Lot Sizing by Profit Factor and Choosing Between Trade Setups
Summary
The discussion compares two ways to allocate capital between trade situations with different estimated profit factors. Scaling lots in proportion to each factor is presented as a way to maximize average profit factor across a fixed number of trials. With the stated estimates of 1.2 and 2.0, this allocation assigns 0.375 of the total to the first situation; the argument does not require the outcomes to be independent or uncorrelated.
It then considers a different decision: whether to take the lower-factor setup now or wait for a better one. If the better setup may arrive before the first position closes, the relevant quantity is its chance of appearing during that window. Under the simplified expected-value model, take the first setup when that chance is below 0.6 and reject it when the chance is above 0.6. These results depend on the stated objective and simplified setup, and do not account for risk, uncertainty in estimated profit factors, or other portfolio constraints.
Key ideas
- Proportional allocation by estimated profit factor maximizes average profit factor for a fixed number of trials in the stated model.
- The allocation result does not depend on independence or correlation between the two setup outcomes.
- When deciding whether to wait for a better setup, compare its arrival probability during the current trade's holding period with the stated threshold.
- The model optimizes expected profit factor and does not include risk or estimation uncertainty.
Tags
Full text
# How should you manage lot sizes in this situation?
# How should you manage lot sizes in this situation?
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 by SBF (score 5, accepted)
https://quant.stackexchange.com/a/1954
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:
- you admit $S_1$ (pr = $\beta$), then pf = $X_1$;
- you reject $S_1$ (pr = $1-\beta$), $S_2$ appears (pr = $p$), then pf = $X_2$;
- 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 $p>0.6$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.