Bayes theorem with unknown model

Question

I have a book on applied statistics that uses a result I don't understand. The example in the book begins with Bayes' theorem applied to hypothesis $H$ and data $D$:

$$P(H|D) = \frac{P(D|H) P(H)}{P(D|\neg H) P(\neg H) + P(D|H) P(H)} \tag{1}$$

In the example it is given that $P(H) = P(\neg H) = 0.5$. This simplifies the equation for $P(H|D)$ since $P(H)$ and $P(\neg H)$ cancel:

$$P(H|D) = \frac{P(D|H)}{P(D|\neg H) + P(D|H)} \tag{2}$$

It is also given that we do not have a model for $P(D|H)$ so we are to assume that each value of $P(D|H)$ is equally likely (i.e., "the unknown value for $P(D|H)$ is uniformly distributed over the interval $[0,1]$").

Given the uniform distribution of $P(D|H)$, the following equation for $P(H|D)$ is stated:

$$P(H|D) = 1-P(D|\neg H) * ln(\frac{1 + P(D|\neg H)}{P(D|\neg H)}) \tag{3}$$

In the example, $P(D|\neg H)$ is known so equation $(3)$ can be used to solve for $P(H|D)$). This concludes the book's example.

However, I do not understand the justification for going from equation $(2)$ to $(3)$. Can anyone enlighten me as to how assuming $P(D|H)$ is uniformly distributed over $[0,1]$ allows for result $(3)$?

Answer 1

I figured it out.

Since $P(D|H)$ is uniformly distributed over $[0,1]$, you can calculate $P(H|D)$ by taking the integral from $0$ to $1$:

(for ease of reading, I substitute $x = P(D|\neg H)$ and $y = P(D|H)$)

\begin{align} P(H|D) & = \int_0^1 \frac{y}{x+y} dy \\ & = [y - x \cdot \ln(y+x)] \Big|_{y=0}^{y=1} \\ & = 1 - x \cdot \ln(1 + x) + x \cdot \ln x \\ & = 1 - x \cdot \big( \ln(1+x) - \ln x \big) \\ & = 1 - x \cdot \ln \frac{1+x}{x} \end{align}

Replace $x = P(D|\neg H)$ and you have result $(3)$ in my question above:

$$P(H|D) = 1 - P(D|\neg H) \cdot \ln(\frac{1+P(D|\neg H)}{P(D|\neg H)})$$

I hope this helps someone else.