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We know that $E\left( I\left\lbrace X_{(i)} > x\right\rbrace\right) = P \left( X_{(i)}> x \right)$. Where, $X_{(1)}\leq X_{(2)}\leq,\ldots,X_{(n)}$ are order statistics of either independent $X_i$ or not independent $X_i$.

We also know that $P \left( X_{(1)}> x \right) = \prod_{i=1}^{n} P \left( X_i> x \right) =\left( 1 - F(x) \right)^n$ for independent $X_i$.

What is the value of $E\left( I\left\lbrace X_{(i)} > x\right\rbrace\right)$ in case of independent $X_i$?

Furthermore, is it possible to find out the value of $E\left( I\left\lbrace X_{(i)} > x\right\rbrace\right)$ in case of not independent $X_i$?

mdewey
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user00110014
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  • There are no simple formulas for either of your questions unless you assume the $X_i$ have the same distribution and (in the second case) you assume some particularly nice symmetrical dependence structure. – whuber Oct 12 '16 at 13:58
  • Yes, in the first case $X_i$ have the same distribution, i.e., $X_i$ is i.i.d. – user00110014 Oct 12 '16 at 14:05
  • However, in the second case $X_i = \min \lbrace Y_i, C_i\rbrace$, where $Y_1, Y_2, \ldots, Y_n$ be i.i.d. survival times with distribution function $F$, which are rightly censored by the i.i.d. censoring random variables, $C_1, C_2, \ldots, C_n$ with distribution function $G$ and assumed to be independent of $Y_i$. The observed data $X_i$ has a distribution function $H$ defined by $1-H =(1-F)(1-G)$. So, $X_i$ are not independent and we don't know any structure of that. – user00110014 Oct 12 '16 at 14:08
  • See http://stats.stackexchange.com/questions/225867. You can also [search our site](http://stats.stackexchange.com/search?q=%22order+statistics%22+distribution+-uniform+-exponential+-normal) for additional related information (which is where I found that thread). – whuber Oct 12 '16 at 14:12
  • For i.i.d. case is it $E\left( I\left\lbrace X_{(i)} > T\right\rbrace\right) = 1- \sum_{r=i}^{n} \frac{n!}{r!(n-r)!}\left( F(T) \right)^r \left( 1-F(T) \right)^{n-r} $? – user00110014 Oct 12 '16 at 15:20
  • Provided $F$ is continuous, the event $X\gt T$ is the same as the event $F(X)\gt F(T)$ and $F(X)$ has a uniform distribution, whence this event has probability $1-F(T)$. That reduces the question to a Binomial distribution, as you have indicated. But when $F$ is not continuous, the situation is more complicated (due to the positive chance of ties). – whuber Oct 12 '16 at 15:32

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