Quote from kut2k2:
SPS == k*W
k(A) = .55/407 - .45/433 = .0003121
SPS(A) = k(A)*433 = 0.135
k(B) = .10/50 - .90/1000 = .0011
SPS(B) = k(B)*1000 = 1.1
System B is much more attractive.
That's not the general case, just the special case of exactly two outcomes (only one win amount and only one loss amount). For the general case of multiple outcomes, I prefer the formula I wrote (k*W) using the exact value for k, because the Kelly approximation (= p/|L| - q/W) is always an overestimation, which means you're overtrading. Not good to overtrade.Quote from Visaria:
Took me a few minutes admittedly to figure out, but this "SPS", lol, is just the expected value per dollar risked!!!!!
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SPS(C) = 1.221 actually.Quote from Visaria:
According to SPS, lol, system C (see above) is even more attractive with a SPS of 1.12.
But tbh, i wouldn't want to trade a system where the prob of a winner is 0.1%.
Quote from kut2k2:
... , because the Kelly approximation (= p/|L| - q/W) is always an overestimation, which means you're overtrading. Not good to overtrade.
Great post. Unfortunately I do not have the proof you seek. My conclusion is empirical.Quote from bustermu:
kut2k2,
I made an equivalent claim in:
http://www.elitetrader.com/vb/showthread.php?s=&postid=3028937#post3028937
Any references to and/or proofs of the claim would be greatly appreciated.
The proof I have uses Jensen's Inequality for conditional expectations which I believe to be beyond the mathematical level of most traders.
Thanks,
Jim Murphy