I think this comment is incorrect (in the stated generality). Here is a simple counterexample. Suppose you have a starting endowment of $1, and that you can bet any amount at 0.50001 probability of doubling your bet and 0.49999 probability of losing everything you bet. You can bet whatever amount of your money you want a total of n times. (If you lost everything in some round, we can think of this as you still being allowed to bet 0 in remaining future rounds.) The strategy that maximizes expected linear utility is the one where you bet everything every time.
It depends on n. If n is small, such as n=1, then you should bet a lot. In the limit of n large, you should use the Kelly criterion. The crossover is about n=10^5. Which is why I said that it depends on having many opportunities.
You can prove e.g. by (backwards) induction that you should bet everything every time. With the odds being p>0.5 and 1-p, if the expectation of whatever strategy you are using after n-1 steps is E, then the maximal expectation over all things you could do on the n’th step is p2E (you can see this by writing the expectation as a conditional sum over the outcomes after n-1 steps), which corresponds uniquely to the strategy where you bet everything in any situation on the n’th step. It then follows that the best you can do on the (n-1)th step is also to maximize the expectation after it, and the same argument gives that you should bet everything, and so on.
(Where did you get n=10^5 from? If it came from some computer computation, then I would wager that there was some overflow/numerical issues.)
If you have money x after n-1 steps, then betting a fraction f on the n’th step gives you expected money (1-f)x+f2px. Given p>0.5, this is maximized at f=1, i.e. betting everything, which gives the expectation 2px. So conditional on having money x after n-1 steps, to maximize expectation after n steps, you should bet everything. Letting X_i be the random variable that is the amount of money you have after i steps given your betting strategy. We have E[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x] (one could also write down a continuous version of the same conditioning but it is a bit easier to read if we assume that the set of possible amounts of money after n-1 steps is discrete, which is what I did here). From this formula, it follows that for any given strategy up to step n-1, hence given values for P(X_{n-1}=x), the thing to do on step n that maximizes E[X_n] is the same as the thing to do that maximizes E[X_n|X_{n-1}=x] for each x. So to maximize E[X_n], you should bet everything on the n’th step. If you bet everything, then the above formula gives E[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x]=∑xP(Xn−1=x)2px=2p∑xP(Xn−1=x)x
=2pE[Xn−1].
To recap what we showed so far: we know that given any strategy for the first n-1 steps, the best thing to do on the last step gives E[X_n]=2pE[X_{n-1}]. It follows that the strategy with maximal E[X_n] is the one with maximal 2pE[X_{n-1}], or equivalently the one with maximal E[X_{n-1}].
Now repeat the same argument for step n-1 to conclude that one should bet everything on step n-1 to maximize the expectation after it, and so on.
Or maybe to state a few things a bit more clearly: we first showed that E[X_n|X_{n-1}=x]<=2px, with equality iff we bet everything on step n. Using this, note thatE[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x]≤∑xP(Xn−1=x)2px=2p∑xP(Xn−1=x)x
=2pE[Xn−1], with equality iff we bet everything on step n conditional on any value of X_{n-1}. So regardless of what you do for the first n-1 steps, what you should do on step n is to bet everything, and this gives you the expectation E[X_n]=2pE[X_{n-1}]. Then finish as before.
I think this comment is incorrect (in the stated generality). Here is a simple counterexample. Suppose you have a starting endowment of $1, and that you can bet any amount at 0.50001 probability of doubling your bet and 0.49999 probability of losing everything you bet. You can bet whatever amount of your money you want a total of n times. (If you lost everything in some round, we can think of this as you still being allowed to bet 0 in remaining future rounds.) The strategy that maximizes expected linear utility is the one where you bet everything every time.
It depends on n. If n is small, such as n=1, then you should bet a lot. In the limit of n large, you should use the Kelly criterion. The crossover is about n=10^5. Which is why I said that it depends on having many opportunities.
You can prove e.g. by (backwards) induction that you should bet everything every time. With the odds being p>0.5 and 1-p, if the expectation of whatever strategy you are using after n-1 steps is E, then the maximal expectation over all things you could do on the n’th step is p2E (you can see this by writing the expectation as a conditional sum over the outcomes after n-1 steps), which corresponds uniquely to the strategy where you bet everything in any situation on the n’th step. It then follows that the best you can do on the (n-1)th step is also to maximize the expectation after it, and the same argument gives that you should bet everything, and so on.
(Where did you get n=10^5 from? If it came from some computer computation, then I would wager that there was some overflow/numerical issues.)
Sorry, I’m confused.
I got 10^5 from 1/(p-1/2).
If you have money x after n-1 steps, then betting a fraction f on the n’th step gives you expected money (1-f)x+f2px. Given p>0.5, this is maximized at f=1, i.e. betting everything, which gives the expectation 2px. So conditional on having money x after n-1 steps, to maximize expectation after n steps, you should bet everything. Letting X_i be the random variable that is the amount of money you have after i steps given your betting strategy. We have E[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x] (one could also write down a continuous version of the same conditioning but it is a bit easier to read if we assume that the set of possible amounts of money after n-1 steps is discrete, which is what I did here). From this formula, it follows that for any given strategy up to step n-1, hence given values for P(X_{n-1}=x), the thing to do on step n that maximizes E[X_n] is the same as the thing to do that maximizes E[X_n|X_{n-1}=x] for each x. So to maximize E[X_n], you should bet everything on the n’th step. If you bet everything, then the above formula gives E[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x]=∑xP(Xn−1=x)2px=2p∑xP(Xn−1=x)x
=2pE[Xn−1].
To recap what we showed so far: we know that given any strategy for the first n-1 steps, the best thing to do on the last step gives E[X_n]=2pE[X_{n-1}]. It follows that the strategy with maximal E[X_n] is the one with maximal 2pE[X_{n-1}], or equivalently the one with maximal E[X_{n-1}].
Now repeat the same argument for step n-1 to conclude that one should bet everything on step n-1 to maximize the expectation after it, and so on.
Or maybe to state a few things a bit more clearly: we first showed that E[X_n|X_{n-1}=x]<=2px, with equality iff we bet everything on step n. Using this, note thatE[Xn]=∑xP(Xn−1=x)E[Xn|Xn−1=x]≤∑xP(Xn−1=x)2px=2p∑xP(Xn−1=x)x
=2pE[Xn−1], with equality iff we bet everything on step n conditional on any value of X_{n-1}. So regardless of what you do for the first n-1 steps, what you should do on step n is to bet everything, and this gives you the expectation E[X_n]=2pE[X_{n-1}]. Then finish as before.