Distribution of Maximum Likelihood EstimatorMaximum likelihood of function of the mean on a restricted...
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Distribution of Maximum Likelihood Estimator
Maximum likelihood of function of the mean on a restricted parameter spaceMaximum Likelihood Estimator (MLE)Maximum Likelihood Estimator of the exponential function parameter based on Order StatisticsFind maximum likelihood estimateMaximum Likelihood Estimation in case of some specific uniform distributionsMaximum likelihood estimate for a univariate gaussianFind the Maximum Likelihood Estimator given two pdfsMaximum Likelihood Estimate for a likelihood defined by partsVariance of distribution for maximum likelihood estimatorHow is $P(D;theta) = P(D|theta)$?
$begingroup$
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
New contributor
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
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asked 3 hours ago
Colin HicksColin Hicks
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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 $theta$.
New contributor
$endgroup$
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
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 $theta$.
New contributor
$endgroup$
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
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 $theta$.
New contributor
$endgroup$
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
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 $theta$.
New contributor
$endgroup$
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 $theta$.
New contributor
edited 2 hours ago
New contributor
answered 3 hours ago
dlnBdlnB
3917
3917
New contributor
New contributor
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
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
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
$frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that
$endgroup$
– Colin Hicks
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– dlnB
3 hours ago
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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