### Deterministic updating and the symmetry argument for Conditionalization

According to the Bayesian, when I learn a proposition to which I assign a positive credence, I should update my credences so that my new unconditional credence in a proposition is my old conditional credence in that proposition conditional on the proposition I learned. Thus, if $c$ is my credence function before I learn $E$, and $c'$ is my credence function afterwards, and $c(E) > 0$, then it ought to be the case that $$c'(-) = c(-|E) := \frac{c(-\ \&\ E)}{c(E)}$$ There are many arguments for this Bayesian norm of updating. Some pay attention to the pragmatic costs of updating any other way (Brown 1976; Lewis 1999); some pay attention to the epistemic costs, which are spelled out in terms of the accuracy of the credences that result from the updating plans (Greaves & Wallace 2006; Briggs & Pettigrew 2018); others show that updating as the Bayesian requires, and only updating in that way, preserves as much as possible about the prior credences while still respecting the new evidence (Diaconis & Zabell 1982; Dietrich, List, and Bradley 2016). And then there are the symmetry arguments that are our focus here (Hughes & van Fraassen 1985; van Fraassen 1987; Grove & Halpern 1998).

In a recent paper, I argued that the pragmatic and epistemic arguments for Bayesian updating are based on an unwarranted assumption, which I called

Let's start by laying out the symmetry argument. Suppose $W$ is a set of possible worlds, and $F$ is an algebra over $W$. Then an

Now, van Fraassen proves the following theorem, though he doesn't phrase it like this because he assumes Deterministic Updating in his definition of an updating rule:

The problem is that, while Certainty is entirely uncontroversial and Symmetry is very plausible, there is no particularly good reason to assume Deterministic Updating. But the argument cannot go through without it. To see this, consider the following updating rule:

It is easy to see that $V$ satisfies Certainty, since $v_w(E) = 1$ for each $w$ in $E$. To see that $V$ satisfies Symmetry, the crucial fact is that $f(v_w) = v_{f(w)}$. First, take a credence function in $V^{M'}(f(P), E')$: that is, $v_{w'}$ for some $w'$ in $E'$. Then $f^{-1}(w')$ is in $f^{-1}(E')$ and so $v_{f^{-1}(w')}$ is in $V^M(P, f^{-1}(E')))$. And $f(v_{f^{-1}(w')}) = v_{w'}$, so $v_{w'}$ is in $f(V^M(P, f^{-1}(E')))$. Next, take a credence function in $f(V^M(P, f^{-1}(E')))$. That is, $f(v_w)$ for some $w$ in $f^{-1}(E')$. Then $f(v_w) = v_{f(w)}$ and thus $f(v_w)$ is in $V^{M'}(f(P), E')$, as required.

So $V$ satisfies Certainty and Symmetry, but it is not the Bayesian updating rule.

Now, perhaps there is some further desirable condition that $V$ fails to meet? Perhaps. And it's difficult to prove a negative existential claim. But one thing we can do is to note that $V$ satisfies all the conditions on updating plans on sets of probabilities that Grove & Halpern explore as they try to extend van Fraassen's argument from the case of precise credences to the case of imprecise credences. All, that is, except Deterministic Updating, which they also impose. Here they are:

So, in sum, it seems that van Fraassen's symmetry argument for Bayesian updating shares the same flaw as the pragmatic and epistemic arguments, namely, they rely on Deterministic Updating, and yet that assumption is unwarranted.

In a recent paper, I argued that the pragmatic and epistemic arguments for Bayesian updating are based on an unwarranted assumption, which I called

*Deterministic Updating*. An*updating plan*says how you'll update in response to a specific piece of evidence. Such a plan is*deterministic*if there's a single credence function that it says you'll adopt in response to that evidence, rather than a range of different credence functions that you might adopt in response. Deterministic Updating says that your updating plan for a particular piece of evidence should be deterministic. That is, if $E$ is a proposition you might learn, your plan for responding to receiving $E$ as evidence should take the form:*If I learn $E$, I'll adopt $c'$*

*If I learn $E$, I might adopt $c'$, I might adopt $c^+$, and I might adopt $c^*$*.

Let's start by laying out the symmetry argument. Suppose $W$ is a set of possible worlds, and $F$ is an algebra over $W$. Then an

*updating plan*on $M = (W, F)$ is a function $U^M$ that takes a credence function $P$ defined on $F$ and a proposition $E$ in $F$ and returns the set of credence functions that the updating plan endorses as responses to learning $E$ for those with credence function $P$. Then we impose three conditions on a family of updating plans $U$.**Deterministic Updating**This says that an updating plan should endorse at most one credence function as a response to learning a given piece of evidence. That is, for any $M = (W, F)$ and $E$ in $F$, $U^M$ endorses at most one credence function as a response to learning $E$. That is, $|U^M(P, E)| \leq 1$ for all $P$ on $F$ and $E$ in $F$.**Certainty**This says that any credence function that an updating plan endorses as a response to learning $E$ must be certain of $E$. That is, for any $M = (W, F)$, $P$ on $F$ and $E$ in $F$, if $P'$ is in $U^M(P, E)$, then $P'(E) = 1$.**Symmetry**This condition requires a bit more work to spell out. Very roughly, it says that the way that an updating plan would have you update should not be sensitive to the way the possibilities are represented. More precisely: Let $M = (W, F)$ and $M' = (W', F')$. Suppose $f : W \rightarrow W'$ is a surjective function. That is, for each $w'$ in $W'$, there is $w$ in $W$ such that $f(w) = w'$. And suppose for each $X$ in $F'$, $f^{-1}(X) = \{w \in W | f(w) \in X\}$ is in $F$. Then the worlds in $W'$ are coarse-grained versions of the worlds in $W$, and the propositions in $F'$ are coarse-grained versions of those in $F$. Now, given a credence function $P$ on $F$, let $f(P)$ be the credence function over $F'$ such that $f(P)(X) = P(f^{-1}(X))$. Then the credence functions that result from updating $f(P)$ by $E'$ in $F'$ using $U^{M'}$ are the image under $f$ of the credence functions that result from updating $P$ on $f^{-1}(E')$ using $U^M$. That is, $U^{M'}(f(P), E') = f(U^M(P, f^{-1}(E')))$.Now, van Fraassen proves the following theorem, though he doesn't phrase it like this because he assumes Deterministic Updating in his definition of an updating rule:

**Theorem (van Fraassen)***If $U$ satisfies Deterministic Updating, Certainty, and Symmetry, then $U$ is the conditionalization updating plan. That is, if $M = (W, F)$, $P$ is defined on $F$ and $E$ is in $F$ with $P(E) > 0$, then $U^M(P, E)$ contains only one credence function $P'$ and $P'(-) = P(-|E)$.*The problem is that, while Certainty is entirely uncontroversial and Symmetry is very plausible, there is no particularly good reason to assume Deterministic Updating. But the argument cannot go through without it. To see this, consider the following updating rule:

- If $0 < P(E) < 1$, then $V^M(P, E) = \{v_w | w \in W\ \&\ w \in E\}$, where $v_w$ is the credence function on $F$ such that $v_w(X) = 1$ if $w$ is in $X$, and $v_w(X) = 0$ is $w$ is not in $X$ ($v_w$ is sometimes called the
*valuation function*for $w$, or the*omniscience credence function*at $w). - If $P(E) = 1$, then $V^M(P, E) = P$.

It is easy to see that $V$ satisfies Certainty, since $v_w(E) = 1$ for each $w$ in $E$. To see that $V$ satisfies Symmetry, the crucial fact is that $f(v_w) = v_{f(w)}$. First, take a credence function in $V^{M'}(f(P), E')$: that is, $v_{w'}$ for some $w'$ in $E'$. Then $f^{-1}(w')$ is in $f^{-1}(E')$ and so $v_{f^{-1}(w')}$ is in $V^M(P, f^{-1}(E')))$. And $f(v_{f^{-1}(w')}) = v_{w'}$, so $v_{w'}$ is in $f(V^M(P, f^{-1}(E')))$. Next, take a credence function in $f(V^M(P, f^{-1}(E')))$. That is, $f(v_w)$ for some $w$ in $f^{-1}(E')$. Then $f(v_w) = v_{f(w)}$ and thus $f(v_w)$ is in $V^{M'}(f(P), E')$, as required.

So $V$ satisfies Certainty and Symmetry, but it is not the Bayesian updating rule.

Now, perhaps there is some further desirable condition that $V$ fails to meet? Perhaps. And it's difficult to prove a negative existential claim. But one thing we can do is to note that $V$ satisfies all the conditions on updating plans on sets of probabilities that Grove & Halpern explore as they try to extend van Fraassen's argument from the case of precise credences to the case of imprecise credences. All, that is, except Deterministic Updating, which they also impose. Here they are:

**Order Invariance**This says that updating first on $E$ and then on $E \cap F$ should result in the same posteriors as updating first on $F$ and then on $E \cap F$. This holds because, either way, you end up with $$U^M(P, E \cap F) = \{v_w : w \in W\ \&\ w \in E \cap F\}$$.**Stationarity**This says that updating on $E$ should have no effect if you are already certain of $E$. That is, if $P(E) = 1$, then $U^M(P, E) = P$. The second clause of our definition of $V$ ensures this.**Non-Triviality**This says that there's some prior that is less than certain of the evidence such that updating it on the evidence leads to some posteriors that the updating plan endorses. That is, for some $M = (W, F)$, some $P$ on $F$, and some $E$ in $F$, $U^M(P, E) \neq \emptyset$. Indeed, $V$ will satisfy this for any $P$ and any $E \neq \emptyset$.So, in sum, it seems that van Fraassen's symmetry argument for Bayesian updating shares the same flaw as the pragmatic and epistemic arguments, namely, they rely on Deterministic Updating, and yet that assumption is unwarranted.

## References

- Briggs, R. A., & Pettigrew, R. (2018). An accuracy-dominance argument for conditionalization.
*Noûs*. https://doi.org/10.1111/nous.12258

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