I'm working on finding the asymptotic variance of an MLE using Fisher's information.
The distribution is a Pareto distribution with density function $f(x|x_0, \theta) = \theta \cdot x^{\theta}_0 \cdot x^{-\theta - 1}$.
There are two steps I don't get, namely step 3 and 5.
(step 1) We have that
$1 = \int^{\infty}_{-\infty} f(x|x_0, \theta)$
(Step 2) We take derrivative wrt $\theta$:
$0 = \int^{\infty}_{-\infty} \frac{\partial f(x|x_0, \theta)}{\partial \theta} dx$
(Step 3) We can (according to my textbook) write the above as:
$0 = \int^{\infty}_{-\infty} \frac{\partial f(x|x_0, \theta)/\partial \theta }{f(x|x_0,\theta)} f(x|x_0,\theta) dx$
(Step 4) This can again be written as an expectation:
$E[\frac{\partial \log f(x|x_0,\theta)}{\partial \theta}] = 0$
(Step 5) According to my textbook we can differentiate again and get:
$0 = \int^{\infty}_{-\infty} \frac{\partial^2 \log f(x|x_0, \theta)}{\partial \theta^2} f(x|x_0, \theta) dx + \int^{\infty}_{- \infty} \frac{\partial \log f(x|x_0,\theta)}{\partial \theta} \frac{\partial \log f(x|x_0,\theta)}{\partial \theta} f(x|x_0, \theta) dx$
I can't figure if I have just stared at these steps for too long, or if there is something fundamental about differentiation and integration I have missed.
I would be very grateful if someone could explain the steps 3 and 5 to me in layman's manner?
(Textbook is Hogg & Craig "introduction to mathematical statistics").
Many of your calculations are redundant. Assuming $x_0$ as a known parameter and considering the MLE of $\theta$
First observe that,
$$\sqrt{nI_X(\theta)}\left[\hat{\theta}_{ML}-\theta \right]\xrightarrow{\mathcal{L}}N(0;1)$$
Thus its variance is
$$\frac{1}{nI_X(\theta)}$$
Where
$$nI_X(\theta)=-n\mathbb{E}\left[\frac{\partial^2}{\partial\theta^2}\log f(x|\theta) \right]$$
Now let's focus on the log density
$$\log f(x|\theta)=\log \theta+\theta\log x_0-\theta \log x-\log x$$
It is self evident that, without many efforts, the only addend that will be not zero after derivating 2 times is $\log \theta$
Thus
$$nI_X(\theta)=\frac{n}{\theta^2}$$
giving the asymptotic requested variance as
$$\frac{\theta^2}{n}$$