Imagine that prior to entering the market you know beforehand the profit factor of similar situations.

For example:

trades similar to TRADE 1 have yielded a 1.2 pf 
trades similar to TRADE 2 have yielded a 2.0 pf

What is the best way to manage lot sizes for the two different trades relative to each other?

The obvious thing to do is

TRADE 1 Lots = X*1.2
TRADE 2 Lots = X*2.0

but I am not sure if this is the smartest move.

  • 1
    $\begingroup$ You need to also consider (1) the variation in returns of each trading scenario, and (2) the covariance among the trading scenarios. $\endgroup$ – chrisaycock Sep 17 '11 at 4:12
  • $\begingroup$ @chrisaycock I had considered this, but I cannot think of a good way to apply those concepts that ends with lot management. $\endgroup$ – Mike Furlender Sep 17 '11 at 5:13
  • $\begingroup$ Did you define a maximum global and per position sharpe ratio and exposition you want to achieve ? $\endgroup$ – Lliane Sep 17 '11 at 6:13
  • $\begingroup$ @Lliane I am not sure how I can do this because my entries/exists are dynamic. I can get an "average profit (or loss)" but not "maximum profit (or loss)" - but say I used that, where do I go from there? $\endgroup$ – Mike Furlender Sep 17 '11 at 6:22

Maybe not really an answer, but a justification of your approach. It's likely that your results can be expresses as $$ \mathsf EX_1 = 1.2\text{ and }\mathsf EX_2 = 2 $$ where $X_i$ for $i=1,2$ is a random pf of a situation in a class $i$ (we denote it $S_i$). Your method solves the following problem: given a fixed number of trials we would like to maximize the average pf: $$ \mathsf E(\alpha X_1+(1-\alpha)X_2)\to\max_{\alpha\in[0,1]} $$ which certainly has a solution $\alpha = \frac{1.2}{1.2+2} = 0.375$ without any assumptions on independence or correlation of $X_1$ and $X_2$.

So your answer fits exactly this problem. On the other hand you can also think about the following model: given that you encounter $S_1$ let $p$ be the probability that the $S_2$ will appear before you close your $S_1$ position. So, $p$ is the probability that you will lose $S_2$ if you admit $S_1$.

Once you encounter $S_1$ you should decide if admit it, or wait for the possible better situation $S_2$. Let us admit $S_1$ with a probability $\beta$ - then what is the optimal $\beta$? The possible outcomes are:

  1. you admit $S_1$ (pr = $\beta$), then pf = $X_1$;

  2. you reject $S_1$ (pr = $1-\beta$), $S_2$ appears (pr = $p$), then pf = $X_2$;

  3. you reject $S_1$ (pr = $1-\beta$), $S_2$ does not appear (pr = $1-p$), then pf = $0$;

so expected pf is $$ \mathsf E(\beta X_1+(1-\beta)p X_2)\to\max_{\beta\in [0,1]} $$ so you have in the right-hand side $(1.2-2p)\beta+2p$ that is you always admit $S_1$ if $p<0.6$ and always reject $S_1$ if $p>0.6$.


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