Thomas Dall
2019-03-26
Risk on this trade:
\[ \begin{aligned} \mathrm{risk} = & 11559 - 11549 \\ = & 10 \mathrm{EUR/cfd} \\ = & \mathrm{R} \end{aligned} \]
“R” is the normalized Risk Unit.
Result of this trade:
+2R
Make lots of trades
\( \rightarrow \) R-distribution
\( \bar{R} = 0.04 \)
This is “mean-R”
But: Reality is not normally distributed!
\( \bar{R} = 0.08 \)
\( f_\mathrm{win} = 0.55 \)
\( \bar{R} = 0.1 \)
\( f_\mathrm{win} = 0.59 \)
System Type:
Mean reversion Swing
\( \bar{R} = 0.29 \)
\( f_\mathrm{win} = 0.34 \)
System type:
Trend-following Break-outs
Usual definition of break-even:
\[ N_\mathrm{win} \bar{P} - N_\mathrm{loss} \bar{L} = 0 \]
First step: divide by \( N_\mathrm{total} \):
\[ f_\mathrm{win} = \frac{N_\mathrm{win}}{N_\mathrm{total}} \]
Second step: Use average R instead of “real” amounts.
Normalized definition of break-even:
\[ f_\mathrm{win} \bar{R}_\mathrm{pos} - f_\mathrm{loss} \bar{R}_\mathrm{neg} = 0 \]
where: \( f_\mathrm{win} + f_\mathrm{loss} = 1 \)
Which means:
\[ \mathrm{PF} = \frac{\bar{R}_\mathrm{pos}}{\bar{R}_\mathrm{neg}} = \frac{1-f_\mathrm{win}}{f_\mathrm{win}} \]
Where PF is the Normalized Profit Factor
And the Big One:
Investopedia:
Monte Carlo simulations are used to model the probability of different outcomes in a process that cannot easily be predicted due to the intervention of random variables.
Select Sample Size and Number of Samples
Typically I use Number of Samples = 1000
The Sample Size depends on reporting time frame etc…
t.test(rBO$R,rRSI$R)
Welch Two Sample t-test
data: rBO$R and rRSI$R
t = 1.0735, df = 182.32, p-value = 0.2845
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.1555376 0.5267527
sample estimates:
mean of x mean of y
0.2897303 0.1041227
Problem: Requires high numbers for accuracy