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Let $X_{1},\ldots,X_{n}$ be independent RVs with common CDF $F_X$. Let $X_{(1)}, ... ,X_{(n)}$ be the associated order statistics. How can show that $$ \sum\limits_{i=1}^{n}Pr(X_{(i)}\leq t)=nF_X(t) $$

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Hint: The idea is that $X_{(n)}$ is less than or equal to t if and only if all $X_i$ are less than or equal to t and keep in mind that the samples are independent with the same probability $F_X(t)$. So for the nth order statistic that probability is multiplied n times. Work out the appropriate computation for each order statistic and sum to get the answer.

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  • $\begingroup$ If all the $X_i$ are no larger than $t$, then of course $X_{(1)}$ is no larger than $t$. But if $X_{(1)}$ is no larger than $t$, why on earth does it follow that all the $X_i$ are smaller than $t$? That is, I am questioning your use of "if and only if" in the first sentence of your answer. $\endgroup$ – Dilip Sarwate Apr 1 '17 at 15:09
  • $\begingroup$ @Dilip Sarwate I did not say the false statement you are attributing to me. All I said was to work out the appropriate computation for the other terms. It involves sum terms being less than or equal to t and others greater than or equal to t multiplied by the appropriate combinatorial term. The question is self-study. So the idea is to provide hints for the OP and not give an explicit complete solution. $\endgroup$ – Michael Chernick Apr 1 '17 at 16:06
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    $\begingroup$ Do you understand the meaning of "if and only if"? You did not explicitly make the statement "If all the $X_i$ are less than or equal to $t$ then $X_{(01)}$ is less than or equal to $t$ but you nonetheless implied it when you used "if and only if" in your first sentence. Writing $A \iff B$ is equivalent to making a_pair_ of statements (i) $A\implies B$ and (ii) $B\implies A$, and I am contending that one of (i) and (ii) is a false statement. -1 pending correction. $\endgroup$ – Dilip Sarwate Apr 1 '17 at 16:43
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    $\begingroup$ @MichaelChernick, $X_{(1)}$ is the minimum. The minimum being $\leq t$ does not mean all of the $X_i$ are. Your statement is true about the maximum, not the minimum. It looks like you're just thinking about the order statistics in reverse (i.e. is descending rather than ascending order). I don't know why Dilip didn't just say that instead of being opaque and implying you don't know what if and only if means. $\endgroup$ – gammer Apr 1 '17 at 18:15
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    $\begingroup$ I have removed my down vote and also corrected $X_{n}$ to $X_{(n)}$. @gammer I didn't even consider the possibility that Michael's notation numbered order statistics in descending order and that $X_{(1)}$ denoted the maximum rather than the minimum. The assertion that I incorrectly accused Michael of making a statement (that $X_{(1)}\leq t$ implies that all the $X_i$ are $\leq t$) that he never made when in fact he had said _" $X_{(1)}\leq t$ if and only if all the $X_i$ are $\leq t$"_was what led me to snidely ask whether Michael understood the meaning of if and only if. Apologies all around. $\endgroup$ – Dilip Sarwate Apr 2 '17 at 21:55
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I can solve this question. We know $$F_{X_{(i)}}(x)=\sum\limits_{j=i}^{n}\binom{n}{j}(F(x))^j(1-F(x))^{n-j} $$ So $$ \sum\limits_{i=1}^{n}Pr(X_{(i)}\leq x) \\ =\sum\limits_{i=1}^{n}\sum\limits_{j=i}^{n}\binom{n}{j}(F(x))^j(1-F(x))^{n-j}\\ =\sum\limits_{j=1}^{n}j\binom{n}{j}(F(x))^j(1-F(x))^{n-j}\\ =nF_X(x)\sum\limits_{j=1}^{n}\binom{n-1}{j-1}(F(x))^{j-1}(1-F(x))^{n-j} =nF_X(x) $$

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  • $\begingroup$ Haven't you assumed $F$ is continuous in applying the initial formula? $\endgroup$ – whuber Apr 1 '17 at 19:02
  • $\begingroup$ @whuber No, this answer is general. $\endgroup$ – amin roshani Apr 1 '17 at 22:38
  • $\begingroup$ Your calculations above are correct, but the premise , that the CDF of $X_{(i)}$ is given by the formula that you display, is applicable only the case of continuous $F$ when the distribution of $X$ has no atoms in it. So you have in effect assumed what @whuber said. $\endgroup$ – Dilip Sarwate Apr 3 '17 at 14:38

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