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.