Here is the result on which the argument is based:

(I) Suppose your credences violate the Principal Principle but satisfy Probabilism. Then there is a book of bets and a price such that: (i) you consider that price favourable for that book -- that is, your subjective expectation of the total net gain is positive; (ii) every possible objective chance function considers that price unfavourable -- that is, the objective expectation of the total net gain is guaranteed to be negative.

(II) Suppose your credences satisfy both the Principal Principle and Probabilism. Then there is no book of bets and a price such that: (i) you consider that price favourable for that book; (ii) every possible objective chance function considers that price unfavourable.

Put another way:

(I') Suppose your credences violate the Principal Principle. There are two actions $a$ and $b$ such that: you prefer $b$ to $a$, but every possible objective chance function prefers $a$ to $b$.

(II') Suppose your credences satisfy the Principal Principle. For any two actions $a$ and $b$: if every possible objective chance function prefers $a$ to $b$, then you prefer $a$ to $b$.

To move from (I) and (II) to (I') and (II'), let $a$ be the action of accepting the bets in $B$ and let $b$ be the action of rejecting them.

The proof splits into two parts:

(1) First, we note that a credence function $c$ satisfies the Principal Principle iff $c$ is in the closed convex hull of the set of possible chance functions.

(2) Second, we prove that:

(2I) If a probability function $c$ lies outside the closed convex hull of a set of probability functions $\mathcal{X}$, then there is a book of bets and a price such the expected total net gain from that book at that price by the lights of $c$ is positive, while the expected total net gain from that book at that price by the lights of each $p$ in $\mathcal{X}$ is negative.

(2II) If a probability function $c$ lies inside the closed convex hull of a set of probability functions $\mathcal{X}$, then there is no book of bets and a price such the expected total net gain from that book at that price by the lights of $c$ is positive, while the expected total net gain from that book at that price by the lights of each $p$ in $\mathcal{X}$ is negative.

Here's the proof of (2), which I lift from my recent justification of linear pooling -- the same technique is applicable since the Principal Principle essentially says that you should set your credences by applying linear pooling to the possible objective chances.

First:

- Let $\Omega$ be the set of possible worlds
- Let $\mathcal{F} = \{X_1, \ldots, X_n\}$ be the set of propositions over which our probability functions are defined. So each $X_i$ is a subset of $\Omega$.

- We represent a probability function $p$ defined on $\mathcal{F}$ as a vector in $\mathbb{R}^n$, namely, $p = \langle p(X_1), \ldots, p(X_n)\rangle$.
- Given a proposition $X$ in $\mathcal{F}$ and a stake $S$ in $\mathbb{R}$, we define the bet $B_{X, S}$ as follows: $$B_{X, S}(\omega) = \left \{ \begin{array}{ll}

S & \mbox{if } \omega \in X \\

0 & \mbox{if } \omega \not \in X

\end{array}

\right.$$ So $B_{X, S}$ pays out $S$ if $X$ is true and $0$ if $X$ is false. - We represent the book of bets $\sum^n_{i=1} B_{X_i, S_i}$ as a vector in $\mathbb{R}^n$, namely, $S = \langle S_1, \ldots, S_n\rangle$.

**Lemma 1**

If $p$ is a probability function on $\mathcal{F}$, the expected payoff of the book of bets $\sum^n_{i=1} B_{X_i, S_i}$ by the lights of $p$ is $$S \cdot p = \sum^n_{i=1} p(X_i)S_i$$

**Lemma 2**

Suppose $c$ is a probability function on $\mathcal{F}$, $\mathcal{X}$ is a set of probability functions on $\mathcal{F}$, and $\mathcal{X}^+$ is the closed convex hull of $\mathcal{X}$. Then, if $c \not \in \mathcal{X}^+$, then there is a vector $S$ and $\varepsilon > 0$ such that, for all $p$ in $\mathcal{X}$, $$S \cdot p < S \cdot c - \varepsilon$$

*Proof of Lemma*

*2*. Suppose $c \not \in \mathcal{X}^+$. Then let $c^*$ be the closest point in $\mathcal{X}^+$ to $c$. Then let $S = c - c^*$. Then, for any $p$ in $\mathcal{X}$, the angle $\theta$ between $S$ and $p - c$ is obtuse and thus $\mathrm{cos}\, \theta < 0$. So, since $S \cdot (p - c) = ||S||\, ||x - p|| \mathrm{cos}\, \theta$ and $||S||, ||p - c|| > 0$, we have $S \cdot (p - c) < 0$. And hence $S \cdot p < S \cdot c$. What's more, since $\mathcal{X}^+$ is closed, $p$ is not a limit point of $\mathcal{X}^+$, and thus there is $\delta > 0$ such that $||p - c|| > \delta$ for all $p$ in $\mathcal{X}$. Thus, there is $\varepsilon > 0$ such that $S \cdot p < S \cdot c - \varepsilon$, for all $p$ in $\mathcal{X}$.

We now derive (2I) and (2II) from Lemmas 1 and 2:

Let $\mathcal{X}$ be the set of possible objective chance functions. If $c$ violates the Principal Principle, then $c$ is not in $\mathcal{X}^+$. Thus, by Lemma 2, there is a book of bets $\sum^n_{i=1} B_{X_i, S_i}$ and $\varepsilon > 0$ such that, for any objective chance function $p$ in $\mathcal{X}$, $S \cdot p < S \cdot c - \varepsilon$. By Lemma 1, $S \cdot p$ is the expected payout of the book of bets by the lights of $p$, while $S \cdot c$ is the expected payout of the book of bets by the lights of $c$. Now, suppose we were to offer an agent with credence function $c$ the book of bets $\sum^n_{i=1} B_{X_i, S_i}$ for the price of $S \cdot c - \frac{\varepsilon}{2}$. Then this would have positive expected payoff by the lights of $c$, but negative expected payoff by the lights of each $p$ in $\mathcal{X}$. This gives (2I).

(2II) then holds because, when $c$ is in the closed convex hull of $\mathcal{X}$, its expectation of a random variable is in the closed convex hull of the expectations of that random variable by the lights of the probability functions in $\mathcal{X}$. Thus, if the expectation of a random variable is negative by the lights of all the probability functions in $\mathcal{X}$, then its expectation by the lights of $c$ is not positive.

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P versus NP is considered as one of the most important open problems in computer science. This consists in knowing the answer of the following question: Is P equal to NP? This question was first mentioned in a letter written by John Nash to the National Security Agency in 1955. A precise statement of the P versus NP problem was introduced independently in 1971 by Stephen Cook and Leonid Levin. Since that date, all efforts to find a proof for this problem have failed. Another major complexity classes are LOGSPACE and NLOGSPACE. Whether LOGSPACE = NLOGSPACE is another fundamental question that it is as important as it is unresolved. SAT is easier if the number of literals in a clause is limited to at most 2, in which case the problem is called 2SAT. This problem can be solved in polynomial time, and in fact is complete for the complexity class NLOGSPACE. If additionally all OR operations in literals are changed to XOR operations, the result is called exclusive-or 2-satisfiability, which is a problem complete for the complexity class LOGSPACE. Given an instance of exclusive-or 2-satisfiability and a positive integer K, the problem maximum exclusive-or 2-satisfiability consists in deciding whether this Boolean formula has a truth assignment with at leat K satisfiable clauses. We prove that maximum exclusive-or 2-satisfiability is in NLOGSPACE. Moreover, we demonstrate this problem is NP-complete. To attack the P versus NP question the concept of NP-completeness has been very useful. If any single NP-complete problem can be solved in polynomial time, then every NP problem has a polynomial time algorithm. Since every language in the class NLOGSPACE is in P, then we show that our problem is in P and NP-complete and thus, P = NP.

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https://www.academia.edu/36915996/P_versus_NP

Given an instance of $\textit{XOR 2SAT}$ and a positive integer $2^{K}$, the problem exponential exclusive-or 2-satisfiability consists in deciding whether this Boolean formula has a truth assignment with at leat $K$ satisfiable clauses. We prove exponential exclusive-or 2-satisfiability is in $QP$ and $\textit{NP-complete}$. In this way, we show $QP \subseteq NP$.

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https://www.academia.edu/36993851/QP_versus_NP

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