Two weeks ago, Dr. Winston Ewert announced at Uncommon Descent a kind of *open mike*. He put up a page at Google Moderator and asked for questions. Unfortunately, not many took advantage of this offer, but I added three questions from the top of my head. The experience made me revisit the paper A General Theory of Information Cost Incurred by Successful Search again, and when I tried - as usual - to construct simple examples, I run into further questions - so, here is another one:

In their paper, the authors W. Dembski, W. Ewert, and R. Marks (DEM) talk about something they call the *natural probability*:

Processes that exhibit stochastic behavior arise from what may be called a *natural probability*. The natural probability characterizes the ordinary stochastic behavior of the process in question. Often the natural probability is the uniform probability. Thus, for a perfect cube with distinguishable sides composed of a rigid homogenous material (i.e., an ordinary die), the probability of any one of its six sides landing on a given toss is 1/6. Yet, for a loaded die, those probabilities will be skewed, with one side consuming the lion’s share of probability. For the loaded die, the natural probability is not uniform.

This

*natural probability* on the search space translates through their idea of lifting to the space of measures $\mathbf{M}(\Omega)$:

As the natural probability on $\Omega$, $\mu$ is not confined simply to $\Omega$ lifts to $\mathbf{M}(\Omega)$, so that its lifting, namely $\overline{\mu}$, becomes the natural probability on $\mathbf{M}(\Omega)$ (this parallels how the uniform probability $\mathbf{U}$, when it is the natural probability on $\Omega$, lifts to the uniform probability $\overline{\mathbf{U}}$ on $\mathbf{M}(\Omega)$, which then becomes the natural probability for this higher-order search space).

As usual, I look at an easy example: a loaded coin which always shows head. So $\Omega=\{H,T\}$ and $\mu=\delta_H$ is the natural measure on $\Omega$. What happens on $\mathbf{M}(\Omega)= \{h\cdot\delta_H + t\cdot\delta_T|0 \le h,t \le 1; h+t=1 \}$? Luckily,
$$(\mathbf{M}(\{H,T\}),\mathbf{U}) \cong ([0,1],\lambda).$$
Let's jump the hoops:

- The Radon-Nikodym derivative of $\delta_H$ with respect to $\mathbf{U}$ is $f(H) = \frac{d\delta_H}{d\mathbf{U}}(H) = 2$, $f(T) = \frac{d\delta_H}{d\mathbf{U}}(T) = 0$
- Let $\theta \in \mathbf{M}(\{H,T\})$, i.e., $\theta= h\delta_H + t\delta_T$. Then$$\overline{f}{(\theta)} = \int_{\Omega} f(x)d\theta(x)$$ $$=f(H)\cdot\theta(\{H\}) + f(T) \cdot\theta(\{T\})$$ $$=2 \cdot h$$

Here, I have the density of my natural measure on $\mathbf{M}(\Omega)$ with regard to $\overline{\mathbf{U}}$,
$$d\overline{\delta_H}(h\cdot\delta_H + t\cdot\delta_T) = 2 \cdot h \cdot d\overline{\mathbf{U}}(h\cdot\delta_H + t\cdot\delta_T).$$
But what is it good for? For the uniform probability, DEM showed the identity $$\mathbf{U}=\int_{\mathbf{M}(\Omega)}\theta d\overline{\mathbf{U}} .$$ Unfortunately, for $\int_{\mathbf{M}(\Omega)}\theta d\overline{\delta_H}$, I get nothing similar:
$$\int_{\mathbf{M}(\Omega)}\theta d\overline{\delta_H} = \frac{2}{3}\delta_H + \frac{1}{3}\delta_T$$

So, again, what does this mean? Wouldn't the Dirac delta function be a more *natural* measure on $\mathbf{M}(\Omega)$?

I hope that Dr. Winston Ewert reacts to all of the questions before *Google Moderator* shuts down for good on June 30, 2015...