## Saturday, June 27, 2015

### On evidence

What is evidence?

There are several definitions. One appealing definition is an event is evidence for another event if observing the first event increases the prior probability of the second, unobservable, event: $B$ is evidence for $A$ if $P(A|B)>P(A)$. However, while this is not a terrible definition of evidence, it has some deficiencies.

The first deficiency is that in common language, saying that $B$ is evidence for $A$ is to imply that $B$ is a good reason to believe $A$. However, if $P(A) = 1×10^{-6}$, and $P(A|B) = 2×10^{-6}$, then observing $B$ has doubled the probability of $A$, but we still don't have a good reason to believe $A$. If we're careful, we can avoid this equivocation, but we do have to be careful. However, translating from math to common language is always fraught with peril, so by itself, the possibility of equivocation is not dispositive. Still, a definition of evidence that is less easily equivocated would be better.

A more mathematical deficiency is that using evidence in the above manner absolutely requires that $A$ be constructed before observing $B$. The problem is that all individual events have very low probability; if we assume a continuous distribution, then all individual events have zero probability. If I observe $B$, and then construct $A$ in terms of $B$, it's always possible to construct $A$ such that $P(A|B)≫P(A)$. I have to at least be able to construct $A$ without knowing $B$ beforehand; ideally I want to construct $A$ before actually knowing $B$.

This problem is especially relevant when we are looking at evidence for intention or planning. Suppose you hand me a deck of cards, and I want to determine if you shuffled them or stacked the deck. I observe the order of the deck, $B$, and note that the probability of getting that specific order is $1/52!=1.24×10^{-68}$. I therefore construct $A$ to be that you intentionally arranged the cards in order $B$. Therefore $P(B|A)=1$ and $P(A|B)={P(B|A)P(A)}/{P(B)}={P(A)}/{P(B)}≫P(A)$: Whatever the prior probability of $A$ is, I've raised the posterior probability by 68 orders of magnitude. Because I am constructing $A$ knowing $B$, any observation $B$ at all "raises" the probability of $A$ by as many orders of magnitude as there are in the sample space of $B$.

A counter-objection is that in science, we construct theories based on our prior observations all the time. We observe, for example, that on the surface of the Earth, things fall when we drop them, so we construct a theory of gravity. We would never have thought about gravity had we not already observed things falling. The rebuttal is that we test (or at least try to test) scientific theories with new observations. We predict a new, as yet unobserved, event, and then observe it.

Although it's true that $P(A|B)≫P(A)$, this calculation might be completely meaningless. Because we don't know the the prior probability of $A$, just that $P(A|B)≫P(A)$ does not tell us whether $P(A)>0.5$ (i.e. the simplest mathematical expression that we have good reasons to believe $A$). But the case is actually worse: because we have narrowed the definition of $A$, we have actually reduced its prior probability, and we may have reduced it relative to the broader definition, the definition that does not include $B$, which I'll label as $A'$, by more than the probability of $B$ improves the probability of $A$. In other words, it may the be the case that even given $P(A|B)≫P(A)$ it could still be true that $P(A|B)<P(A')$.

For example, let's assume that there are a quadrillion ($10^{15}$) ways to order a deck of cards such that the ordering is "apparent": all cards in suit and rank order (A♠-2♠,A♥-2♥,...2♣), rank and suit order (A♠-A♣,K♠-K♣,...2♣), all ranks together in random order of suit, all suits together in random order of rank, all even and odd cards together, ordered such that if we played 5 card draw, you would get four of a kind or a straight flush and I would have a full house, etc. Let us assume that if you intentionally order the deck, you will order it in one of the quadrillion apparent ways. Furthermore, let us assume that the prior probability you will choose to order the deck vs. shuffling it randomly is 0.5: you flip a coin: heads, you order the deck; tails, you shuffle it.

$A': \text " you flipped heads, and ordered the deck in an 'apparently' intentional order"$
$A: \text " you flipped heads and intentionally ordered the deck in order " B$
$B: \text " the deck appears in a specific order, which might or might not be 'apparently' ordered"$
$C: \text " the deck appears in one of the 'apparently' intentional orders"$
$P(A')=0.5$
$P(B)=1.24×10^{-68}$
$P(C)=10^{15}/{52!}=1.24x10^{-53}$
$P(C|A')=1$
$P(B|A)=1$
Therefore,
$$P(A'|C)={P(C|A')P(A')}/{P(C|A')P(A')+P(C|~A')P(~A')}≈0.99...9$$

Thus, if I see the deck in an "apparent" order, I'm pretty sure you flipped heads and intentionally ordered the deck.

However, constructing $A$ in terms of $B$, we get:
$$P(A|B)={P(B|A)P(A)}/{P(B|A)P(A)+P(B|~A)P(~A)}={P(A)}/{P(A)+1.24×10^{-68}P(~A)}$$
To calculate this probability, we have to calculate some prior probability for $A$. However, we have narrowed the definition of $A$ with respect to $A'$: $A'$ says that you order the deck in an "apparent" ordering; $A$ says that you ordered the deck in order $B$, which can include apparent or non-apparent orderings, i.e. any possible ordering. Therefore, it's arguable that the prior probability of $A$ is $0.5 * 1.24×10^{-68}$ (you flip a coin; heads, you intentionally order the deck in a specific possible order that might or might not be "apparent"; tails, you shuffle the deck into a specific possible ordering that might or might not be "apparent". Therefore, $P(A|B)=0.5$ exactly: even though $P(A|B)=1≫P(A)=1.24×10^{-68}$, observing the ordering has given me zero information about whether you intentionally ordered the deck. This conclusion holds even if the deck is "apparently" ordered: the probability that had you flipped heads, you would have picked an apparent ordering is exactly equal to the probability that an apparent ordering would appear by chance.

Worse yet, if we include even a single case where if you flip heads, you might do something other than give me a deck in some order (e.g. there is a $1:1.24×10^{68}+1$ chance you might have just kept the deck if you flipped heads), then $P(A|B)<0.5$ (only slightly, but smaller nonetheless): just receiving a deck in some order means that the posterior probability that you flipped heads is (slightly) lower than the prior probability.

Therefore, simply increasing the posterior probability relative to the prior probability is not always meaningful. (It can, of course, sometimes be meaningful, but at least the principle of prior prediction, or something with the same effect, has to hold, so we can calculate the prior probability without reference to the actual observation itself.)

Thus, we can modify our definition: an event is evidence relative to a pair of mutually exclusive hypotheses if observing the event distinguishes between those hypotheses, i.e. Given $P(A|B)=P(B|A)=0$, ${P(A|C)}/{P(B|C)}>{P(A)}/{P(B)}$. It is still the case that observing $C$ might not be a good enough reason to believe $A$ or $B$, but if we calculate the priors correctly, then constructing $A$ or $B$ in terms of an already observed $C$ entails that ${P(A|C)}/{P(B|C)}≤{P(A)}/{P(B)}$.

For example, given that we have observed that life exists, we want to distinguish between the mutually exclusive hypotheses that life exists by chance, and life exists by design. If we include the prior probabilities correctly, then given
$$A: \text " the universe exists by design"$$
$$B: \text " the universe exists by chance"$$
$$C: \text " life exists"$$
Then
$${P(A|C)}/{P(B|C)}={P(A)}/{P(B)}$$
And the existence of life is just not evidence for these pairs of hypotheses. Note that even if we adjust only hypothesis $A$ to include " the universe exists by a designer who designed life" *$A'$), then even though it is true that $P(A'|C)≫P(A')$, because we have to lower the prior probability of $A'$, it is still the case that ${P(A'|C)}/{P(B|C)}≤{P(A')}/{P(B)}$.

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