Given the distribution:

$f(x;\theta) = \frac{3}{\theta}x^2e^{-x^3/\theta}$ if $x>0$

the MLE for $\theta$ is $\frac{1}{n}\sum_{i=1}^n x_i^3$. It's an unbiased estimator with variance $\theta^2/n$. The Fisher information number is $1/\theta^2$.

Now, I am supposed to check whether or not this estimator shows asymptotic normality. The problem here is that my textbook has a condition which I can't find in any other source:

"$\frac{d^2}{d\theta^2} \ln f(x;\theta)$ is a continuous function in $\theta$, uniformly in $x$."

It's this uniformly continuity that I can't seem to find anywhere else. This seems to be the only condition not met for asymptotic normality. I've found that $\frac{d^2}{d\theta^2} ln f(x;\theta) = \frac{1}{\theta^2} - 2\frac{x^3}{\theta^3}$. Since uniform continuity is equivalent with the boundedness of the derivative, I derived this to $x$ to find $ - 6\frac{x^2}{\theta^3}$, which is not bounded for x.

My textbook also mentions that "these conditions are in particular fulfilled for (identifiable) members of the exponential class". So, my question is, is this really a necessary condition, and am I right that this MLE is not asymptotically efficient?


  • 3
    $\begingroup$ Hint: The estimator you've given is a mean of iid random variables. Ignore the regularity conditions. How can you connect means of iid random variables to asymptotic normality? $\endgroup$
    – cardinal
    Oct 17, 2012 at 18:31
  • $\begingroup$ I suppose you mean the use of the central limit theorem, which implies that $\sqrt{n}(\hat{\theta_n} - \theta) \rightarrow N(0,1/\theta^2)$, which is also the case here as far as I know. $\endgroup$
    – Ezueneok
    Oct 17, 2012 at 18:42
  • $\begingroup$ Yes, though the variance written in the comment doesn't look quite right. $\endgroup$
    – cardinal
    Oct 17, 2012 at 18:46
  • $\begingroup$ Ah, indeed, it should be $\theta^2$. So basically you're saying that, if this convergence happens, the MLE is asymptotically efficient regardless of the regularity conditions? $\endgroup$
    – Ezueneok
    Oct 17, 2012 at 19:04
  • 2
    $\begingroup$ Yes, just recall the definition of asymptotic efficiency in this context: All it says is that the (centered and rescaled) estimator has an asymptotic normal distribution with zero mean and variance of $1/I(\theta)$. So, if you prove that, you're done. The regularity conditions are sufficient to guarantee this (in conjunction with the assumptions on how the sample was obtained). (+1 to your question) $\endgroup$
    – cardinal
    Oct 17, 2012 at 20:05

1 Answer 1


So that the question does not remain open:
Set $X^3 = Z \Rightarrow X = Z^{1/3}$. Then

$$f_Z(z) = \left|\frac {\partial X}{\partial Z}\right|\cdot f_X(z^{1/3}) = \frac 13z^{-2/3}\cdot \frac{3}{\theta}z^{2/3}e^{-z/\theta} = \frac 1{\theta}e^{-z/\theta}$$

So we see that $Z=X^3$ is a random variable that follows the exponential distribution with $E(Z) = \theta$ and $\operatorname {Var}(Z) = \theta^2$.


$\hat \theta_{MLE} = \frac{1}{n}\sum_{i=1}^n z_i$ is the sample mean of $n$ (i.i.d. as clarified in the comments) exponential random variables, with $E[\hat \theta_{MLE}]=\theta$ and $\operatorname {Var}(\hat \theta_{MLE}) = \theta^2/n$ indeed.
The variance of each r.v. involved is bounded and so the Lindeberg-Levy CLT applies, meaning

$$\sqrt n(\hat \theta_{MLE}-\theta) \rightarrow_d N(0,\theta^2)$$

and the MLE is asymptotically normal.

The Fisher Information of $\theta$ is, indeed,

$$\mathcal{I}(\theta)=\operatorname{E} \left[\left. \left(\frac{\partial}{\partial\theta} \log f_Z(z;\theta)\right)^2\right|\theta \right] = \operatorname{E} \left[\left. (-\frac 1{\theta} + \frac {z}{\theta^2} \right)^2\right] = \frac 1{\theta^2}$$

So $$\operatorname {Avar}(\hat \theta_{MLE}) = \frac 1{\mathcal{I}(\theta)}$$

and the MLE is asymptotically efficient.


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