## The Messiah Problem

### Introduction

In this essay I am going to argue that the probability of a messianic religion being true is approximately zero.

The argument doesn't focus on the specific claims being made by a given religion but rather the type of the evidence we have for those religions versus the rarity of messianic figures.

The central problem is this: Messiahs (if they exist at all) have to be rare. The kind of evidence used to justify messiahs tends to be much more common than the messiahs themselves. This means the number of false positives would dominate the number of true positives, allowing us to doubt any such claim.

This is not a new idea. This was a problem examined by Hume in his work "On Miracles". The presentation is slightly different. Hume's argument appeals to probabilities but presents no raw mathematics to justify his point. This is one of those cases where doing the maths really adds power to the argument outlined by Hume . I present the maths inside this essay.

I have picked a slightly easier case then Hume. The argument presented is one surrounding messiahs rather than miracles. The number of supposed miracles is much larger than the number of messiahs. The lower rate of claimed messiahs means the quality of the evidence in favour of them needs to be much higher.

Functionally, the two claims have the same effect in trying to argue that messianic religions have a low probability of being true.

We can think of the evidence for a messiah as being a test that tries to establish whether a given person is in fact really a messiah or not. Messiahs are so rare that any test we devise to identify them needs to have an unrealistically low false-positive rate in order to work reliably. In the calculations that follow, we find even under very charitable assumptions the sort of false positive rate needs to be around < 0.001% in order to give us a > 50% chance of the person under examination actually being a real messiah.

### Bayes' law and conditional probability

A police officer pulls over a driver for speeding. What is the probability that the driver was in fact speeding given that the speed gun indicated he was speeding?

This is an example of conditional probability. The probability that he was speeding depends on the quality of the evidence provided by the speed gun.

Bayes' theorem gives us the mathematical machinery required to make sense of these sorts of problems. One form of Bayes' theorem is given below:

$$P(A \mid B) = \frac{P(A) \cdot P(B \mid A)}{ P(B \mid A) \cdot P(A) + P(B \mid \neg A) \cdot P(\neg A)}$$

This can look quite daunting but with a bit of explanation it is easy to work with.

When we write $$P(A)$$ we are asking for the probability that $$A$$ is true. When we write $$P(A \mid B)$$ we are asking for the probability that $$A$$ occurs given $$B$$ has occurred. Likewise, when we write $$P(B \mid A)$$ we are asking for the probability that $$B$$ occurred given that $$A$$ has occurred. When we write $$P(\neg A)$$ we are asking the probability that $$A$$ did not occur. Finally, when we write $$P(B \mid \neg A)$$ we are asking for the probability that $$B$$ occurred given that $$A$$ did not occur.

What are $$A$$ and $$B$$ though? For our example, we define:

• $$A$$ as being: "The probability that a given car is speeding (given no other evidence)"
• $$B$$ as being: "A speed gun test result over the speed limit"
• $$\neg A$$ as being: The probability that a given car is not speeding (given no other evidence)

In this example, we're going to put some made up numbers in to the machine and see what comes out.

Let's say that 10% of cars that go past the police officer are actually speeding. This is our $$P(A)$$ quantity. The probability that a car going past him is not speeding must be 90%. This is our $$P(\neg A)$$ quantity.

Then we need to ask: What is the probability of a failed speed trap test given a person is actually speeding? Let's say the test never gives a false-negative. If a person is speeding, it always detects it. This is our $$P(B \mid A)$$ quantity.

Next we ask the complementary question. What is the probability of a failed speed trap test given a person isn't speeding? This is the false positive rate. Let's say this is 5%. So of hundred non-speeding people, on average 5% will be incorrectly classified as speeding. This value is our $$P(B \mid \neg A)$$ quantity.

So next we plug in these quantities in to the equation above:

$$\begin{eqnarray} P(A \mid B) &=& \frac{0.10 \cdot 1}{ 0.10 \cdot 1 + (0.05 \cdot 0.9)} \\ &=& \frac{0.1}{ 0.1 + 0.045} \\ &=& \frac{0.1}{ 0.145 } \\ &=& 69\% \end{eqnarray}$$

This result is surprising to people who are new to Bayes' law. Despite a 100% true positive rate and a 5% false positive rate we are just 69% confident in the result!

The problem is the low rate of people are speeding in the first place. If just one-in-ten are speeding, then the one-in-twenty false positive rate is in the same ball park. That weakens the power of the test to the point where we expect 30% of the people to not be speeding at all.

The Messiah Problem now comes in to focus. Given the very low base rate of messiahs, just how good must the test be to have confidence a person is actually a messiah? To that question, we now turn.

### The Messiah problem

What is the probability that a person is a messiah given the evidence we have for them? This will be different for each prospective messiah and those claims would have to judged individually on their merits. A huge amount of debate would surround the parametrization of Bayes' law and what each of the values should be.

Let's simplify the problem. For the sake or argument let's say we have a messiah detector. A device that we can point at somebody and it shows a green light if they are in fact the messiah and a red light if not. Let's assume that our messiah detector always correctly identifies a person if they are in fact a messiah. However, when presented with a person who is not in fact a messiah it very occasionally makes a mistake and reports them as a messiah.

The first interesting problem to examine is the connection between the base rate of messiahs and the false-positive rate of the machine. We want to know what false-positive rate do we need to have to detect a given rarity of messiah with a probability greater than 50%.

We can find this relationship by setting $$P(A \mid B)$$ to be equal to 50% and then solving for the false-positive term $$P(B \mid A)$$. We assume that the true-positive rate is always 100%, as per the definition of the machine above.

$$\begin{eqnarray} 0.5 &=& \frac{P(A) \cdot 1}{ (P(A) \cdot 1) + (P(B \mid \neg A) \cdot (1 - P(A)))} \\[10pt] 0.5 &=& \frac{P(A)}{ P(A) + (P(B \mid \neg A) \cdot (1 - P(A)))} \\[10pt] P(A) &=& 0.5 \cdot (P(A) + (P(B \mid \neg A) \cdot (1 - P(A)))) \\[10pt] P(A) &=& (0.5 \cdot P(A)) + (0.5 \cdot P(B \mid \neg A) \cdot (0.5 - 0.5 \cdot P(A)))) \\[10pt] P(A) - (0.5 \cdot P(A)) &=& (0.5 \cdot P(B \mid \neg A) \cdot (0.5 - 0.5 \cdot P(A)))) \\[10pt] (0.5 \cdot P(A)) &=& (0.5 \cdot P(B \mid \neg A) \cdot (0.5 - 0.5 \cdot P(A)))) \\[10pt] 0.5 \cdot P(B \mid \neg A) &=& \frac{(0.5 \cdot P(A))} {(0.5 - 0.5 \cdot P(A))} \\[10pt] 0.5 \cdot P(B \mid \neg A) &=& 0.5 \cdot \frac{P(A)} {(1 - P(A))}\\[10pt] P(B \mid \neg A) &=& \frac{P(A)} {1 - P(A)} \end{eqnarray}$$

As we can see from this that when the base rate of messiahs is small, that the quantity in the denominator is approximately 1. Therefore, we can say in our analysis that the following relation holds:

$$P(\text{Positive test} \mid \text{Person is not a messiah}) \approx P(\text{Person is a messiah, given no other evidence})$$

So in summary, if messiahs are a one in a million event, you need a detector with a false-positive rate of about one in a million to detect that messiah with better than a 50% strike rate.

### How rare are messiahs?

Messiahs have to be rare. A one in a million chance of an individual being a messiah would mean every major city would have at least one messiah. A one in a billion chance of an individual being a messiah would mean approximately 7 should be alive in the world today. A one in a hundred billion chance of being a messiah would mean approximately 1 messiah in all of human history. This last case is the one normally advocated by messianic religion.

### How good is our evidence for messiahs?

Messianic religions usually start as oral traditions first. Then some time later, as the religion grows, people start to write down stories about the messiah and the code of practice this messiah wanted his followers to adopt.

If you go back far enough with any religion there comes a point where everything that anyone could possibly know about the religion was communicated orally. There was a time before books, letters, priests and temples. This was the time of ground zero in the religion; the core of everything that followed in that religion starts from this point.

Now the question is this: Is this oral evidence sufficiently powerful to overcome The Messiah Problem? If you were there to hear it, would it convince you? Is there more than a million to one shot, a billion to one shot or a 100 billion to one shot chance that at this crucial stage this evidence is faulty?

I would have to say yes. This is obviously true. Oral evidence has a very good chance of being faulty. It would never be powerful enough to overcome The Messiah Problem.

I think we can say all messianic religions, by the very nature of how they start, must very probably be false.

1. 2018-01-07 21:38:00 GMT
2. #Math #Philosophy