-07-05    0.000000
2015-07-06    0.073099
2015-07-07    0.035088
2015-07-08    0.002924
dtype: float64

d = D / (D + value)
d

2015-07-03         NaN
2015-07-04    0.000000
2015-07-05    0.000000
2015-07-06    0.067935
2015-07-07    0.032609
2015-07-08    0.002717
dtype: float64

MDD = D.max()
MDD

0.07309941520467844

mdd =d.max()
mdd
# 对应的最大回撤率值为

0.06793478260869572

# 采用ffn库计算收益率累积最大回撤
ffn.calc_max_drawdown(value)

-0.06793478260869568

from empyrical import max_drawdown

# 使用 empyrical 计算收益率序列最大回撤
max_drawdown(r)

-0.06793478260869572

3. Java版本的最大回测

import java.util.Arrays;
import java.util.List;import lombok.extern.slf4j.Slf4j;
@Slf4j
public class MaxDrawdownService {/*** 计算最大回撤:某段时间内连续收益率之和最小的值为最大回撤* * @Param rates：收益率*/@SuppressWarnings("all")public static double caculateMaxDrawdown(List<Double> rates) {double s = 0;double e = 0;double max = 0;double temp = 0;double ts = 0;// 收益率不为空if (!rates.isEmpty()) {for (int i = 0; i < rates.size(); i++) {// 获得收益率double r = rates.get(i);temp = temp + r;if (temp > 0) {ts = i + 1;e = i + 1;temp = 0;} else {if (temp < max) {s = ts;e = i;max = temp;}}}}log.info("最大回撤计算结果：maxsum={},start={}，end={}", max, s, e);return max;}/*** 按照资金计算最大回撤* @param equityValues* @return*/public static double calMaxDrawdown(List<Double> equityValues) {if (equityValues == null || equityValues.size() < 2)return 0;double maxDrawdown = 0; // 最大回撤double maxEquityValue = equityValues.get(0); // 当日之前的最大资产净值for (int i = 1; i < equityValues.size(); i++) {double currentEquityValue = equityValues.get(i); // 当日资产净值double drawDown = (1 - currentEquityValue / maxEquityValue);maxDrawdown = Math.max(maxDrawdown, drawDown);maxEquityValue = Math.max(currentEquityValue, maxEquityValue);}log.info("calMaxDrawdown最大回撤计算结果：{}", maxDrawdown);return maxDrawdown;}public static void main(String[] args) {Double[] c = {0.026316,0.048433,-0.067935,0.037901,0.030899};caculateMaxDrawdown(Arrays.asList(c));Double[] c2 = {3.42,3.51,3.68,3.43,3.56,3.67};calMaxDrawdown(Arrays.asList(c2));}
}