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This is just a hypothetical question that I had (I made this question myself, just am curious about something). What if we were trying to find the $20\text{th}$ percentile in a discrete probability distribution from $0 \leq x \leq 5$, where

$F(0) = 0.1$

$F(1) = 0.3$

$F(2) = 0.5$

$F(3) = 0.6$

$F(4) = 0.8$

$F(5) = 1$

$F(x)$ is the cdf of course.

Would the $20\text{th}$ percentile be $0$ or $1$ or something else entirely?

This is where the $20\text{th}$ percentile is the value above or equal to $20%$ of the data

Can anyone give me the algorithm to find percentile in a discrete probability distribution if there is one?

Thank you for any help/comments.

  • What is your definition of the $20$th percentile? – Henry Oct 08 '20 at 01:18
  • 20th percentile is to me, the value that is above or equal to 20% of the data –  Oct 08 '20 at 01:19
  • $10%$ of the data is $0$ or less so $90%$ is $1$ or more; $30%$ of the data is $1$ or less so $70%$ is $2$ or more, ... – Henry Oct 08 '20 at 01:21
  • so you're saying that it's okay to go over 20% percent of the data in this specific case, as long as we're not BELOW the percentile (20%). So the correct answer in this case would be 1, not 2, 3, 4, or 5, because 1 is the minimum value that exceeds the percentile set (20%) correct? If we use 0, we're below it so that's not right. So it's okay to go over in this case with picking the 20th percentile value to be 1 because there's no way to be above or equal to exactly 20% of the data right? –  Oct 08 '20 at 01:23
  • I think you are looking for the value $k$ where $P(X < k) \le 0.2 \le P(X \le k)$. For $k=0$ you have $P(X < 0) =0$ and $P(X \le 0)=0.1$; for $k=1$ you have $P(X < 1) =0.1$ and $P(X \le 1)=0.3$; for $k=2$ you have $P(X < 2) =0.3$ and $P(X \le 2)=0.5$; one of those works and the other two do not – Henry Oct 08 '20 at 01:30
  • Why doesn't $1$ work? shouldn't $0.1 \leq 0.2 \leq 0.3$ be correct? (I edited my comment from earlier to fix the mistake of using 2, I should have written 1 to be the 20th percentile value, sorry!) –  Oct 08 '20 at 01:34
  • $1$ does work, as you say – Henry Oct 08 '20 at 01:36
  • Ah perfect. Thanks a lot. But I have a question about using this algorithm to check my answer $P(X<k) \leq p \leq P(X \leq k) $. Does this algorithm always work? Also is my intuitive thinking of "since we can't be at exactly p, just go the next closest x value that has a F(x) above p" wrong? (Where p is the percentile) –  Oct 08 '20 at 01:39
  • It works in general though there may be more than one solution. If in your example you looked for the median ($50%$) then $k$ could be $2$ or $3$ or any number in between them such as $2.5$ – Henry Oct 08 '20 at 01:47
  • If there's more then one solution. How would I know which number to pick then as my percentile? The smallest one? –  Oct 08 '20 at 02:04
  • All the solutions would then be correct, so it is an arbitrary choice - some people take the middle of the range of values, but that too is arbitrary – Henry Oct 08 '20 at 06:54
  • Got it, thanks so much for your help. –  Oct 08 '20 at 18:10

1 Answers1

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From comments: I think you are looking for the value $k$ where $P(X<k)≤0.2≤P(X≤k)$

  • for $k=0$ you have $P(X<0)=0$ and $P(X≤0)=0.1$
  • for $k=1$ you have $P(X<1)=0.1$ and $P(X≤1)=0.3$
  • for $k=2$ you have $P(X<2)=0.3$ and $P(X≤2)=0.5$
  • $\cdots$

so the $20\%$ percentile is at the value $1$ for this distribution.

This approach works in general, but though there may be more than one solution. If in your example you had been looking for the median $(50\%)$ then $k$ could be $2$ or $3$ or any number in between; all these solutions would then be correct, so it is an arbitrary choice - some people conventionally take the middle of the range of correct values so here $2.5$ for the median, but that too is arbitrary.

Henry
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