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# 物理代写|统计力学代写Statistical Mechanics代考|PH635 Cox’ “Axioms” and Theorem

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## 物理代写|统计力学代写Statistical Mechanics代考|Cox’ “Axioms” and Theorem

As an aside, let us mention also that, in 1946, Cox [90], inspired by previous ideas of Keynes [190], gave a foundation to the “subjective” approach to probability based on reasonings about the plausibility of propositions (see Jaynes [183] for an extensive discussion of this approach). Instead of assigning probabilities to events, as in elementary probabilities, or to sets, as in the mathematical version (see Appendix 2.A), Cox gives a numerical value to the plausibility $\mathcal{P}(p \mid q)$ of a proposition $p$ given that another proposition $q$ is true. ${ }^8$

Then, Cox imposes some rules of rationality on those plausibility assignments and derive from them, for a given proposition $\mathrm{r}$, the sum rule:
$\mathcal{P}(p$ or $q \mid r)=\mathcal{P}(p \mid r)+\mathcal{P}(q \mid r)-\mathcal{P}(p$ and $q \mid r)$,
and the product rule:
$\mathcal{P}(p$ and $q \mid r)=\mathcal{P}(p \mid q$ and $r) \mathcal{P}(q \mid r)=\mathcal{P}(q \mid p$ and $r) \mathcal{P}(p \mid r)$

## 物理代写|统计力学代写Statistical Mechanics代考|Bayesian Updating

Suppose that we have a certain number of hypotheses $H_1, H_2, \ldots, H_n$ and that we have assigned probabilities $P\left(H_i\right)$ to each of them, probabilities that exhaust all possibilities and are mutually exclusive:
$$\sum_{i=1}^n P\left(H_i\right)=1 .$$
Those probabilities are called the prior probabilities.
Now, we collect new data (D) and we want to know how to change our assignments of probabilities to those various hypotheses. We will write $P\left(H_i \mid D\right)$ for the (updated) probability of hypothesis $H_i$, given $D$.

We assume that we know enough about the system to compute the probabilities of the data, for each hypothesis: $P\left(D \mid H_i\right), i=1, \ldots, n$. Those probabilities are called the likelihoods.
Then we simply use Bayes’ formula:
$$P\left(H_i \mid D\right)=\frac{P\left(D \mid H_i\right) P\left(H_i\right)}{P(D)},$$
where $P(D)=\sum_{i=1}^n P\left(D \mid H_i\right) P\left(H_i\right)$; this implies that the new probabilities still add up to one:
$$\sum_{i=1}^n P\left(H_i \mid D\right)=1$$

## 物理代写|统计力学代写Statistical Mechanics代考|Cox’ “Axioms” and Theorem

$\mathcal{P}(p$ 或者 $q \mid r)=\mathcal{P}(p \mid r)+\mathcal{P}(q \mid r)-\mathcal{P}(p$ 和 $q \mid r)$ ，

$$\mathcal{P}(p \text { 和 } q \mid r)=\mathcal{P}(p \mid q \text { 和 } r) \mathcal{P}(q \mid r)=\mathcal{P}(q \mid p \text { 和 } r) \mathcal{P}(p \mid r)$$

## 物理代写|统计力学代写Statistical Mechanics代考|Bayesian Updating

$$\sum_{i=1}^n P\left(H_i\right)=1$$

$$P\left(H_i \mid D\right)=\frac{P\left(D \mid H_i\right) P\left(H_i\right)}{P(D)},$$

$$\sum_{i=1}^n P\left(H_i \mid D\right)=1$$

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