Bayes' rule: Functional form

Discuss the wikitag on this page. Here is the place to ask questions and propose changes.
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Eliezer Yudkowsky I think there should be a small change here? Variable f becomes x and back to f, and I believe it should just be f?

Or, writing out the whole operation from scratch:

Invalid LaTeX $\mathbb P(f\mid e\!=\!\textbf {THT}) = \dfrac{\mathcal L(e\!=\!\textbf{THT}\mid f) \cdot \mathbb P(f)}{\mathbb P(e\!=\!\textbf {THT})} = **\dfrac{(1 - x) \cdot x \cdot (1 - x) \cdot 1}{\int_0^1 (1 - x) \cdot x \cdot (1 - x) \cdot 1 \** \operatorname{d}\!f} = 12 \cdot f(1 - f)^2: TeX parse error: Undefined control sequence \*

The formula uses "x"s, but should it use "f"s instead?

If you're going to start using probability density functions instead of just probability functions, I'd introduce their type and consider using a different symbol (e.g. lowercase p) for PDFs. I expect some people to get fairly confused when you use but drop in a in out of nowhere -- "how did a tiny distance come into this!?" cries a reader-model.

Also, I'm not sure it's standard to include the delta in the density -- I'm more familiar with notation saying that the uniform PDF is 1 everywhere, and the cumulative distribution is the integral times .