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Apr 8, 2019 at 12:09 comment added aagold Thanks so much for your response and edits, stats_model. This was my first question on stack exchange... This seems like a great community!
S Apr 8, 2019 at 11:21 history suggested stats_model CC BY-SA 4.0
Formatting with TeX to make question more readable; changed wording to be more mathematically correct (in particular, a distribution and a random variable are different)
Apr 8, 2019 at 4:19 comment added stats_model By continuous mapping theorem, however, it is easy enough to show that if your asymptotics fix $N$ and let $M \to \infty$, your proposed estimator will be consistent (you don't even need normality for this).
Apr 8, 2019 at 4:12 comment added stats_model Your proposed estimator certainly won't be unbiased in general, as demonstrated with the following counterexample. Consider $N = 2$ and suppose WLOG that $\mu_1 > \mu_2$. Then $E[m_{i_{max}}] = E[E[\max(m_1, m_2) | m_2]] >E[\max(E[m_1 | m_2], m_2)] = E[\max(\mu_1,m_2)] >\max(\mu_1, E[m_2]) = \mu_1$ where I used convexity of $\max$ and Jensen's inequality.
Apr 8, 2019 at 3:56 review Suggested edits
S Apr 8, 2019 at 11:21
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Apr 9, 2019 at 22:17
Apr 8, 2019 at 1:40 review First posts
Apr 8, 2019 at 2:38
Apr 8, 2019 at 1:35 history asked aagold CC BY-SA 4.0