I have $X_1, X_2,\ldots,X_n$ be a random sample of size n from a distribution with probability density function: $$p(x) = \theta^2xe^{-\theta x}I (x > 0).$$ How can I find an asymptotically normal estimator?

  • $\begingroup$ Maximum likelihood will fill the bill. $\endgroup$
    – utobi
    Mar 15 at 19:16
  • $\begingroup$ @utobi can u give more details please $\endgroup$ Mar 15 at 19:20

1 Answer 1


You can use the maximum likelihood estimator, which, in regular cases as this one has limiting normal distribution. This is defined as

$$ \hat\theta = \text{arg max}_{\theta\in\Theta} L(\theta), $$

where $L(\theta) = \prod_{i=1}^n f(x_i;\theta)$, is the likelihood function and $f$ is your density function; I am assuming that the sample is independently and identically distributed.

From a computational perspective, it is much easier to maximise the log-likelihood function $\ell(\theta) = \log L(\theta)$ instead of maximising $L(\theta)$.

Replacing $f$ with your density will deliver the required estimator; I leave this detail to you. As for the limiting distribution, the theory of maximum likelihood tells us that

$$\sqrt{n}(\hat\theta-\theta) \overset{n\to\infty}{\to} N\left(0, I(\theta)^{-1}\right),$$

where $I(\theta)$ is the expected Fisher information for a single observation.


The likelihood function is $$L(\theta) = \prod_{i=1}^n \theta^2 X_i e^{-\theta X_i} = \theta^{2n}e^{-\theta \sum_i X_i}\prod_i X_i.$$

The log-likelihood function is

$$\ell(\theta) = 2n\log\theta - \theta\sum_{i} X_i + \log \prod_i X_i.$$

Next, solve $d \ell(\theta)/d\theta = 0$ in $\theta$ to get a stationary point. Lastly, check the second derivative of $\ell(\theta)$ to make sure the stationary point you found is a maximum. If it is, that's the maximum likelihood estimator for $\theta$.

  • $\begingroup$ thx u! i know general method but have problem with compute sum $L(\theta)$ $\endgroup$ Mar 15 at 19:41
  • $\begingroup$ @KarlosMargaritos I added some further details for you. $\endgroup$
    – utobi
    Mar 15 at 19:52
  • $\begingroup$ thx u! now understand! $\endgroup$ Mar 15 at 19:53
  • $\begingroup$ am I right that $dl(\theta)\ d\theta$ equal to $2n/\theta$ in my case $\endgroup$ Mar 15 at 21:27
  • 1
    $\begingroup$ and then we have $2n / sum x_i$ $\endgroup$ Mar 15 at 21:55

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