0
$\begingroup$

Let's say we have N distributions $\mathcal N(\mu_i, \sigma_i)$, each with unknown mean $\mu_i$ and unknown standard deviation $\sigma_i$, $i=0,...,N-1$.

For each $i$, $M$ independent random samples are drawn from from the distributions above $x_{ij} \sim N(\mu_i,\sigma_i)$, $i=0....N-1, j=0,...,M-1$ with the draws being jointly independent of each other.

Compute $N$ sample means $m_i = \frac1M\sum_j {x_{ij}}$ and $N$ sample variances $v_i = \frac{1}{M-1}\sum_j{(x_{ij} - m_i)^2}$.

Search for index $i_{max}$ such that $m_{i_{max}} = \max_i\, m_i$. That is, $i_{max}$ is the index of the random variable with the maximum sample mean.

What is the best unbiased estimate of true mean $\mu_{i_{max}}$? Is it simply $m_{i_{max}}$?

Improve this question
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – stats_model
    Commented Apr 8, 2019 at 4:12
  • $\begingroup$ 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). $\endgroup$
    – stats_model
    Commented Apr 8, 2019 at 4:19
  • $\begingroup$ Thanks so much for your response and edits, stats_model. This was my first question on stack exchange... This seems like a great community! $\endgroup$
    – aagold
    Commented Apr 8, 2019 at 12:09

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.