# How to find asymptotically normal estimator if I know probability density function [closed]

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?

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

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.

Hint

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 $$\theta$$.

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