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What is the probability that the highest order statistic of a random sample of size n from any continuous distribution is below the median ( population median ) of that distribution.

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closed as off-topic by Did, Daniel, Claude Leibovici, user91500, colormegone Jun 22 '15 at 5:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Daniel, Claude Leibovici, user91500, colormegone
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  • $\begingroup$ What is the distribution? Uniform? Normal? Exponential? What? $\endgroup$ – Graham Kemp Nov 26 '14 at 0:15
  • $\begingroup$ From any continuous distribution. (Question edited) $\endgroup$ – moksha Nov 26 '14 at 0:18
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    $\begingroup$ no, we'd need to know what kind of distribution. The median and the probability of being below it, varies. $\endgroup$ – Graham Kemp Nov 26 '14 at 0:29
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    $\begingroup$ This question is OK, nothing unclear about it. Should not be closed. $\endgroup$ – kjetil b halvorsen Jun 14 '15 at 12:05
  • $\begingroup$ @kjetilbhalvorsen The question is clear and totally lacking context. $\endgroup$ – Did Jun 15 '15 at 6:30
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If the maximum is less than the median, then all points in the sample have to be less than the median. A sampled point will be less than the median with probability $1 \over 2$, so if $Y$ is the maximum and $m$ is the population median, then $$P[Y \le m] = {\left({1} \over {2} \right) }^n$$

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  • $\begingroup$ ... assuming the distribution is absolutely continuous, yes. $\endgroup$ – kjetil b halvorsen Jun 14 '15 at 11:50
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The probability that the highest order statistic in a sample of $n$ iid continuos random variables is below the median $m$ is:

$$\begin{align} \mathsf P(\max\{X_i\}_n<m) & = \mathsf F_X(m)^n \\[1ex] & = {\big(\tfrac 1 2\big)}^n & \text{by definition of the median} \end{align}$$

Here, $F_X$ is the cumulative probability function for the distribution, and by definition of what a median is.

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  • $\begingroup$ thank you for your time and response. $\endgroup$ – moksha Nov 26 '14 at 0:29
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    $\begingroup$ Wrong! If the distribution is absolutely continuous, then $F_X(m)=1/2$, by definition of median. $\endgroup$ – kjetil b halvorsen Jun 14 '15 at 11:51
  • $\begingroup$ Why not delete? $\endgroup$ – Did Jun 15 '15 at 6:29

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