Is consecutively-observed evidence stronger in inference?

Question

I'm new to Bayesian theories, and the question is about modification on 'sunrise problem'.

The question is that, if I observed sun rise in n consecutive days, will the probability of 'sun rises tomorrow' be higher than that derived from if I observed sun in n non-consecutive days (i.e. on some day I observe, and I do not on other days so that I don't know about if the sun did rise on those days). (Intuitively, I suppose the first is a stronger condition, however, I have no clue on that.)

Similarly, under same condition from above, will the probability of sun will rise tomorrow be different from sun will rise every day? It's confusing to me since if sun would rise tomorrow, the probability for the day after tomorrow will change, isn't it? (i.e. it somehow works like a iterative method?)

To further expose the contradiction, if the probability of 'sun will rise' is a finite number less than 1, then, obviously, the probability of 'sun will rise every day' will be 0.

Answer 1

Suppose you observed the sun rise on July 1st, 5th, 7th and 20th. Four non-consecutive days. I think that for your problem you are assuming that there is no data for the other days, you don't know if the sun did or did not rise.

Alternatively, you could have observed the sun rise on July 17th, 18th, 19th and 20th. Four consecutive days.

By itself the consecutive days show no stronger evidence of the sun rising tomorrow. The updating of prior probabilities will only be affected if your probability model includes some dependence on the previous day's sun. If you assume that the probability of the sun rising on one day is independent from the sun rising on previous days then the prior probabilities are updated the same way in both cases.

Answer 2

The answer is basically "it depends on a lot of things".

There are two situations. I'm going to assume you know for certain in advance that on each day, either the sun rises or it doesn't, and you have the option of checking the sun once per day (so that you know when you've missed a day):

• You've seen the sun rise on $n$ consecutive days, and never seen the sun fail to rise;
• You've seen the sun rise on $n$ days, and never the sun fail to rise, but some of the $n$ days are separated from each other. (The other days, you just didn't get out of bed.)

The probability of "sun rises tomorrow" may be different between the two cases; it depends on your priors. The first situation reinforces most priors that say "the sun will rise very often in long blocks". The second situation reinforces priors that say "the sun will rise very often in certain patterns". Of course, "the sun always rises" is reinforced in both cases, but remember that you can only reinforce a hypothesis by discounting other hypotheses.

For a concrete example, suppose your priors are 50% on "the sun always rises", and 50% on "the sun rises exactly on alternate days"; and suppose you only check every other day. Then there is no update of probabilities possible.

On the other hand, suppose your priors are 50% on "the sun always rises", and 50% on "the sun rises once every third day", and you check every other day. Then you will update your probabilities on day 3.

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The following paragraph makes me slightly queasy, because I'm not very sure about Bayesianism in infinite cases, but here goes.

The probability of "the sun will rise every day forever" is $0$ or $1$ by the Kolmogorov $0/1$ law; since it is possible for the sun not to rise, it has to be $0$.