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Distribution of Maximum Likelihood Estimator

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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 $P(D;theta) = P(D|theta)$?

1

$begingroup$

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

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asked 4 hours ago

Colin HicksColin Hicks

1353

New contributor

Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

1

$begingroup$

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

share|cite|improve this question

asked 4 hours ago

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

share|cite|improve this question

asked 4 hours ago

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.

$endgroup$

share|cite|improve this question

asked 4 hours ago

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.

share|cite|improve this question

asked 4 hours ago

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 4 hours ago

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 4 hours ago

Colin HicksColin Hicks

1353

1 Answer
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3

$begingroup$

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 $theta$.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

dlnBdlnB

3917

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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome :)
  $endgroup$
  – dlnB
  3 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

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 $theta$.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

dlnBdlnB

3917

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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome :)
  $endgroup$
  – dlnB
  3 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 $theta$.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

dlnBdlnB

3917

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
  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 $theta$.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

dlnBdlnB

3917

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.

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

dlnBdlnB

3917

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.

answered 4 hours ago

dlnBdlnB

3917

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.

answered 4 hours ago

dlnBdlnB

3917