Show $Eleft[left(frac{partial}{partialtheta} ln f(X)right)^2right]=-Eleft[frac{partial^2}{partialtheta^2} ln...
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!
probability-distributions statistical-inference expected-value
add a comment |
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!
probability-distributions statistical-inference expected-value
See Fisher information.
– StubbornAtom
11 hours ago
Thank you very much!
– Chen
2 hours ago
add a comment |
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!
probability-distributions statistical-inference expected-value
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
probability-distributions statistical-inference expected-value
edited 11 hours ago
StubbornAtom
4,65411136
4,65411136
asked yesterday
Chen
213
213
See Fisher information.
– StubbornAtom
11 hours ago
Thank you very much!
– Chen
2 hours ago
add a comment |
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
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994742%2fshow-e-left-left-frac-partial-partial-theta-ln-fx-right2-right-e-l%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
See Fisher information.
– StubbornAtom
11 hours ago
Thank you very much!
– Chen
2 hours ago