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> Average wealth increases; but average log(wealth) decreases What’s the mathematical rationale (or intuition) for looking at the log values? I don’t quite un

> Average wealth increases; but average log(wealth) decreases What’s the mathematical rationale (or intuition) for looking at the log values? I don’t quite understand what they represent.

joncrocks 3y ago
It's reducing the impact of the outliers. e.g. x = 60, 10, 10, 10, 10 avg. x = 20 avg log(x) = 1.1556 (which is 10^1.1556 = 14.3)
twobitshifter 3y ago
Log is useful for repeated trials, since it’s a way of looking at exponential growth. If you write out the formula for a repeated trial, you’ll get an exponential function. Log tells you the value of the exponent a in e^at.
kgwgk 3y ago
Multiplication becomes addition after you take logs. That makes many things easier to understand and manipulate. Exponential trends become linear. The geometric random walk where you multiply by X or divide by Y with equal probability becomes an additive random walk where you add log(X) or substract log(Y) and you can represent that as a standard random walk (where the up and down steps have the same magnitude and probability) plus a trend.
n2d4 3y ago
The other people gave good intuition on why the logarithm is good but not on why the logarithm is the best choice. For example, max(sqrt(EV)) would also reduce outliers. There is a more mathematical way akin to the original derivation of the Kelly criterion [1]. Let's compare two strategies, max(EV) and max(log(EV)). Our game is coin-toss with heads=+50%/tails=-40% payouts and infinite repetitions; max(EV) would always go all-in, while max(log(EV)) would only bet a fraction of its current wealth at each turn. - After the first turn, there are two cases; if heads, max(EV) has a higher payout, if tails, max(log(EV)) has a higher payout. On average, max(EV) has a higher payout because it went all-in, and there's a positive EV per toss. - After two turns, there are four cases; in one of them (heads-heads), max(EV) has a higher payout, and in the other three, max(log(EV)) does. The latter is hence the "safer" choice. However, in the 25% chance that max(EV) wins, it has significantly more money than max(log(EV)), so the average (not median) payout is higher for max(EV). - This repeats, and after more and more repetitions, the chance that max(log(EV)) has a higher payout converges towards 1. Note that the payout for max(EV) in the (very!) rare case that it gets lucky skyrockets at an even faster speed, so on average, max(EV) still has a higher payout. - Turns out this is not just true for max(EV) and max(log(EV)), but in fact max(log(EV)) always maximizes the chance of higher payouts, given enough repetitions. So, in other words, max(EV) maximizes the average payout, while max(log(EV)) maximizes the chance of having a higher payout than any other strategy. The original Kelly paper has more precise maths, but I hope this helps to guide you. [1] https://en.wikipedia.org/wiki/Kelly_criterion https://en.wikipedia.org/wiki/Kelly_criterion
nzealand 3y ago
If you lose half your wealth, you have to double your wealth to get back to where you were. The log change intuitively shows this.... log (0.5) = -0.3 & log (2) = 0.3 If you lose three quarters of your wealth, you have to quadruple your wealth to get back to where you where. Again, log intuitively shows this.... log (0.25) = -0.6 & log (4) =0.6
kevinwang 3y ago
I think my explanation is more intuitive than every other comment so far. An intuition is that log(wealth) represents utility. An intuition for why log(wealth) represents utility is that wealth has decreasing marginal returns. The logarithm is a function that fits this. If you play around with it a bit you might see that it conforms to your intuitions: increasing your net worth from $100k to $200k increases your utility significantly, maybe about as much as increasing your net worth from $1m to $2m, and much more than increasing your net worth from $1m to $1.1m.

> Average wealth increases; but average log(wealth) decreases What’s the mathematical rationale (or intuition) for looking at the log values? I don’t quite understand what they represent.

joncrocks 3y ago

It's reducing the impact of the outliers. e.g. x = 60, 10, 10, 10, 10 avg. x = 20 avg log(x) = 1.1556 (which is 10^1.1556 = 14.3)

twobitshifter 3y ago

Log is useful for repeated trials, since it’s a way of looking at exponential growth. If you write out the formula for a repeated trial, you’ll get an exponential function. Log tells you the value of the exponent a in e^at.

kgwgk 3y ago

Multiplication becomes addition after you take logs. That makes many things easier to understand and manipulate. Exponential trends become linear. The geometric random walk where you multiply by X or divide by Y with equal probability becomes an additive random walk where you add log(X) or substract log(Y) and you can represent that as a standard random walk (where the up and down steps have the same magnitude and probability) plus a trend.

n2d4 3y ago

The other people gave good intuition on why the logarithm is good but not on why the logarithm is the best choice. For example, max(sqrt(EV)) would also reduce outliers. There is a more mathematical way akin to the original derivation of the Kelly criterion [1]. Let's compare two strategies, max(EV) and max(log(EV)). Our game is coin-toss with heads=+50%/tails=-40% payouts and infinite repetitions; max(EV) would always go all-in, while max(log(EV)) would only bet a fraction of its current wealth at each turn. - After the first turn, there are two cases; if heads, max(EV) has a higher payout, if tails, max(log(EV)) has a higher payout. On average, max(EV) has a higher payout because it went all-in, and there's a positive EV per toss. - After two turns, there are four cases; in one of them (heads-heads), max(EV) has a higher payout, and in the other three, max(log(EV)) does. The latter is hence the "safer" choice. However, in the 25% chance that max(EV) wins, it has significantly more money than max(log(EV)), so the average (not median) payout is higher for max(EV). - This repeats, and after more and more repetitions, the chance that max(log(EV)) has a higher payout converges towards 1. Note that the payout for max(EV) in the (very!) rare case that it gets lucky skyrockets at an even faster speed, so on average, max(EV) still has a higher payout. - Turns out this is not just true for max(EV) and max(log(EV)), but in fact max(log(EV)) always maximizes the chance of higher payouts, given enough repetitions. So, in other words, max(EV) maximizes the average payout, while max(log(EV)) maximizes the chance of having a higher payout than any other strategy. The original Kelly paper has more precise maths, but I hope this helps to guide you. [1] https://en.wikipedia.org/wiki/Kelly_criterion https://en.wikipedia.org/wiki/Kelly_criterion

nzealand 3y ago

If you lose half your wealth, you have to double your wealth to get back to where you were. The log change intuitively shows this.... log (0.5) = -0.3 & log (2) = 0.3 If you lose three quarters of your wealth, you have to quadruple your wealth to get back to where you where. Again, log intuitively shows this.... log (0.25) = -0.6 & log (4) =0.6

kevinwang 3y ago

I think my explanation is more intuitive than every other comment so far. An intuition is that log(wealth) represents utility. An intuition for why log(wealth) represents utility is that wealth has decreasing marginal returns. The logarithm is a function that fits this. If you play around with it a bit you might see that it conforms to your intuitions: increasing your net worth from $100k to $200k increases your utility significantly, maybe about as much as increasing your net worth from $1m to $2m, and much more than increasing your net worth from $1m to $1.1m.