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0 pointstsimionescu5mo ago0 comments

> "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.

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froobius5mo ago

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.

tsimionescuOP5mo ago

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.

froobius5mo ago

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.

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froobius5mo ago

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

tsimionescuOP5mo ago

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.

froobius5mo ago

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.

1 more reply

j / k navigate · click thread line to collapse