Hmm, I’ve thought about this some more and I actually still don’t understand it. I might just be being dense, but I feel like you’ve made a very interesting claim here that would be important if true, so I’d really like to understand it. Perhaps others can benefit as well.
Here’s what I was able to work out for myself. Given that log(X+d) ~ log(X) + d/X, then:
So maximizing E[d/X] should be approximately equivalent to maximizing E[log(1 + d/X)]. This is looking closer to what you said, but there are two things I still don’t understand.
It looks like you’ve multiplied the second term (E[log(1+d/X)]) through by an X. Can you do that within an expectation, given that X isn’t a constant?
Even once you have E[log(d+X)] as a maximization target, I’d describe that as maximizing the log of the sum of your wealth and the world’s. And it seems like a quite different goal from maximizing the log-wealth of the world. Is there another step I’m missing?
It looks like you’ve multiplied the second term (E[log(1+d/X)]) through by an X. Can you do that within an expectation, given that X isn’t a constant?
You’re multiplying by X inside the log. That amounts to adding log(X), and an expectation of a sum is just the sum of the expectation. But this does seem to change exactly what you’re maximising.
Even once you have E[log(d+X)] as a maximization target, I’d describe that as maximizing the log of the sum of your wealth and the world’s. And it seems like a quite different goal from maximizing the log-wealth of the world. Is there another step I’m missing?
My derivation goes: Let Y denote the wealth of the world not controlled by you. By assumption Y is an independent variable (note: this assumption seems questionable, and without it the two conclusions definitely come apart, as the investor may have opportunities which increase the wealth of the rest of the world but not the wealth of the investor).
So X = Y + d.
So E(d/X)
= E(d/(Y+d))
~ E(log(1 + d/(Y+d)))
= E(log((Y + d + d)/Y))
= E(log(1+2d)/Y))
~ E(2d/Y)
= 2 E(d/Y)
= 2 E(log(1+d/Y))
= 2 E(log(Y + d) - log(Y))
= 2 E(log(X) - log(Y))
= 2 E(log(X)) − 2 E(log(Y))
Now since Y is independent, E(log(Y)) is constant, so maximising this is equivalent to maximising E(log(X)).
Edit: how did you do the code? I had difficulty with formatting, hence the excess line breaks.
How did you get from one of these steps to the other? Shouldn’t the second be E[log((Y+d+d)/(Y+d))]?
Yeah, it should. Hmm. Also, looking at my derivation as a whole it’s implausible that E(d/X) ~ E(2d/Y) -- the second should be almost twice as big.
So I think I made a mistake there. I suspect that to get the result we need to use the approximation d/X ~ d/Y (roughly right since d is very small in comparison to Y).
Hmm, I’ve thought about this some more and I actually still don’t understand it. I might just be being dense, but I feel like you’ve made a very interesting claim here that would be important if true, so I’d really like to understand it. Perhaps others can benefit as well.
Here’s what I was able to work out for myself. Given that log(X+d) ~ log(X) + d/X, then:
So maximizing E[d/X] should be approximately equivalent to maximizing E[log(1 + d/X)]. This is looking closer to what you said, but there are two things I still don’t understand.
It looks like you’ve multiplied the second term (E[log(1+d/X)]) through by an X. Can you do that within an expectation, given that X isn’t a constant?
Even once you have E[log(d+X)] as a maximization target, I’d describe that as maximizing the log of the sum of your wealth and the world’s. And it seems like a quite different goal from maximizing the log-wealth of the world. Is there another step I’m missing?
You’re multiplying by X inside the log. That amounts to adding log(X), and an expectation of a sum is just the sum of the expectation. But this does seem to change exactly what you’re maximising.
My derivation goes: Let Y denote the wealth of the world not controlled by you. By assumption Y is an independent variable (note: this assumption seems questionable, and without it the two conclusions definitely come apart, as the investor may have opportunities which increase the wealth of the rest of the world but not the wealth of the investor).
So X = Y + d.
So E(d/X)
= E(d/(Y+d))
~ E(log(1 + d/(Y+d)))
= E(log((Y + d + d)/Y))
= E(log(1+2d)/Y))
~ E(2d/Y)
= 2 E(d/Y)
= 2 E(log(1+d/Y))
= 2 E(log(Y + d) - log(Y))
= 2 E(log(X) - log(Y))
= 2 E(log(X)) − 2 E(log(Y))
Now since Y is independent, E(log(Y)) is constant, so maximising this is equivalent to maximising E(log(X)).
Edit: how did you do the code? I had difficulty with formatting, hence the excess line breaks.
Good catch, my bad.
Add four spaces at the beginning of a line to make it appear as code.
How did you get from one of these steps to the other? Shouldn’t the second be E[log((Y+d+d)/(Y+d))]?
Yeah, it should. Hmm. Also, looking at my derivation as a whole it’s implausible that E(d/X) ~ E(2d/Y) -- the second should be almost twice as big.
So I think I made a mistake there. I suspect that to get the result we need to use the approximation d/X ~ d/Y (roughly right since d is very small in comparison to Y).