Distribution of Maximum Likelihood Estimator
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Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)
Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$
Then after taking sample of size n
$$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$
And we want to find $theta_{max}$ such that $L(theta)$ is maximized and $theta_{max}$ is our estimate (once a sample has actually been selected)
Since $theta_{max}$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$
where
$$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$
Taking the derivative with respect to $theta$
$$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$
$theta_{max}$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?
probability distributions normal-distribution estimation sampling
asked yesterday
Colin HicksColin Hicks
1353
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add a comment |
$begingroup$
Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)
Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$
Then after taking sample of size n
$$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$
And we want to find $theta_{max}$ such that $L(theta)$ is maximized and $theta_{max}$ is our estimate (once a sample has actually been selected)
Since $theta_{max}$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$
where
$$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$
Taking the derivative with respect to $theta$
$$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$
$theta_{max}$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?
probability distributions normal-distribution estimation sampling
asked yesterday
Colin HicksColin Hicks
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)
Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$
Then after taking sample of size n
$$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$
And we want to find $theta_{max}$ such that $L(theta)$ is maximized and $theta_{max}$ is our estimate (once a sample has actually been selected)
Since $theta_{max}$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$
where
$$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$
Taking the derivative with respect to $theta$
$$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$
$theta_{max}$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?
probability distributions normal-distribution estimation sampling
asked yesterday
Colin HicksColin Hicks
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)
Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$
Then after taking sample of size n
$$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$
And we want to find $theta_{max}$ such that $L(theta)$ is maximized and $theta_{max}$ is our estimate (once a sample has actually been selected)
Since $theta_{max}$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$
where
$$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$
Taking the derivative with respect to $theta$
$$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$
$theta_{max}$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?
probability distributions normal-distribution estimation sampling
probability distributions normal-distribution estimation sampling
asked yesterday
Colin HicksColin Hicks
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Colin HicksColin Hicks
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Colin HicksColin Hicks
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Colin HicksColin Hicks
1353
asked yesterday
Colin HicksColin Hicks
1353
1353
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac{ f'(x;theta)}{f(x;theta)}.$ Then ${g(x_i;theta)}_{i=1}^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
$$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
$$bar{g}_n(theta)=0.$$
It follows that the ML estimator is given by
$$hat{theta}=bar{g}_n^{-1}(0).$$
So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.
answered yesterday
dlnBdlnB
5689
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dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
add a comment |
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$begingroup$
MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac{ f'(x;theta)}{f(x;theta)}.$ Then ${g(x_i;theta)}_{i=1}^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
$$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
$$bar{g}_n(theta)=0.$$
It follows that the ML estimator is given by
$$hat{theta}=bar{g}_n^{-1}(0).$$
So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.
answered yesterday
dlnBdlnB
5689
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
add a comment |
$begingroup$
MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac{ f'(x;theta)}{f(x;theta)}.$ Then ${g(x_i;theta)}_{i=1}^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
$$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
$$bar{g}_n(theta)=0.$$
It follows that the ML estimator is given by
$$hat{theta}=bar{g}_n^{-1}(0).$$
So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.
answered yesterday
dlnBdlnB
5689
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
add a comment |
$begingroup$
MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac{ f'(x;theta)}{f(x;theta)}.$ Then ${g(x_i;theta)}_{i=1}^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
$$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
$$bar{g}_n(theta)=0.$$
It follows that the ML estimator is given by
$$hat{theta}=bar{g}_n^{-1}(0).$$
So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.
answered yesterday
dlnBdlnB
5689
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac{ f'(x;theta)}{f(x;theta)}.$ Then ${g(x_i;theta)}_{i=1}^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
$$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
$$bar{g}_n(theta)=0.$$
It follows that the ML estimator is given by
$$hat{theta}=bar{g}_n^{-1}(0).$$
So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.
answered yesterday
dlnBdlnB
5689
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
answered yesterday
dlnBdlnB
5689
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answered yesterday
dlnBdlnB
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answered yesterday
dlnBdlnB
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5689
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$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
add a comment |
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
$begingroup$
You're welcome :)
$endgroup$
– dlnB
yesterday
add a comment |
Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.
Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.
Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.
Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.
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