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Let $X$ be a sample space. and $p$ be a pdf(pmf). Let supp(p)=$\{x:p(x)>0\}$.Letting $X$ be redefined as supp(p), is equivalent to assuming that $p(x;\xi)>0$ holds for all $\xi\in E$ and for all $x\in X$.This means that $S$ is a subset of $$P(X)=\{p:X\to\mathbb{R}:p(x)>0(\forall x \in X),\int p(x)dx=1\}$$

The Coefficient of the Fisher information matrix is defined $$g_{ij}(\xi)=E_{\xi}[\partial_{i}l_{\xi}\partial_{j}l_{\xi}].............(1)$$ where, $E_{\xi}$ denotes the expectation with respect to the distribution $p_{\xi}.$ It is also possible to write $g_{ij}(\xi)$ as $$g_{ij}(\xi)=-E_{\xi}[\partial_{i}\partial_{j}l_{\xi}]...............(2)$$ and we also have $E_{\xi}[\partial_{i}l_{\xi}]=0.............(3)$

Now lets us define $S=\{p_{\xi}/\xi\in E\}$ as a subset of $$\bar{P}(X)=\{p:X\to\mathbb{R}/p(x)>0 \forall x\in X,\int p(x)dx<\infty\}$$

Then in this case the Fisher metric is the same as defined in equation $(1)$, But equation $(2)$ and $(3)$ does not hold. I didn't get why $(2)$ and $(3)$ does not hold for this case.Someone please explain. Thanks

Jean-Claude Arbaut
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Andyale
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1 Answers1

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Expression 3 needs the following to be true: $$ 0 = \frac{\partial}{\partial \zeta_i} \int_X p_x(x,\zeta) dx = \int_X \frac{\partial}{\partial \zeta_i} p_x(x,\zeta) dx$$

Expression 2 needs the following to be true: $$ 0 = \frac{\partial}{\partial \zeta_i} \frac{\partial}{\partial \zeta_j} \int_X p_x(x,\zeta) dx = \int_X \frac{\partial}{\partial \zeta_i} \frac{\partial}{\partial \zeta_j} p_x(x,\zeta) dx$$

This swapping makes both these integrals evaluate to $0$ since $\int_X p_X(x,\zeta) dx = 1$ in the original set, making the derivative $0$ and make equation/expression 2,3 to be true.

This swapping of integral and derivative need not be true in general in the set of $p_X(x,\zeta)$ you are defining as new set. Even if the swapping of integral and derivative is true, we need not have $\int_X p_X(x,\zeta) dx = constant$ to have the derivatives of integrals become $0$. So the above derivatives of integrals are not $0$. So expression 2,3 are not true.

Check when can we interchange integration and differentiation for information on when u can interchange derivative and integrals.

  • What if assume this assumption is true for the new set. – Andyale Dec 05 '22 at 07:26
  • Actually, I am reading "Information geometry by Shun-Ichi-Amari" There he said that all the regularity conditions are true for this new set and $2,3$ are not true. – Andyale Dec 05 '22 at 07:30
  • Check each step in the derivation of equation 2,3. The above mentioned will be the only issue possible. If you find some other problem in the derivation, let me know. Cheers ! –  Dec 05 '22 at 07:42
  • @Andyale I got it. The issue is even if we can swap the integral and derivative sign, its not $0$ since $\int_X p_X(x) dx \neq constant$ like in ur old set. I updated my answer. See if u r convinced. –  Dec 05 '22 at 07:57
  • okay so you meant to say that $\int_{X}p(x)dx$ is vary for each $p(x)$ and in earlier case for all $p(x)$ the integral was 1. Am I right? – Andyale Dec 05 '22 at 08:20
  • Yes, now $\int_X p_X(x,\zeta) \neq constant, \forall \zeta$ and varies with $\zeta$. –  Dec 05 '22 at 08:43
  • @Andyale if u r convinced on my answer, mark it as correct answer and useful answer so that i know. –  Dec 06 '22 at 11:44
  • o yes yes sorry I forgot it.Thanks it was helpful. – Andyale Dec 06 '22 at 11:50