I have been researching about the Cramer-Rao bound and I have found two inequalities:

$$\text{Var}\left(\hat{\theta}\right)\geq\frac{1}{\text{E}\left[\left[\frac{\partial}{\partial\theta}\ln f(X;\theta)\right]^2\right]}$$

$$\text{Var}\left(\hat{\theta}\right)\geq\frac{1}{n\text{E}\left[\left[\frac{\partial}{\partial\theta}\ln f(X;\theta)\right]^2\right]}$$

It is easy to see that the second is more general than the first since:

$$\frac{1}{\text{E}\left[\left[\frac{\partial}{\partial\theta}\ln f(X;\theta)\right]^2\right]}\geq\frac{1}{n\text{E}\left[\left[\frac{\partial}{\partial\theta}\ln f(X;\theta)\right]^2\right]}$$

I would like to know why there are two bounds. In other words I want to know when I can't use the first one and have to use the second bound.

  • $\begingroup$ Well, first notice that when $n$ = 1, the two expressions for the CR bound are equivalent. Also, the expressions are written using log probabilities base $e$ (via the pdf) instead of the natural log likelihood. The natural log of the product becomes the sum of the individual natural logs for the variables/parameters making up the pdf. So, in the end, the answer to your question depends on whether you are working with a probability or a likelihood. $\endgroup$ Mar 14, 2021 at 5:40

1 Answer 1


They are really equivalent. The leftmost expression assumes a single realization $x$ distributed according to $X$. The rightmost expression is appropriate to "convert" from the probability scale to the likelihood scale.

That is,

$L(\theta | X) = L(\theta | x_1,...x_n) = \prod_{i=1}^n{f(x_i | \theta)}$.

Taking natural logs (ln) on both sides of the above expression gives

$l(\theta | X)$ = ln$\left(\prod_{i=1}^n{f(x_i | \theta)}\right)$ = $\sum_{i=1}^n{lnf(x_i | \theta)}$

Note: As an aside, for the Cramer Rao lower bound, we typically like to work with the quantity

$-\mathbb{E}\left[\frac{\partial^2}{\partial \theta^2}lnf(x_i | \theta)\right]$

instead of

$\mathbb{E}\left[(\frac{\partial}{\partial \theta}lnf(x_i | \theta))^2\right]$

  • $\begingroup$ Thanks a lot. I finally understand my mistake. $\endgroup$
    – KKLK
    Mar 15, 2021 at 3:42

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