Show $Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln...











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I encountered a question given $theta$ in a random sample of size $n$ and also



$$-ntimes Eleft[frac{partial^2}{partial theta^2}times ln f(X)right]$$



, where $f(x)$ is the p.d.f. at $x$, provided that the extremes of the region for which $f(x) neq0$ do not depend on $theta$. The derivation of this formula take the following steps:



(a) Differentiating the expressions on both sides of
$int f(x),dx=1$
with respect to $theta$, show that



$$int frac{partial}{partialtheta} ln f(x)times f(x),dx=0$$



by interchanging the order of integration and differentiation.



Now I am having trouble understanding what this question was asking me to do. I used



$$int f(x),dx=1$$



and then,



$$int frac{partial}{partialtheta} f(x), dx = 0$$



But did not know how to proceed to get the answer.



In case anyone want to know, the (b) is:
Differentiating again with respect to $theta$, show that.



$$Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln f(X)right]$$



Thank you very much for your reading and any assistance would be appreciated!! Thanks!










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  • See Fisher information.
    – StubbornAtom
    11 hours ago










  • Thank you very much!
    – Chen
    2 hours ago















up vote
0
down vote

favorite












I encountered a question given $theta$ in a random sample of size $n$ and also



$$-ntimes Eleft[frac{partial^2}{partial theta^2}times ln f(X)right]$$



, where $f(x)$ is the p.d.f. at $x$, provided that the extremes of the region for which $f(x) neq0$ do not depend on $theta$. The derivation of this formula take the following steps:



(a) Differentiating the expressions on both sides of
$int f(x),dx=1$
with respect to $theta$, show that



$$int frac{partial}{partialtheta} ln f(x)times f(x),dx=0$$



by interchanging the order of integration and differentiation.



Now I am having trouble understanding what this question was asking me to do. I used



$$int f(x),dx=1$$



and then,



$$int frac{partial}{partialtheta} f(x), dx = 0$$



But did not know how to proceed to get the answer.



In case anyone want to know, the (b) is:
Differentiating again with respect to $theta$, show that.



$$Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln f(X)right]$$



Thank you very much for your reading and any assistance would be appreciated!! Thanks!










share|cite|improve this question
























  • See Fisher information.
    – StubbornAtom
    11 hours ago










  • Thank you very much!
    – Chen
    2 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I encountered a question given $theta$ in a random sample of size $n$ and also



$$-ntimes Eleft[frac{partial^2}{partial theta^2}times ln f(X)right]$$



, where $f(x)$ is the p.d.f. at $x$, provided that the extremes of the region for which $f(x) neq0$ do not depend on $theta$. The derivation of this formula take the following steps:



(a) Differentiating the expressions on both sides of
$int f(x),dx=1$
with respect to $theta$, show that



$$int frac{partial}{partialtheta} ln f(x)times f(x),dx=0$$



by interchanging the order of integration and differentiation.



Now I am having trouble understanding what this question was asking me to do. I used



$$int f(x),dx=1$$



and then,



$$int frac{partial}{partialtheta} f(x), dx = 0$$



But did not know how to proceed to get the answer.



In case anyone want to know, the (b) is:
Differentiating again with respect to $theta$, show that.



$$Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln f(X)right]$$



Thank you very much for your reading and any assistance would be appreciated!! Thanks!










share|cite|improve this question















I encountered a question given $theta$ in a random sample of size $n$ and also



$$-ntimes Eleft[frac{partial^2}{partial theta^2}times ln f(X)right]$$



, where $f(x)$ is the p.d.f. at $x$, provided that the extremes of the region for which $f(x) neq0$ do not depend on $theta$. The derivation of this formula take the following steps:



(a) Differentiating the expressions on both sides of
$int f(x),dx=1$
with respect to $theta$, show that



$$int frac{partial}{partialtheta} ln f(x)times f(x),dx=0$$



by interchanging the order of integration and differentiation.



Now I am having trouble understanding what this question was asking me to do. I used



$$int f(x),dx=1$$



and then,



$$int frac{partial}{partialtheta} f(x), dx = 0$$



But did not know how to proceed to get the answer.



In case anyone want to know, the (b) is:
Differentiating again with respect to $theta$, show that.



$$Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln f(X)right]$$



Thank you very much for your reading and any assistance would be appreciated!! Thanks!







probability-distributions statistical-inference expected-value






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edited 11 hours ago









StubbornAtom

4,65411136




4,65411136










asked yesterday









Chen

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213












  • See Fisher information.
    – StubbornAtom
    11 hours ago










  • Thank you very much!
    – Chen
    2 hours ago


















  • See Fisher information.
    – StubbornAtom
    11 hours ago










  • Thank you very much!
    – Chen
    2 hours ago
















See Fisher information.
– StubbornAtom
11 hours ago




See Fisher information.
– StubbornAtom
11 hours ago












Thank you very much!
– Chen
2 hours ago




Thank you very much!
– Chen
2 hours ago















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