4 ms·

While I agree, that conclusion is wrong because you forgot to look at the win-win and loss-loss cases. Even with two coin tosses, as in your example, you'll win

by n2d4 3y ago

While I agree, that conclusion is wrong because you forgot to look at the win-win and loss-loss cases. Even with two coin tosses, as in your example, you'll win on average: 1.5 * 1.5 = 2.25 0.6 * 1.5 = 0.9 1.5 * 0.6 = 0.9 0.6 * 0.6 = 0.36 => 1.1025x total It's counter-intuitive because even though you almost always lose, you still win (linear) wealth on average (but not median). The difference is that if you have unlimited tosses available, you don't care about maximizing EV after X tosses. Instead you care about minimizing your risk of losing it all.

deleted 3y ago
[deleted]
hgsgm 3y ago
Why do you agree with something plainly wrong than you disproved?
- svnt 3y ago
  They are saying parent is correct that plus 50% and minus 50%, while appearing to be balanced, is a losing proposition for the individual. They are also saying parent is not correct in saying that it also loses on a population basis, showing that the average across a population is still winning.
  - n2d4 3y ago
    Yes, that's what I meant. You probably don't want to take unnecessary risk by going +50%/-50% markets, but it's not for the reason the parent mentioned in the conclusion.
    - cubefox 3y ago
      No the parent was right, I commented.
cubefox 3y ago
You should maximize the geometric EV, not the (usual) arithmetic EV. In your average calculation you took the arithmetic mean, but the geometric mean is relevant. (2.25 * 0.9 * 0.9 * 0.36)^(1/4) =0.9 Which is below 1! To test the geometric mean with the earlier example ("+100% is equivalent to -50%"): (2.0 * 0.5)^(1/2) =1, as expected.
- n2d4 3y ago
  Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic). Instead, you calculated the geometric growth rate, which the Kelly criterion would then maximize; but it doesn't maximize your average outcome. Try it; run a simulation, and you'll see what I mean! (Some guy already posted his results [1]) If you're curious, I wrote a fairly lengthy explanation in another comment [2] about why your calculation is preferable in many cases, but to maximize average wealth, you really do want to go all-in. [1] https://news.ycombinator.com/item?id=36982560 https://news.ycombinator.com/item?id=36982560 [2] https://news.ycombinator.com/item?id=36986147 https://news.ycombinator.com/item?id=36986147
  - cubefox 3y ago
    > Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic). That depends on how you define it. The arithmetic mean is usually presupposed when talking about expected values, but the arithmetic mean is not the only mean, and in fact it is inappropriate to use in some cases, so arithmetic expected value is likewise not the only "expected value", and in fact inappropriate sometimes. So yes, talking about non-arithmetic EV should be just as common as talking about non-arithmetic mean, even though it isn't. > Instead, you maximized the geometric growth rate, which is what the Kelly criterion does; but it doesn't maximize your average outcome. Well, it doesn't maximize the arithmetic average, but it does maximize the geometric average. :P There is no point in assuming we have only one kind of average and one kind of expectation. A suboptimal common usage of terms can in fact be a hindrance to thinking straight. You use logarithms to keep talking about the arithmetic mean/average, but in my opinion logarithms just obfuscate the fact that we are in geometric mean territory.

by n2d4 3y ago

deleted 3y ago

[deleted]

hgsgm 3y ago

Why do you agree with something plainly wrong than you disproved?

svnt 3y ago
They are saying parent is correct that plus 50% and minus 50%, while appearing to be balanced, is a losing proposition for the individual. They are also saying parent is not correct in saying that it also loses on a population basis, showing that the average across a population is still winning.
- n2d4 3y ago
  Yes, that's what I meant. You probably don't want to take unnecessary risk by going +50%/-50% markets, but it's not for the reason the parent mentioned in the conclusion.
  - cubefox 3y ago
    No the parent was right, I commented.

cubefox 3y ago

You should maximize the geometric EV, not the (usual) arithmetic EV. In your average calculation you took the arithmetic mean, but the geometric mean is relevant. (2.25 * 0.9 * 0.9 * 0.36)^(1/4) =0.9 Which is below 1! To test the geometric mean with the earlier example ("+100% is equivalent to -50%"): (2.0 * 0.5)^(1/2) =1, as expected.

n2d4 3y ago
Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic). Instead, you calculated the geometric growth rate, which the Kelly criterion would then maximize; but it doesn't maximize your average outcome. Try it; run a simulation, and you'll see what I mean! (Some guy already posted his results [1]) If you're curious, I wrote a fairly lengthy explanation in another comment [2] about why your calculation is preferable in many cases, but to maximize average wealth, you really do want to go all-in. [1] https://news.ycombinator.com/item?id=36982560 https://news.ycombinator.com/item?id=36982560 [2] https://news.ycombinator.com/item?id=36986147 https://news.ycombinator.com/item?id=36986147
- cubefox 3y ago
  > Not quite; there's no such thing as a geometric EV (EV is by definition always arithmetic). That depends on how you define it. The arithmetic mean is usually presupposed when talking about expected values, but the arithmetic mean is not the only mean, and in fact it is inappropriate to use in some cases, so arithmetic expected value is likewise not the only "expected value", and in fact inappropriate sometimes. So yes, talking about non-arithmetic EV should be just as common as talking about non-arithmetic mean, even though it isn't. > Instead, you maximized the geometric growth rate, which is what the Kelly criterion does; but it doesn't maximize your average outcome. Well, it doesn't maximize the arithmetic average, but it does maximize the geometric average. :P There is no point in assuming we have only one kind of average and one kind of expectation. A suboptimal common usage of terms can in fact be a hindrance to thinking straight. You use logarithms to keep talking about the arithmetic mean/average, but in my opinion logarithms just obfuscate the fact that we are in geometric mean territory.