So let's say you have a distribution where X is the 16% quantile. Then you take the log of all the values of the distribution. Would log(X) still be the 16% quantile in the log distribution?
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asked Sep 23, 2010 at 3:57
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2 Answers
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Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then $$ \text{Pr}(Y\le q_{0.16}) = \text{Pr}(\log(Y)\le\log(q_{0.16})) = 0.16. $$ Generally, if $f$ is monotonic and increasing then $$ \text{Pr}(Y\le q_{\alpha}) = \text{Pr}(f(Y)\le f(q_{\alpha})) = \alpha. $$
answered Sep 23, 2010 at 4:30
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3$\begingroup$ Monotonic and increasing, of course. $\endgroup$– whuber ♦Commented Sep 23, 2010 at 13:32
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Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number greater than X is greater than log(X), so the ordering is not changed.
answered Sep 23, 2010 at 18:45