Mixing
Intuition for the vanishing expectation of score? $E[S(theta)] =0$?
$begingroup$
Suppose $f$ is a probability density on $y in mathbb{R}$ and $theta in mathbb{R}$ is some parameter.
Define $$S(theta) := log big( frac{partial f(y; theta)}{partial theta} big)$$
Then I'm trying to show that $E[S(theta)] = 0$. The derivation shown in my class is first noticing that, when $L(theta; y) = f(y; theta)$ and then seeing that:
begin{align}
0 &= frac{partial 1}{partial theta}dy \
&= frac{partial}{partial theta} int f(y; theta) dy \
&= int f(y; theta) frac{ frac{ partial L(theta; y) }{partial theta} }{ L(theta; y) } dy \
&=int f(y; theta) frac{ partial log(L(theta; y)) }{ partial theta} dy\
&= E[S(theta)]
end{align}
I think that my lack of intuition for this derivation may come from a weak background in analysis, that is, I'm not so intuitively clear on what conditions are required for interchanging differentiation and integration and so perhaps without this intuition I don't see why this result is obvious.
So, I was trying to gain more intuition by proceeding more directly, starting with $E[S(theta)]$ and then unpacking the result until I arrived at $$E[S(theta)] = int frac{ partial f(y; theta) }{partial theta } dy$$
and from here, it is not obvious to me that I should be able to interchange integration and differentiation and so I don't know how to proceed directly.
Thanks in advance.
statistics statistical-inference
asked Jan 25 at 18:42
libbylibby
1137
$endgroup$
add a comment |
$begingroup$
Suppose $f$ is a probability density on $y in mathbb{R}$ and $theta in mathbb{R}$ is some parameter.
Define $$S(theta) := log big( frac{partial f(y; theta)}{partial theta} big)$$
Then I'm trying to show that $E[S(theta)] = 0$. The derivation shown in my class is first noticing that, when $L(theta; y) = f(y; theta)$ and then seeing that:
begin{align}
0 &= frac{partial 1}{partial theta}dy \
&= frac{partial}{partial theta} int f(y; theta) dy \
&= int f(y; theta) frac{ frac{ partial L(theta; y) }{partial theta} }{ L(theta; y) } dy \
&=int f(y; theta) frac{ partial log(L(theta; y)) }{ partial theta} dy\
&= E[S(theta)]
end{align}
I think that my lack of intuition for this derivation may come from a weak background in analysis, that is, I'm not so intuitively clear on what conditions are required for interchanging differentiation and integration and so perhaps without this intuition I don't see why this result is obvious.
So, I was trying to gain more intuition by proceeding more directly, starting with $E[S(theta)]$ and then unpacking the result until I arrived at $$E[S(theta)] = int frac{ partial f(y; theta) }{partial theta } dy$$
and from here, it is not obvious to me that I should be able to interchange integration and differentiation and so I don't know how to proceed directly.
Thanks in advance.
statistics statistical-inference
asked Jan 25 at 18:42
libbylibby
1137
$endgroup$
$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05
add a comment |
$begingroup$
Suppose $f$ is a probability density on $y in mathbb{R}$ and $theta in mathbb{R}$ is some parameter.
Define $$S(theta) := log big( frac{partial f(y; theta)}{partial theta} big)$$
Then I'm trying to show that $E[S(theta)] = 0$. The derivation shown in my class is first noticing that, when $L(theta; y) = f(y; theta)$ and then seeing that:
begin{align}
0 &= frac{partial 1}{partial theta}dy \
&= frac{partial}{partial theta} int f(y; theta) dy \
&= int f(y; theta) frac{ frac{ partial L(theta; y) }{partial theta} }{ L(theta; y) } dy \
&=int f(y; theta) frac{ partial log(L(theta; y)) }{ partial theta} dy\
&= E[S(theta)]
end{align}
I think that my lack of intuition for this derivation may come from a weak background in analysis, that is, I'm not so intuitively clear on what conditions are required for interchanging differentiation and integration and so perhaps without this intuition I don't see why this result is obvious.
So, I was trying to gain more intuition by proceeding more directly, starting with $E[S(theta)]$ and then unpacking the result until I arrived at $$E[S(theta)] = int frac{ partial f(y; theta) }{partial theta } dy$$
and from here, it is not obvious to me that I should be able to interchange integration and differentiation and so I don't know how to proceed directly.
Thanks in advance.
statistics statistical-inference
asked Jan 25 at 18:42
libbylibby
1137
$endgroup$
Suppose $f$ is a probability density on $y in mathbb{R}$ and $theta in mathbb{R}$ is some parameter.
Define $$S(theta) := log big( frac{partial f(y; theta)}{partial theta} big)$$
Then I'm trying to show that $E[S(theta)] = 0$. The derivation shown in my class is first noticing that, when $L(theta; y) = f(y; theta)$ and then seeing that:
begin{align}
0 &= frac{partial 1}{partial theta}dy \
&= frac{partial}{partial theta} int f(y; theta) dy \
&= int f(y; theta) frac{ frac{ partial L(theta; y) }{partial theta} }{ L(theta; y) } dy \
&=int f(y; theta) frac{ partial log(L(theta; y)) }{ partial theta} dy\
&= E[S(theta)]
end{align}
I think that my lack of intuition for this derivation may come from a weak background in analysis, that is, I'm not so intuitively clear on what conditions are required for interchanging differentiation and integration and so perhaps without this intuition I don't see why this result is obvious.
So, I was trying to gain more intuition by proceeding more directly, starting with $E[S(theta)]$ and then unpacking the result until I arrived at $$E[S(theta)] = int frac{ partial f(y; theta) }{partial theta } dy$$
and from here, it is not obvious to me that I should be able to interchange integration and differentiation and so I don't know how to proceed directly.
Thanks in advance.
statistics statistical-inference
statistics statistical-inference
asked Jan 25 at 18:42
libbylibby
1137
asked Jan 25 at 18:42
libbylibby
1137
asked Jan 25 at 18:42
libbylibby
1137
asked Jan 25 at 18:42
libbylibby
1137
asked Jan 25 at 18:42
libbylibby
1137
1137
$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05
add a comment |
$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05
$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05
add a comment |
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$begingroup$
The fact that the integral and derivative can be interchanged is part of the assumptions ("regularity conditions").
$endgroup$
– StubbornAtom
Jan 25 at 18:48
$begingroup$
Sure, but why do we need this assumption?
$endgroup$
– libby
Jan 25 at 20:05