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I recently learned that to find the pdf of the median of say $X_1,X_2, X_3$, you first find the Cdf via $ P(M \le x) =P(\text{at least 2 are}\, \le x) = P( \text{exactly 2 are}\, \le x) + P(\text{all 3 are} \le x)$ where $M$ denotes the median. Finaly you differentiate to get the required pdf.

My questions:

  1. How does one find the cdf/pdf of an arbitrary number of order statistics?
  2. Is there a generalized formula?

Thanks

Edit:

Let's suppose the $X_i$'s are iid.

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    @guy Thanks. ${}{}{}{}{}{}$2012-07-24

1 Answers 1

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Suppose (capital) $F$ is the cumulative probability distribution function of any of the three observations $X_1,X_2,X_3$, so that $F(x) = \Pr(X_1\le x) = \Pr(X_2\le x) = \Pr(X_3\le x)$. Then $ \Pr(\text{exactly two}\le x) $ is the probability of exactly two successes in three independent trials with probability $F(x)$ of success on each trial. In other words, the probability distribution of the number of observations that are $\le x$ is a binomial distribution with $n=3$ and $p=F(x)$. So we have $ \Pr(\text{exactly two}\le x) = {3\choose 2} F(x)^2(1-F(x))^1. $

More generally $ \Pr(\text{exactly $k$ of $n$ observations are}\le x) = {n\choose k} F(x)^k(1-F(x))^{n-k}. $

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    @John : Did you mean "How do I find the median?" or "How do I find the distribution of the median?"? In the latter case, the equality after the words "More generally" will do it if you pick n and $k$ correctly.2012-07-24