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Suppose that we have $X_1,\ldots,X_n\overset{\textrm{iid}}{\sim}N(\mu,\sigma^2)$, where $\mu$ is unknown and $\sigma^2 = 1. $ Is the following estimator:

$$\hat{\mu} = \frac{1}{n-1}\sum{x_i}$$

consistent but not asymptotically efficient?

There are different definitions of what it means to be asymptotically efficient and here, it means that if we have the form:

$$\sqrt{n}(\hat{\mu}-\mu)\sim N(0,v(\mu)), $$

and $v(\mu) \geq \frac{1}{I_1(\mu)}$ where $I_1(\mu)$ is the Fisher information of one observation,

then $\hat{\mu}$ is asymptotically efficient if $$v(\hat \mu) = \frac{1}{I_1(\mu)}.$$

I tried to rearrange $\hat{\mu}$ but I cannot arrange it into the form above.

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  • $\begingroup$ "Rearranging" $\hat\mu$ does not seem relevant: you need to compute $v(\hat\mu)$ (not $v(\mu)$, which is zero) and $I_1(\mu)$ in order to test the inequality. How far do you get when you do this? $\endgroup$
    – whuber ♦
    Feb 27, 2014 at 1:12
  • $\begingroup$ So I ended up getting that: $µ ˜ N[ (n/(n-1))*µ, (n/(n-1))^2 (1 /n) ]$ However, I still cant get it into the form above. $\endgroup$
    – user123276
    Feb 27, 2014 at 1:14
  • $\begingroup$ Asymptotically you therefore obtain $\sqrt{n}\hat\mu\sim N(\mu,1).$ So what do you compute for $I_1(\mu)$? $\endgroup$
    – whuber ♦
    Feb 27, 2014 at 15:19

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