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Okay, so I sent a confession of love to a girl detailing several reasons and coming up with a 90% probability that I will be with her together for my whole life.

But I then thought that I was supposed to be a frequentist and not a Bayesian, because I cannot start up 100 samples of me and her in a virtual machine and build prediction intervals for breakup time.

Is this true? I always thought that probability is utterly useless if it cannot allow you to, say, claim that a coin you never flipped has a probability 1/6. On the other hand it seems wrong to suppose, say, that my probability of being forever with my beloved would decrease due to a revelation that she is a spy for the NSA.

What exactly is the difference between frequentists and Bayesians on probabilities of unknown events? My frequentist stat teacher simply handwaves about it and say that we just call it a 90% "confidence" without daring to call it a probability and violating Þe Olde Frequentist School. That seems so unrigorous, and besides when calculating my 90% "confidence", I used formulas with, guess what, $Pr(...)$ in them.

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I think you might be mixing some concepts here. Say you have a bag containing 100 coins, 95 of which are balanced and 5 are unbalanced. The probability that you randomly pick a balanced coin is 95 %. The probability that a picked coin is balanced is either 0 or 1 - either it is or it isn't. You can, however, say that a coin you have picked is balanced with 95 % certainty.

For simplicity, let's say that you will, sadly, die in 2060 (before your girlfriend). Let $X$ denote the random variable that is your relationship. Thus, your statement to her was that:

$$ Pr(X>2060|X>2013)=0.9 $$

That is, your relationship has survived 2013, so the probability that it will survive until 2060 is 0.9. In order not to speak of probabilities in this example as a frequentist, you would need to have already broken up. Say you started dating in 2012 and break up in 2016. Then:

$$ Pr(X>2060|2016>X>2012)=0 $$

under the assumption that once you break up you will not start dating again. This is analogous to the coin example: when you condition on something having already happen, it is useless to speak of probabilities.

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  • $\begingroup$ So actually after I throw the coin it is meaningless to say "Oh that coin had a 0.5 probability of being the other side"? $\endgroup$ – ithisa Aug 6 '13 at 12:12
  • $\begingroup$ @Eric Dong: I think the distinction is more that the probability that the flip you just made landed heads is 1, but the parameter in the underlying Bernoulli distribution, $\theta$, is - and was - 0.5. "That coin had" means we do not condition on the actual outcome.. $\endgroup$ – hejseb Aug 6 '13 at 12:18
  • $\begingroup$ Actually, when I think about it, it feels like the whole question is based on some misunderstanding. If we assume your relationship is not predetermined to end at some point, it can be considered a random variable. You are just interested in the probability that the realization of this r.v. (i.e. a breakup) exceeds the death of one of you. I think the whole ambiguity arises because she's not thinking of it as a random variable. $\endgroup$ – hejseb Aug 6 '13 at 12:25
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This entirely depends on how you formulate the problem:

The frequentist may assume that there is an underlying random variable with true, yet unknown, parameters.
In your example the random variable could answer "Yes" or "No" to the question of whether or not you guys are still together. Looking at the answer at any point in time would be a realization of this random variable and since you can repeat this - you have a process in time which is in the frequentist domain. However, this inheretly assumes a couple of things, namely a certain "independence" of these realizations. Because you assume that while maybe the realizations are influenced by outside factors or even past realizations, there is one underlying mechanic to this issue. Every day you spend together is a new run of your experiment of living together.

Your girl, the Bayesian in your relationship, now says something different: There is not one variable with many realizations, but instead your whole life together is one single random experiment and the "realizations" are in truth just data from one single event. And this event, obviously, is unique and happens only once.
And then, only if one starts to assume properties of this process, such as ergodicity, it is possible to infer things about the process. And this assumption means that during the one run you can observe, the observable parameters behave during their run much like they would if the experiments could be repeated and one would measure the mean of those runs. This is obviously a pretty strong assumption and it usually entails the interpretation that there are indeed realizations of a random process consisting of a series of random variables in time... However your girl will have nothing of that, rejecting the idea that you can gain insight about the total likelihood of your guys future by merely adding the probability that you guys are together at certain days.

I don't think there is solution to this. Have a nice dinner maybe.

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