Isn't log($) arbitrary?

Posted by Thicko on July 15, 1997 at 04:02:09: In Reply to: Re: Reasons for Kelly posted by Pete Moss on July 14, 1997 at 17:10:05:

Pete, you wrote:
"To maximize that product, it is sufficient to maximize its logarithm, because log() is a monotonically increasing function."

Does that mean any monotonically increasing function would do?

Bisser's point seems to be a good one - that using log() is equivalent to saying your utility for $ is log($). Log($) does sound fairly reasonable, as Bisser says, but it's still arbitrary. Log($) is concave, which is OK, since it's fair to suppose that investors are risk averse. But why shouldn't they be a bit more or less risk averse than what log($) implies? Is there really something special about log() in this context?

If you chose a different monotonically increasing, strictly concave function, how would that alter the result? Is it just that it's a lot easier to work out if you use log, since we know the expansion series?

Regards,


Mathematically challenged



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