Document optimal strategy for project 6

manual_strategy/indicators.py

@@ -1,5 +1,6 @@
 import pandas as pd
 import datetime as dt
+import matplotlib.pyplot as plt
 import TheoreticallyOptimalStrategy as tos
 from util import plot_data, get_data
 from marketsim.marketsim import compute_portvals
@@ -14,7 +15,6 @@ def test_policy():
     symbol = "JPM"
     start_value = 100000
     sd = dt.datetime(2008, 1, 1)
-    # ed = dt.datetime(2008, 1, 30)
     ed = dt.datetime(2009, 12, 31)
 
     orders = tos.testPolicy(symbol=symbol, sd=sd, ed=ed, sv=start_value)
@@ -55,8 +55,9 @@ def test_policy():
     portvals["Bench"] = bench["Portval"]
     portvals.drop(columns=["Portval"], inplace=True)
 
-    #XXX: Save to file instead.
+    ax = portvals.plot(title="Optimal strategy versus 1000 shares of JPM")
-    plot_data(portvals)
+    plt.savefig('figure_5.png')
+
 
 def normalize(timeseries):
     return timeseries / timeseries.iloc[0]

manual_strategy/manual_strategy.md

@@ -0,0 +1,54 @@
+# Indicators
+
+
+
+
+# Optimal Strategy
+
+The optimal strategy works by applying every possible buy/sell action to the
+current positions. A position is cash value, the current amount of shares, and
+previous transactions.
+
+The algorithm then starts with a single initial position with the initial cash
+amount, no shares, and no transactions. The algorithm first executes all
+possible trades (sell 2000, sell 1000, do nothing, buy 1000, buy 2000). Next, it
+filters the invalid holdings, i.e., the other ones where shares are not equal to
+-1000, 0, or 1000. Finally, the algorithm keeps the highest cash value for each
+amount of shares.
+
+That is the critical insight that makes the algorithm work. Since the same
+amount of shares has the same future potential for creating cash, the only
+differentiating factor is the current cash value. Accordingly, for each day, the
+algorithm ends up with three new positions. The algorithm could drop the
+position with zero shares because buying and selling 2000 shares leverages
+future price movements optimally.
+
+After each trading day, the positions have the same value. I did not expect
+that, but it makes sense because the algorithm only keeps the best position from
+the previous day. In other words, we can simplify the algorithm by selecting the
+best position and then compute the two different positions. Since we already
+established that holding zero shares does not make sense, the only two
+possibilities are buying or selling 2000 depending on the current amount of
+shares.
+
+![Optimal strategy versus holding 1000 shares](figure_5.png)
+
+```
+Date Range: 2008-01-02 00:00:00 to 2009-12-31 00:00:00
+
+Sharpe Ratio of Optimal Strategy: 13.322769848217233
+Sharpe Ratio of bench: 0.1569184064240322
+
+Cumulative Return of Optimal Strategy: 5.7861
+Cumulative Return of bench: 0.012299999999999978
+
+Standard Deviation of Optimal Strategy: 0.004547823197907996
+Standard Deviation of bench: 0.01700436627121376
+
+Average Daily Return of Optimal Strategy: 0.0038167861508578214
+Average Daily Return of bench: 0.00016808697819094233
+
+Final Portfolio Value Optimal: 678610.0
+Final Portfolio Value Bench: 101230.0
+```
+