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I have the following data, where $+$ denotes right censoring.

$$ 2, 6, 7^+, 8^+, 10, 11, 14^+, 16 $$ I would like to find an unbiased estimator for $S(6)S(10)$, where $S(\cdot)$ mean survival function.


My try

KM estimate for $\hat{S}(6) = \frac{7}{8} \times \frac{6}{7} = \frac{3}{4} $, $\hat{S}(10) = \frac{7}{8} \times \frac{6}{7} \times \frac{3}{4} = \frac{9}{16} $, where $\hat{S}(\cdot)$ denotes Kaplan-Meier estimator.

But $\hat{S}(6)\hat{S}(10)$ would never work, without their independence, and I have heard that KM estimator is biased, so I'm stuck at how I should proceed.

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    $\begingroup$ We estimate the parameters, not the statistics, such as $\hat S(6)\hat S(10). $\endgroup$
    – user158565
    Oct 26, 2018 at 1:36
  • $\begingroup$ @a_statistician, thx for the comment, edited. $\endgroup$
    – moreblue
    Oct 26, 2018 at 1:40
  • $\begingroup$ What is the meaning of $S(6)S(10)$? One person survivals to 6 and 10, or two persons, one survivals to 6 another survivals to 10? $\endgroup$
    – user158565
    Oct 26, 2018 at 1:47
  • $\begingroup$ @a_statistician it's just $P(X > 6) P(X > 10)$, where $X$ is the lifetime. $\endgroup$
    – moreblue
    Oct 26, 2018 at 1:50
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    $\begingroup$ Need two conditions: 1)KM is unbiased. 2)person 1's survival is independent from person 2's survival. 2) should be true. I do not know where you get the idea that 1) is not true. $\endgroup$
    – user158565
    Oct 26, 2018 at 2:03

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