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froobius | 4 months ago
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|4 months ago
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|4 months ago
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.