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For me it seems that it does. Here is my reasoning, please correct me if I'm wrong:

We have 3 definitions of a percentile, let's check if the 0th exists for each of them.

The exclusive definition: suppose we have set of numbers {10,20,30,40}. If we take 10, then we will discover that 0% of datapoints have lesser value then 10, thus making it the 0th percentile. Meaning that there is 0th percentile.

The inclusive definition: suppose we have set of numbers {10,20,30,40}. If we take 10, then we will discover that 25% of datapoints are lesser or equal to 10 (namely, it's equal to itself, so we have 1/4 of datapoints equal to 10). Hypothetically a number below 10 would be 0th percentile, but it would be outside our model because our smallest number is 10 (although if we collect statistic from a random variable it's possible that our statistical model isn't yet accurate enough to model said variable. So theoretically we could get, say, 8 and it would blow our model. But if we have decided to work with a model [as opposed to throwing it out of a window], then it means that we already assumed that we won't go outside limits of our model. Model-wise it's just meaningless to say that number can be less than 10). Meaning that there is no the 0th percentile.

The third definition of a percentile takes weighted mean of values of percentiles that we got in previous two definition. But the value of the 0th percentile for the inclusive definition doesn't exist, so we can't even apply the third definition here, meaning that there is no the 0th percentile for this definition.

Conclusion: We can have the 0th percentile! But if and only if we use the exclusive definition of a percentile.

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I assume that I'm right because there seem to be no reasons to think otherwise.

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  • $\begingroup$ Yes, you just proved yourself that if a number would be less than ten in the exclusive definition then it is 0th percentile. $\endgroup$ – BeastCoder2 Jun 2 at 14:05

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