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+ | **Simple Example Two: Continuous R.V.: Estimate mean of Gaussian with Known <math>\Sigma</math> | ||
+ | <math>\rho(\vec{x}|\vec{\mu})=N(\vec{\mu},\Sigma)</math>, where <math>\mu</math> is | ||
+ | unknown, and <math>Sigma</math> is known. | ||
+ | <math>\rho(D|\vec{\mu}) = \prod\limits_{j=1}^{N}\rho(x_j|\vec{\mu})</math> | ||
+ | Observe the MLE <math>\in \hat{\theta}</math>, also maximize | ||
+ | <math>log\rho_D(D|\theta)</math> since log is monotonic | ||
+ | = <math>\sum\limits_{j=1}^{N}ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}})</math> | ||
+ | <math>\exp^{-\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2}}</math> | ||
+ | = <math>\sum\limits_{j=1}^{N}ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}})</math> | ||
+ | <math>-\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2}</math> | ||
− | + | which is <math>\infty</math> many times differentiable for <math>\vec{\mu}</math>, | |
− | + | so local max are where <math>\nabla=0</math> | |
− | + | ||
− | [ | + | compute <math>\nabla</math>, <math>\nabla_{\vec{\mu}}ln\rho_{D}(D|\vec{\mu})</math> |
− | + | ||
+ | =<math>\sum\limits_{j=1}^{N}\nabla_{\vec{\mu}} (ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}})</math> | ||
+ | <math>-\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2})</math> | ||
+ | |||
+ | =<math>-1/2\sum\limits_{j=1}^{N}\nabla_{\vec{\mu}}[(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})]</math> | ||
+ | |||
+ | =<math>-1/2\sum\limits_{j=1}^{N} | ||
+ | \begin{bmatrix} | ||
+ | \frac{\partial}{\partial\mu_1} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | \frac{\partial}{\partial\mu_2} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | \vdots \\ | ||
+ | \frac{\partial}{\partial\mu_n} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | But <math> \frac{\partial}{\partial\mu_1} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})</math> | ||
+ | |||
+ | =<math>(\frac{\partial}{\partial \mu_i}(x_j-\mu)^T)\Sigma^{-1}</math> | ||
+ | <math>(x_j-\mu)+(x_j-\mu)^T\Sigma^{-1}\frac{\partial}{\partial \mu_i}(x_j-\mu)</math> | ||
+ | |||
+ | =<math>2\frac{\partial}{\partial \mu_i}(x_j-\mu)^T)\Sigma^{-1}(x_j-\mu)</math> | ||
+ | |||
+ | =<math>2(0,0,0,...,-1,0,...,0)\Sigma^{-1}(x_j-\mu)</math> | ||
+ | |||
+ | =<math>-2\vec{e_i}^{T}\Sigma^{-1}(x_j-\mu)</math> | ||
+ | |||
+ | so, <math>\nabla{ln\rho_D(D|\mu)} = -1/2\sum\limits_{j=1}^{N}</math> | ||
+ | <math>\begin{bmatrix} | ||
+ | -2\vec{e_1}^{T}\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | -2\vec{e_2}^{T}\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | \vdots \\ | ||
+ | -2\vec{e_n}^{T}\Sigma^{-1}(x_j-{\mu})\\ | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | =<math>\sum\limits_{j=1}^{N}</math> | ||
+ | <math>\begin{bmatrix} | ||
+ | -2\vec{e_1}^{T}\\ | ||
+ | -2\vec{e_2}^{T}\\ | ||
+ | \vdots \\ | ||
+ | -2\vec{e_n}^{T}\\ | ||
+ | \end{bmatrix}</math> | ||
+ | <math>\Sigma^{-1}(x_j-\mu)</math>, the vector of <math>\vec{e_i}</math> is the space | ||
+ | domain of feature | ||
+ | |||
+ | =<math>\sum\limits_{j=1}^{N}\Sigma^{-1}(x_j-\mu)</math> | ||
+ | |||
+ | =<math>\Sigma^{-1}\sum\limits_{j=1}^{N}(x_j-\mu)</math> set to be 0 | ||
+ | |||
+ | <math>\Rightarrow \Sigma\Sigma^{-1}\sum\limits_{j=1}^{N}\Sigma^{-1}(x_j-\mu) = \Sigma \cdot 0</math> | ||
+ | |||
+ | <math>\Rightarrow \sum\limits_{j=1}^{N}(x_j-\mu) = 0</math> | ||
+ | |||
+ | <math>\Rightarrow \frac{1}{N}\sum\limits_{j=1}^{N}x_j = \mu</math> | ||
+ | |||
+ | <math>\rightarrow</math> the sample mean is the maximum likelihood | ||
+ | estimate for <math>\mu</math> | ||
Revision as of 21:27, 5 May 2014
Expected Value of MLE estimate over standard deviation and expected deviation
A slecture by ECE student Zhenpeng Zhao
Partly based on the ECE662 Spring 2014 lecture material of Prof. Mireille Boutin.
1. Motivation
- Most likely converge as number of number of training sample increase.
- Simpler than alternate methods such as Bayesian technique.
2. MLE as a Parametric Density Estimation
- Statistical Density Theory Context
- Given c classes + some knowledge about features $ x \in \mathbb{R}^n $ (or some other space)
- Given training data, $ x_j\sim\rho(x)=\sum\limits_{i=1}^n\rho(x|w_i) Prob(w_i) $, unknown class $ w_{ij} $ for $ x_j $ is know, $ \forall{j}=1,...,N $ (N hopefully large enough)
- In order to make decision, we need to estimate $ \rho(x|w_i) $, $ Prob(w_i) $ $ \rightarrow $ use Bayes rule, or $ \rho(x|w_i) $ $ \rightarrow $ use Neyman-Pearson Criterion
- To estimate the above two, use training data.
- The parametric pdf|Prob estimation problem
- Let $ D={x_1,x_2,...,x_N} $, $ x_j $ is drown independently from some probability law.
- Choose parametric from $ \rho(x|\theta) $ for the pdf of x or $ Prob(x|\theta) $ for the probability of x $ \rightarrow $ an unknown parametric vector
- Use $ D $ to estimate $ \theta $
- Definition: The maximum likelihood estimate of $ \theta $ is the value $ \hat{\theta} $ that maximize $ \rho_D(D|\theta) $, if x is continuous R.V., or $ Prob(D|\theta) $, if x is discrete R.V.
- Observation: By independence, $ \rho(D|\theta)=\rho(x_1,x_2,...,x_N|\theta) $ = $ \prod\limits_{j=1}^n\rho(x_j|\theta) $
- Simple Example One:
Those to estimate the priors: $ Prob(w_1), Prob(w_2) $ for $ c=2 $ classes.
Let $ Prob(w_1)=P $, $ \Rightarrow $ $ Prob(w_2)=1-P $, as an unknown parameter ($ \theta=P $)
Let $ w_j $ be the class of some $ x_j $, ($ j\in{1,2,...N} $)
$ Prob(D|P) $ = $ \prod\limits_{j=1}^n Prob(w_{ij}|P) $, $ x\sim \rho(x) $
=$ \prod\limits_{j=1}^{N_1} Prob(w_{ij}|P)\prod\limits_{j=1}^{N_2}Prob(w_{ij}|p) $
=$ P^{N_1}\dot(1-P)^{N-N_1} $
, the first $ w_{ij}=w_1 $ and the second $ w_{ij}=w_2 $,
$ N1 $= number of sample from class 1 Then, we $ \infty $ differentiate P $ (Prob(D|P)) $, so local max is where derivative = 0.
$ \frac{d}{dP} Prob(D|P)=\frac{d}{dP} P^{N_1}(1-P)^{N-N_1} $
$ =N_1P^{N_1-1}(1-P)^{N-N_1}-(N-N_1)P^{N_1}(1-p) $
$ =p^{N_1-1}(1-P)^{N-N_1-1}[N_1(1-P)-(N-N_1)P]=0 $
$ \Rightarrow $ So either P=0 or P=1 $ \rightarrow N_1(1-P) $
$ \Leftrightarrow P=\frac{N_1}{N} $
- Simple Example Two: Continuous R.V.: Estimate mean of Gaussian with Known $ \Sigma $
$ \rho(\vec{x}|\vec{\mu})=N(\vec{\mu},\Sigma) $, where $ \mu $ is
unknown, and $ Sigma $ is known.
$ \rho(D|\vec{\mu}) = \prod\limits_{j=1}^{N}\rho(x_j|\vec{\mu}) $
Observe the MLE $ \in \hat{\theta} $, also maximize $ log\rho_D(D|\theta) $ since log is monotonic
= $ \sum\limits_{j=1}^{N}ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}) $ $ \exp^{-\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2}} $
= $ \sum\limits_{j=1}^{N}ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}) $ $ -\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2} $
which is $ \infty $ many times differentiable for $ \vec{\mu} $, so local max are where $ \nabla=0 $
compute $ \nabla $, $ \nabla_{\vec{\mu}}ln\rho_{D}(D|\vec{\mu}) $
=$ \sum\limits_{j=1}^{N}\nabla_{\vec{\mu}} (ln(\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}) $ $ -\frac{(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})}{2}) $
=$ -1/2\sum\limits_{j=1}^{N}\nabla_{\vec{\mu}}[(x_j-\vec{\mu})^T\Sigma^{-1}(x_j-\vec{\mu})] $
=$ -1/2\sum\limits_{j=1}^{N} \begin{bmatrix} \frac{\partial}{\partial\mu_1} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ \frac{\partial}{\partial\mu_2} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ \vdots \\ \frac{\partial}{\partial\mu_n} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu})\\ \end{bmatrix} $
But $ \frac{\partial}{\partial\mu_1} (x_j-{\mu})^T\Sigma^{-1}(x_j-{\mu}) $
=$ (\frac{\partial}{\partial \mu_i}(x_j-\mu)^T)\Sigma^{-1} $ $ (x_j-\mu)+(x_j-\mu)^T\Sigma^{-1}\frac{\partial}{\partial \mu_i}(x_j-\mu) $
=$ 2\frac{\partial}{\partial \mu_i}(x_j-\mu)^T)\Sigma^{-1}(x_j-\mu) $
=$ 2(0,0,0,...,-1,0,...,0)\Sigma^{-1}(x_j-\mu) $
=$ -2\vec{e_i}^{T}\Sigma^{-1}(x_j-\mu) $
so, $ \nabla{ln\rho_D(D|\mu)} = -1/2\sum\limits_{j=1}^{N} $ $ \begin{bmatrix} -2\vec{e_1}^{T}\Sigma^{-1}(x_j-{\mu})\\ -2\vec{e_2}^{T}\Sigma^{-1}(x_j-{\mu})\\ \vdots \\ -2\vec{e_n}^{T}\Sigma^{-1}(x_j-{\mu})\\ \end{bmatrix} $
=$ \sum\limits_{j=1}^{N} $ $ \begin{bmatrix} -2\vec{e_1}^{T}\\ -2\vec{e_2}^{T}\\ \vdots \\ -2\vec{e_n}^{T}\\ \end{bmatrix} $ $ \Sigma^{-1}(x_j-\mu) $, the vector of $ \vec{e_i} $ is the space domain of feature
=$ \sum\limits_{j=1}^{N}\Sigma^{-1}(x_j-\mu) $
=$ \Sigma^{-1}\sum\limits_{j=1}^{N}(x_j-\mu) $ set to be 0
$ \Rightarrow \Sigma\Sigma^{-1}\sum\limits_{j=1}^{N}\Sigma^{-1}(x_j-\mu) = \Sigma \cdot 0 $
$ \Rightarrow \sum\limits_{j=1}^{N}(x_j-\mu) = 0 $
$ \Rightarrow \frac{1}{N}\sum\limits_{j=1}^{N}x_j = \mu $
$ \rightarrow $ the sample mean is the maximum likelihood estimate for $ \mu $
(create a question page and put a link below)
Questions and comments
If you have any questions, comments, etc. please post them on https://kiwi.ecn.purdue.edu/rhea/index.php/ECE662Selecture_ZHenpengMLE_Ques.