[tsimionescu]

> "I'll look into it, but without knowing more about it, stastically we have a 50% chance of it being back up in an hour"

This is exactly what I don't think is right. This particular outage has the same a priori chance of being back in 20 minutes, in one hour, in 30 hours, in two weeks, etc.

[froobius]

Ah, that's not correct... That explains why you think it's "trite", (which it isn't).

The distribution is uniform before you get the measurement of time taken already. But once you get that measurement, it's no longer uniform. There's a decaying curve whose shape is defined by the time taken so far. Such that the statement above is correct, and the estimate `time_left=time_so_far` is useful.

[tsimionescu]

Can you suggest some mathematical reasoning that would apply?

If P(1 more minute | 1 minute so far) = x, then why would P(1 more minute | 2 minutes so far) < x?

Of course, P(it will last for 2 minutes total | 2 minutes elapsed) = 0, but that can only increase the probabilities of any subsequent duration, not decrease them.

[froobius]

That's inverted, it would be:

If: P(1 more minute | 1 minute so far) = x

Then: P(1 more minute | 2 minutes so far) > x

The curve is:

P(survival) = t_obs / (t_obs + t_more)

(t_obs is time observed to have survived, t_more how long to survive)

Case 1 (x): It has lasted 1 minute (t_obs=1). The probability of it lasting 1 more minute is: 1 / (1 + 1) = 1/2 = 50%

Case 2: It has lasted 2 minutes (t_obs=2). The probability of it lasting 1 more minute is: 2 / (2 + 1) = 2/3 ≈ 67%

I.e. the curve is a decaying curve, but the shape / height of it changes based on t_obs.

That gets to the whole point of this, which is that the length of time something has survived is useful / provides some information on how long it is likely to survive.

[tsimionescu]

> P(survival) = t_obs / (t_obs + t_more)

Where are you getting this formula from? Either way, it doesn't have the property we were originally discussing - the claim that the best estimate of the duration of an event is the double of it's current age. That is, by this formula, the probability of anything collapsing in the next millisecond is P(1 more millisecond | t_obs) = t_obs / t_obs + 1ms ~= 1 for any t_obs >> 1ms. So by this logic, the best estimate for how much longer an event will take is that it will end right away.

The formula I've found that appears to summarize the original "Copernican argument" for duration is more complex - for 50% confidence, it would say:

  P(t_more in [1/3 t_obs, 3t_obs]) = 50%

That is, if given that we have a 50% chance to be experiencing the middle part of an event, we should expect its future life to be between one third and three times its past life.

Of course, this can be turned on its head: we're also 50% likely to be experiencing the extreme ends of an event, so by the same logic we can also say that P(t_more = 0 [we're at the very end] or t_more = +inf [we're at the very beginning and it could last forever] ) is also 50%. So the chance t_more > t_obs is equal to the chance it's any other value. So we have precisely 0 information.

The bottom line is that you can't get more information out a uniform distribution. If we assume all future durations have the same probability, then they have the same probability, and we can't predict anything useful about them. We can play word games, like this 50% CI thing, but it's just that - word games, not actual insight.

[froobius]

I think the main thing to clarify is:

It's not a uniform distribution after the first measurement, t_obs. That enables us to update the distribution, and it becomes a decaying one.

I think you mistakenly believe the distribution is still uniform after that measurement.

The best guess, that it will last for as long as it already survived for, is actually the "median" of that distribution. The median isn't the highest point on the probability curve, but the point where half the area under the curve is before it, and half the area under the curve is after it.

And the above equation is consistent with that.