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I'm currently reading the textbook "Statistics for Mathematician" from Victor Panaretos. On page 65, the author presents the following equation for the Cramér-Rao Lower Bound (Note: I set the numerator to 1 by assuming an unbiased estimator)

$$\text{Var}(T) \geq \frac{1}{n\int_{\mathcal{X}}(\frac{\partial}{\partial\theta}\text{log }f(x;\theta))²\:f(x;\theta)dx}=\frac{1}{n\mathbb{E}[\frac{\partial}{\partial\theta}\text{log }f(x;\theta) ]^2}$$

However, I don't understand the equality in the denominator, namely why $$\int_{\mathcal{X}}(\frac{\partial}{\partial\theta}\text{log }f(x;\theta))²\:f(x;\theta)dx = \mathbb{E}[\frac{\partial}{\partial\theta}\text{log }f(x;\theta) ]^2$$ holds.

Shouldn't it rather be $\int_{\mathcal{X}}(\frac{\partial}{\partial\theta}\text{log }f(x;\theta))²\:f(x;\theta)dx = \mathbb{E}[(\frac{\partial}{\partial\theta}\text{log }f(x;\theta))^2]$?

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  • $\begingroup$ It appears that your integral for the expectation of the square of the score function in the second-last sentence is equivalent to the integral given in the last sentence. Equivalently, the denominator of the Cramer-Rao lower bound can be written in terms of the negative expectation of the second partial derivative of the score function. $\endgroup$
    – compbiostats
    Commented Dec 29, 2018 at 18:43
  • $\begingroup$ It's the same thing really. You are saying $E[X^2]$ and they are saying $E[X]^2$. $\endgroup$
    – StubbornAtom
    Commented Dec 29, 2018 at 19:49
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    $\begingroup$ I meant $E[X]^2$ is supposed to mean $E[X^2]$ and not $(E[X])^2$, as is quite clear from the context. Putting a parenthesis as you have done certainly removes any ambiguity. $\endgroup$
    – StubbornAtom
    Commented Dec 29, 2018 at 22:28
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    $\begingroup$ @StubbornAtom since that appears to have resolved the question you might post it as an answer. $\endgroup$
    – Glen_b
    Commented Dec 29, 2018 at 23:53
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    $\begingroup$ I would ask OP @DrosoNeuro to post their own answer and accept it. Regarding notation, it is more standard to write the square of expectation as $(E[X])^2$ (see here for a related question). $\endgroup$
    – StubbornAtom
    Commented Dec 30, 2018 at 6:32

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It appears to be simply a notation issue (thanks to @StubbornAtom for pointing this out): The author of the textbook probably used $\mathbb{E}[X]²$ to indicate $\mathbb{E}[X²]$. But $\mathbb{E}[X]²$ could also be understood as $(\mathbb{E}[X])²$, hence my confusion.

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