Quote from DontMissTheBus:
The easiest (and most naive way) to do it is just with conditional probability.
For system 1: prob(t=6, pnl>+10|t=1...5) = 4/5
For system 2: prob(t=6, pnl>+10|t=1...5) = 3/5
The amount of effort to improve on the estimation using this method rises dramatically... and not necessarily better.
).I agree with your PS. Let's try a different approach.Quote from braincell:
This is a question for those good at Math and probability. It has to do with probability distribution and probability mass function.
Now, let's say we have two trading systems; System1 and System2.
Each places trades for 5 days and these are the end of day profit results:
System1:
1: +210
2: +90
3: +20
4: +70
5: -340
Total System1 = +50
Average daily = +10
System2:
1: +20
2: -10
3: +15
4: -5
5: +30
Total System2 = +50
Average daily = +10
--------------------------------
On the face of it, both systems are equally profitable. However, the price distribution on system1 is much wider than system2, indicating
there was an element of luck involved. This means that the true profitability (predicted forward) of system1 should be lower than system2,
if we were to adjust according to the risk and unreliability of the results.
I am using only 5 days for the sake of simplicity, but sample size will be a constraint in real testing also.
Question:
What is the probability that the next (6th) day will be +10 (average daily) for system1 and system2?
I expect that it will be higher for system2 and lower for system1. How would I calculate exactly by how much?
I tried using the probability distribution formulas but i haven't had any luck. I'm overworked and underslept for the last few days and
this just isn't getting through to my brain. I normally solve these things myself but for whatever reason my brain is just stuck on a seemingly simple problem.
Embarrassing.
Any clues will be appreciated.
PS
I'm not even sure i'm approaching this problem correctly, or that it has so much to do with probability. Maybe i'm just using the wrong formulas.
Thanks.
Quote from kut2k2:
I agree with your PS. Let's try a different approach.
Edge = E[x]/sqrt[E[x²]]
System1: E[x]/sqrt[E[x²]] = +10/sqrt[34620] = 0.05374
System2: E[x]/sqrt[E[x²]] = +10/sqrt[330] = 0.55048
System2 is clearly superior to system1.
You're talking out of your ass. What makes you ASSume that there's a stable population being modeled here? I don't follow the common nonsense of treating a nonstationary time series as a probability distribution, I just use the information at hand to collect useful metrics. What you amusingly ASSume is the signal-to-noise ratio is in fact the trader's edge. It's plainly labeled. A signal-to-noise ratio applies only to the underlying, not to a trading system.Quote from ronblack:
You can only use these calculation when there is a sufficient large sample.
If you have ever taken a course in statistics and probability you should have learned that the average is a good estimator of the expected value only for sufficient samples.
Besides, why do you think the signal-to-noise ratio makes sense for a r. variable that takes negative values?
Quote from kut2k2:
You're talking out of your ass. What makes you ASSume that there's a stable population being modeled here? I don't follow the common nonsense of treating a nonstationary time series as a probability distribution, I just use the information at hand to collect useful metrics. What you amusingly ASSume is the signal-to-noise ratio is in fact the trader's edge. It's plainly labeled.
Speak for yourself.Quote from ronblack:
You exhibit a total lack of formal education.
I repeat to you: the average of the values of a r. variable converges to its expected value only for sufficient samples.
Didn't use the expected value notation?
Do you understand the difference between expected value and average?
You definitely do not.
If ever there was any doubt about your psychosis, you've now removed all of it.Quote from ronblack:
If you come back with insults I will contact your school teacher and ask him to put black pepper in your dirty mouth tomorrow, first thing, if you show up in school I guess..

Quote from braincell:
Interesting approach kut2k2... I'd have to try it on other small samples to see how it pans out.
....
Thanks for your input everyone.
Quote from kcgoogler:
Somthing is wrong with the above computations. Intuitively i think the OP is correct. The +10 day happening on the 6th day should be higher for the second system. Intuitively consider this.. say this is a continous distribution (you should technically use pmf's for this but for convinience sake assume its a cont dist); then the 1std dev range for the first system would be something like 10+-50-60ish. For the second system something like 10+-5-10ish. So the probability for the thing hittting exactly 10 should be more for the system with less variability.
Ofcourse, if you do rigorous math my intuition might end up wrong (it has in the past).
-gariki