It would probably be helpful if you considered the concept of variance. Akin to this are the standard deviation and confidence interval. The variance of series I is much larger than series II.
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
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 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?
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.
Watch your language here...These posts stay...If you want to insult people talk to you own family. Not here... You did not use any information. You messed up with the information. You played like a kid plays on the beach with sand. 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 you use the expected value notation? Do you understand the difference between expected value and average? You definitely do not. 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..
Speak for yourself. Expected value = mean = arithmetic average. And what makes you ASSume that trade results are random variables? Just because you can't model them, that doesn't automatically make them random. Go back to remedial math; you're just embarrassing yourself at this point. If ever there was any doubt about your psychosis, you've now removed all of it.
Interesting approach kut2k2... I'd have to try it on other small samples to see how it pans out. The constraint of small sample will always be there, so in a way, the method needed needs to deal as best as possible with this constraint. I don't get what it is about this forum that people are so strongly opinionated about a seemingly concrete discussion (math of all things) that they easily resort to insults of varying severity. If you disagree, don't you think that just "stating it" would be enough of a thorn in the side of the person you're disagreeing with and cause enough of a reaction without the added name calling? It's like all traders think other traders/system devs are their enemies or something. They're not. If there were no other traders, prices wouldn't be making the intraday (even intramonth) moves they are, and all prices would move like they do in non-RTH. But let's not get offtopic. Thanks for your input everyone.
I hope you realize that kut2k2 is a crank, right? He regularly pose as an expert in subjects he has no real understanding of - For example, managed to recreate the sharpe ratio with E[X]/E[X^2] without even realizing it while calling everyone incorrect.
You realize this was in reply to the OP's INITIAL question, "What is the probability that the next (6th) day will be +10 (average daily) for system1 and system2?" - which is not the same as which system is better, right? Further, thanks for recreating the Sharpe Ratio in E[X]/sqrt([E[X*X]]) (that bottom part is just standard deviation assuming zero mean, which is often used) without realizing you've done so - if you bother to read all the posts preceding you, we've already discussed that