Andreas_Giger

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# Wiki Contributions

There used to be a thread on LW that dealt with interesting ways to make small sums of money and ways to reduce expenditure. I think among other things going to Australia for a year was discussed. Does anyone know which thread I'm talking about and can provide me with the link? I can't seem to find it.

The "many more days that include them" is the 3^n part in my expression that is missing from any per day series. This 3^n is the sum of all interviews in that coin flip sequence ("coin flip sequence" = "all the interviews that are done because one coin flip showed up tails", right?) and in the per day (aka per interview) series the exact same sum exists, just as 3^n summands.

In both cases, the weight of the later coin flip sequences increases, because the number of interviews (3^n) increases faster than the probabilistic weight of the coin flip (1/2^n) decreases.

However, this doesn't mean that there exists no Cesàro sum. In fact the existence of such a sum can be proven for my original expression because the quotient of the last two numerators (if we include both odd and even coin flips) of the isomorphic series is always 3:1, regardless of wether the last coin flip was even or odd. (The same thing can be said for the quotient of the last 3^n and 3^(n-1) summands of your series. Basically, the per day series is just a dragged out per coin flip series.)

The reason why my estimation for the Cesàro sum is 0.5 is that if we express that quotient in a way that one coin state is written first, then it alternates between 3:1 and 1:3, which results in 1:1 which is 0.5. Obviously this is not exact maths, but it's a good way for a quick estimation. (Alternatively, you could intuitively infer that if there exists a Cesàro sum it must be 0.5, because whether you look for even or odd coin flips gets increasingly irrelevant as the series approaches infinity.)

Also, since I haven't previously touched upon the subject of the isomorphic series: If we call my original expression f, then we can construct a function g where g(n) = f(n)-f(n-1) with f(-1) = 0, and a series a = g(0) + g(1) + g(2) + ...

Does that all make sense?

I think people have slightly misunderstood what I was referring to with this:

• There exist no problems shown to be possible in real life for which CDT yields superior results.
• There exists at least one problem shown to be possible in real life for which TDT yields superior results.

My question was whether there is a conclusive, formal proof for this, not whether this is widely accepted on this site (I already realized TDT is popular). If someone thinks such a proof is given somewhere in an article (this one?) then please direct me to the point in the article where I can find that proof. I'm very suspicious about this though, since the wiki makes blatantly false claims, e.g. that TDT performs better in one-shot PD than CDT, while in fact it can only perform better if access to source coude is given. So the wiki article feels more like promotion than anything.

Also, I would be very interested to hear about what kind of reaction from the scientific community TDT has received. Like, very very interested.

I take it that my approach was not discussed in the heated debate you had? Because it seems a good exercise for grad students.

Also, I don't understand why you think a per interview series would net fundamentally different results than a per coin toss series. I'd be interested in your reports after you (or your colleagues) have done the math.

I could have said that the beauty was simulated floor(5^x) times where x is a random real between 0 and n

Ah, I see now what you mean. Disregarding this new problem for the moment, you can still formulate my original expression on a per-interview basis, and it will still have the same Cesàro sum because it still diverges in the same manner; it just does so more continuously. If you envision a graph of an isomorphic series of my original expression, it will have "saw teeth" where it alternates between even and odd coin flips, and if you formulate that series on a per-interview basis, those saw teeth just get increasingly longer, which has no impact on the Cesàro sum (because the series alternates between those saw teeth).

Concerning your new problem, it can still be expressed as a series with a Cesàro sum, it's just a lot more complicated. If I were you, I'd first try to find the simplest variant of that problem with the same properties. Still, the fact that this is solvable in an analogous way should be clear, because you can essentially solve the "floor(5^x) times where x is a random real between 0 and n" part with a series for x (similar to the one for the original problem) and then have a series of those series for n. Basically you're adding another dimension (or recursion level), but not doing anything fundamentally different.

What do you mean by "time" in this case? It sounds like you want to interrupt the interviews at an arbitrary point even though Beauty knows that interviews are quantised in a 3^n fashion.

(1/2 3^0 + 1/8 3^2 + ...) / (1/2 3^0 + 1/4 3^1 + 1/8 * 3^2 + ...)

... which can be transformed into an infinite series with a Cesàro sum of 0.5, so that's my answer.

I don't like the notion of using different decision theories depending on the situation, because the very idea of a decision theory is that it is consistent and comprehensive. Now if TDT were formulated as a plugin that seamlessly integrated into CDT in such a way that the resulting decision theory could be applied to any and all problems and would always yield optimal results, then that would be reason for me to learn about TDT. However, from what I gathered this doesn't seem to be the case?