Friday, June 10, 2016

Sleight of hand with the regulator


Just to make explicit the switch from the ordinary mean for the ensemble average to the geometric mean in the time average in Ole Peters paper on the resolution of the St Petersburg paradox (that I talked about yesterday), compare equations (5.2) and (5.7)

$$
\text{(5.2) }\; \langle r \rangle_{N} = \frac{1}{N} \sum_{i = 1}^{N} r_{i}
$$

$$
\text{(5.7) }\; \bar{r}_{T} = \left( \prod_{i = 1}^{T} r_{i} \right)^{1/T}
$$

The probabilities $p_{i}$ of each $r_{i}$ are inserted (there is a tiny subtlety here, but the result is as if one had just mapped $r_{i} \rightarrow N p_{i} r_{i}$ in one case and $r_{i} \rightarrow  r_{i}^{T p_{i}}$ in the other) as one would for an ordinary mean and a geometric mean (absorbing the $1/N$ and $1/T$, respectively):

$$
\text{(5.2a) }\; \langle r \rangle_{N} =  \sum_{i = 1}^{N} p_{i} r_{i}
$$

$$
\text{(5.7a) }\; \bar{r}_{T} =  \prod_{i = 1}^{T} r_{i}^{p_{i}}
$$

There is a notational difference designating the time average, but we could re-write (5.7a) as

$$
\langle r \rangle_{T} =  \prod_{i = 1}^{T} r_{i}^{p_{i}}
$$

Note that $T$ is a dummy index (it is taken to dimensionless infinity later just like $N$), so WOLOG we can take $T \rightarrow N$ and the previous equation can be rewritten:

$$
\langle r \rangle_{N} = \exp \left( \sum_{i = 1}^{N} p_{i} \log r_{i} \right)
$$

Peters takes the logarithm of both (5.2) and (5.7) and $N \rightarrow \infty$ to obtain the final result

$$
\log \langle r \rangle = \log \sum_{i = 1}^{\infty} p_{i} r_{i}
$$

$$
\log \langle r \rangle =  \sum_{i = 1}^{\infty} p_{i} \log r_{i}
$$

The first sum diverges. The second sum converges because the infinity has been regulated

$$
\log r_{i} = \log (w - c + 2^{i-1}) - \log (w)
$$

The regulator entered at the beginning -- by taking the geometric mean.

2 comments:

  1. The arithmetic ('ordinary') mean is vastly more popular than the geometric mean. Probably because it's easier to compute. But just because something is easy to compute doesn't mean that it is also what you *should* compute. :)

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    Replies
    1. Regardless of the relevant mean, you should at least use the same mean if you are trying to argue two quantities are unequal. That the geometric mean is not equal to the arithmetic mean is a fairly trivial observation.

      That one can regulate an infinite quantity is interesting, but not what Peters was trying to show.

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