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Having a dataset with distribution $p$ (e.g. uniform or normal), if we divide the dataset into $n$ parts with equal size, is it valid to say that each part still has distribution $p$?

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I don't really think so. The question here is how you divide the dataset? Theoratically, if you randomly select a sample, of say 50, by Central Limit Theorem, the distribution of the sample follows a normal distribution, assuming you select randomly from the population.

Consider that I have 100 data points of height. These represent the height of 100 students. Let's also consider that they are left-skewed, so the tail is on the left, meaning that a small proportion of students are short.

Now, you want to divide them into 5 parts equally of 20 students. If you pick the last 20 (tallest) students, you can see that the distribution might NOT be left-skewed. It might be uniform (all of them have height 170cm). Clearly this part does not reflect the original distribution.

Hope this helps.

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  • $\begingroup$ well, there is no selection here. Assume the distribution has 100 data points (with a given distribution), and we want to divide them into 5 parts. As the data points are randomly distributed within the dataset, we pick the first 20 data points as the first part, the next 20 data points as second part, etc. Does each part has the original distribution in this case? $\endgroup$
    – HHH
    Feb 8, 2013 at 4:46
  • $\begingroup$ please see the updated ans $\endgroup$ Feb 8, 2013 at 5:34
  • $\begingroup$ Thanks. In your example you make an assumption that the data is sorted based the height, and then you pick the last 20! however if you do not make an assumption on the order of the data points and instead assume that they are randomly distributed, is your justification still valid?! $\endgroup$
    – HHH
    Feb 8, 2013 at 18:19
  • $\begingroup$ Then see first paragraph - you will be picking a sample out of a population. Regardless of the population distribution, the long term result will follow a normal distribution, by CLT. $\endgroup$ Feb 9, 2013 at 3:28

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