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by gowld
3034 days ago
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That's not correct. That's the opposite of the correct reason. Consider a simple (not-exactly-Poisson) set of blocks: 20x1minute intervals, plus 1x10minute interval. The average interval size is (20x1min + 1x10min)/21 = ~1.4min A random moment in time is twice as likely to be in a fast blocks as in the slow blocks. But average wait time (given a uniformly random start times) is (20min0.5min + 10min 10min)/30min = 110min^2/30min = ~3.6min. That per-start-instant average wait time is longer NOT because you are more likely to be in a slow block, but because being in a slow block has a much longer average wait-time than being in any of the fast blocks. |
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First of all, when the GP says "you are more likely to be in a slow block", it means "you are more likely to be in a slow block, relative to how many blocks there are".
In your example, if you pick a block at random, you have 1/21 chance of being in a slow block. If you pick a block by choosing a moment in time at random, you have 1/3 chance.
It is obviously not true that with any distribution you would be absolutely more likely to be a slower block.
Secondly, your last sentence seems to give a reason which doesn't fully make sense, and certainly isn't 'the opposite' of the given reason. Slow blocks of course have longer average wait-times than fast blocks. But this affects both the 'time-weighted' average and the 'block-weighted' average.
If you think that slow blocks have a longer wait-time than fast blocks, but that slow blocks are less likely than fast blocks (in the time-weighted average), shouldn't that make the time-weighted average LOWER?