# Combining causal and identity relations

Suppose I have the causal relation `C causes E`, and the symmetric identity relation `E is (E1 ^ E2 ^ E3)`, which have the following probability functions: `E = bC` and `E = E1 * E2 * E3` where E, C, E1, E2 and E3 are variables and b is a constant. Combining these equations yields `bC = E1 * E2 * E3`.

Assuming that b, E2 and E3 are 1, could one conclude that C causes E1? If not, why? What are the limits for how causal and identity relations can be combined?

• I made some minor edits which you may roll back or continue editing. You can see the versions by clicking on the "edited" link above. Welcome! Commented Sep 2, 2018 at 14:26
• This is fine here, but math.stackexchange.com might be better since it appears to be a purely mathematical question. stats.stackexchange.com is probably overkill
– user935
Commented Sep 2, 2018 at 21:12
• Mathematically I think the answer is a clear yes, i.e. given that everything is 1, then C = E. But I wonder if there additional considerations if the assumption is that E is the same thing as E1 & E2 & E3 in Leibnitz's sense.
– user34796
Commented Sep 4, 2018 at 19:03
• It would make more sense to say that C causes E with probability `1/b`. Actually, you assume a premise that C causes E and not just they both have `1` correlation. On this premise we can say that C causes E1 (like of you jump on the Earth you fall). Commented Sep 5, 2018 at 9:02

## TL;DR

What you can say is that, if E2 and E3 are certain, and C causes E, then C causes E1.

To dissect what is going on, let’s review a little of probability theory, then address the subtleties of your question.

## Probablity theory

You want to know if E is true or not. You know that if C has happened, then there is s chance that E has happened as well. We normally write this as P(E | C) = b for some number b between 0 and 1. You read this equation as “the probability of E given C is b”. So now if you want to know the probability of E happening you just need the probability of \$C\$:

P(E) = P(E | C) P(C) + P(E | not C) ( 1 - P(C))

Note this is Bayes’s theorem and it’s a statement of correlation not of causation. Distinguishing between correlation and causation using probabilities is quite a subtle matter. I will use the two terms loosely, generally implying correlation.

Now there is another way of getting at the probability of \$E\$. You know that \$E\$ is actually the combination of 2 events (you can use three but like this is clearer):

E = E_1 \wedge E_2

We have to distinguish two cases: the cases where the 2 events are dependent and independent. An example of the first case is “I am a woman named Alex” which is the conjunction of the two statements “I am a woman” and “my name is Alex”. Not that these propositions are not independent: if I know your name is Alex, I might assume you are a man, since most people named Alex are men. The probability of your name being a man is not the same as the probability of you being a man, given your name is Alex. We have the following.
An example of independent events is getting heads at my first coin flip and heads at my second one. The probability of getting heads at the second flip is 50% regardless of the ouctome of the first one.
In both cases we have

P(E) = P( E_2 | E_1 ) P(E_1) = P(E_1 | E_2) P(E_2)

But only in the case of independent outcomes we can write

P(E) = P(E_1) P(E_2)

## A couple of issues with your question

You seem to be confusing statements with their probabilities. Statements are sentences while probabilities are numbers. So writing E = bC is meaningless, because even if you can define what multiplying a statement by a number means, it probably won’t meant what you want it to mean, which is instead the perfectly sensible equation between numbers P(E) = b P(C).

Similarly, when you wrote E = E_1 * E_2 you meant P(E) = P(E_1)P(E_2). The second issue here is that this equation only holds if E_1 and E_2 are independent, or if one of the two equals 1. Actually, let's quickly prove that this last clause is true. From earlier we have

P(E_2 | E_1) P(E_1) = P(E_1 | E_2) P(E_2)

But if E_2 is certain, we have P(E_2) = 1. We also have P(E_2 | E_1) = 1 since E_2 happens no matter what! The equation above then reduces to:

P(E_1) = P(E_1|E_2)

and thus:

P(E) = P(E_1|E_2)P(E_2) = P(E_1)P(E_2)