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I was reading Cramér–Rao bound to multiple parameters from Wikipedia page, but I could not follow this line in the article:

Let $\displaystyle {\boldsymbol {T}}(X)$ be an estimator of any vector function of parameters...

I could not understand what $\displaystyle {\boldsymbol {T}}(X)$ and $\displaystyle {\boldsymbol{\psi}}({\boldsymbol {\theta}})$ mean.

Can someone help?

Thanks

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$\displaystyle {\boldsymbol {T}}(X)$ is a vector of estimators and $\displaystyle {\boldsymbol{\psi}}({\boldsymbol {\theta}})$ is its expectation. That is $$ \mathbb{E}\left[\displaystyle {\boldsymbol {T}}(X)\right] = \displaystyle {\boldsymbol{\psi}}({\boldsymbol {\theta}}). $$

As an example, consider a random sample from a univariate Normal distribution. Then ${\boldsymbol {\theta}} = (\mu, \sigma^2)^T$, $\displaystyle {\boldsymbol {T}}(X) = \left(\sum_i X_i , \sum_i X_i^2\right)^T$ and $\displaystyle {\boldsymbol{\psi}}({\boldsymbol {\theta}}) = (n\mu, n\sigma^2 + n\mu^2)^T$.

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  • $\begingroup$ This may sound silly and I am sorry for that. I am interested in deciding a confidence interval for two parameters, then how do I get $\displaystyle {\boldsymbol {T}}(X)$ and $ {\boldsymbol{\psi}}({\boldsymbol {\theta}})$ $\endgroup$ – pkj Mar 14 '18 at 18:07
  • $\begingroup$ yes I understand that, CRLB provides me a confidence interval. My problem: I have optimized a loss function with respect to two parameters and I am interested in obtaining a confidence band for them. I am available with Hessian matrix. $\endgroup$ – pkj Mar 14 '18 at 18:14
  • $\begingroup$ @pkj oh my fault, yeah if you're using maximum likelihood, you can get the "observed information" by looking at the negative Hessian. Then to get the covariance matrix you take the inverse of that; that's the CRLB. $\endgroup$ – Taylor Mar 14 '18 at 18:37
  • $\begingroup$ on the last page of this document stat.umn.edu/geyer/old03/5102/notes/fish.pdf , you have $\mathbf{t}$ and I do not know what is that $\endgroup$ – pkj Mar 14 '18 at 18:49
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    $\begingroup$ Thank you @Taylor, you have been generous enough and spared time to answer my doubts. I thought the $t$s are same. I will give it another try, if I could not succeed, I will draft an another question. Thank you again. $\endgroup$ – pkj Mar 15 '18 at 2:45

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