How should you manage lot sizes in this situation?

Question

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.

Answer 1

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 $p>0.6$.