(21 intermediate revisions by 2 users not shown) | |||
Line 3: | Line 3: | ||
<font size="4">Expected Value of MLE estimate over standard deviation and expected deviation </font> | <font size="4">Expected Value of MLE estimate over standard deviation and expected deviation </font> | ||
− | A [ | + | A [http://www.projectrhea.org/learning/slectures.php slecture] by ECE student Zhenpeng Zhao |
− | Partly based on the [[ | + | Partly based on the [[2014_Spring_ECE_662_Boutin_Statistical_Pattern_recognition_slectures|ECE662 Spring 2014 lecture]] material of [[User:Mboutin|Prof. Mireille Boutin]]. |
</center> | </center> | ||
---- | ---- | ||
Line 17: | Line 17: | ||
− | === <br> 2. | + | === <br> 2. MLE as a Parametric Density Estimation === |
*Statistical Density Theory Context | *Statistical Density Theory Context | ||
**Given c classes + some knowledge about features <math>x \in \mathbb{R}^n</math> (or some other space) | **Given c classes + some knowledge about features <math>x \in \mathbb{R}^n</math> (or some other space) | ||
**Given training data, <math>x_j\sim\rho(x)=\sum\limits_{i=1}^n\rho(x|w_i) Prob(w_i)</math>, unknown class <math>w_{ij}</math> for <math>x_j</math> is know, <math>\forall{j}=1,...,N</math> (N hopefully large enough) | **Given training data, <math>x_j\sim\rho(x)=\sum\limits_{i=1}^n\rho(x|w_i) Prob(w_i)</math>, unknown class <math>w_{ij}</math> for <math>x_j</math> is know, <math>\forall{j}=1,...,N</math> (N hopefully large enough) | ||
− | **In order to make decision, we need to estimate | + | **In order to make decision, we need to estimate <math>\rho(x|w_i)</math>, <math>Prob(w_i)</math> <math>\rightarrow</math> use Bayes rule, or <math>\rho(x|w_i)</math> <math>\rightarrow</math> use Neyman-Pearson Criterion |
+ | **To estimate the above two, use training data. | ||
− | <math>\rho(x| | + | *The parametric pdf|Prob estimation problem |
+ | ** Let <math>D={x_1,x_2,...,x_N}</math>, <math>x_j</math> is drown independently from some probability law. | ||
+ | ** Choose parametric from <math>\rho(x|\theta)</math> for the pdf of x or <math>Prob(x|\theta)</math> for the probability of x <math>\rightarrow</math> an unknown parametric vector | ||
+ | **Use <math>D</math> to estimate <math>\theta</math> | ||
− | + | *Definition: The maximum likelihood estimate of <math>\theta</math> is the value <math>\hat{\theta}</math> that maximize <math>\rho_D(D|\theta)</math>, if x is continuous R.V., or <math>Prob(D|\theta)</math>, if x is discrete R.V. | |
+ | *Observation: By independence, <math>\rho(D|\theta)=\rho(x_1,x_2,...,x_N|\theta)</math> = <math>\prod\limits_{j=1}^n\rho(x_j|\theta)</math> | ||
+ | **Simple Example One: | ||
+ | Those to estimate the priors: <math>Prob(w_1), Prob(w_2)</math> for <math>c=2</math> classes. | ||
− | + | Let <math>Prob(w_1)=P</math>, <math>\Rightarrow</math> <math>Prob(w_2)=1-P</math>, as an unknown parameter (<math>\theta=P</math>) | |
− | + | ||
− | + | ||
− | + | ||
− | + | ||
+ | Let <math>w_j</math> be the class of some <math>x_j</math>, (<math>j\in{1,2,...N}</math>) | ||
+ | <math>Prob(D|P)</math> = <math>\prod\limits_{j=1}^n Prob(w_{ij}|P)</math>, <math>x\sim \rho(x)</math> | ||
+ | |||
+ | |||
+ | =<math>\prod\limits_{j=1}^{N_1} Prob(w_{ij}|P)\prod\limits_{j=1}^{N_2}Prob(w_{ij}|p)</math> | ||
+ | |||
+ | =<math>P^{N_1}\dot(1-P)^{N-N_1}</math> | ||
+ | |||
+ | , the first <math>w_{ij}=w_1</math> and the second <math>w_{ij}=w_2</math>, | ||
+ | |||
+ | <math>N1</math>= number of sample from class 1 | ||
+ | Then, we <math>\infty</math> differentiate P <math>(Prob(D|P))</math>, so local max is where derivative = 0. | ||
+ | |||
+ | <math>\frac{d}{dP} Prob(D|P)=\frac{d}{dP} P^{N_1}(1-P)^{N-N_1}</math> | ||
+ | |||
+ | <math>=N_1P^{N_1-1}(1-P)^{N-N_1}-(N-N_1)P^{N_1}(1-p)</math> | ||
+ | |||
+ | <math>=p^{N_1-1}(1-P)^{N-N_1-1}[N_1(1-P)-(N-N_1)P]=0</math> | ||
+ | |||
+ | <math>\Rightarrow</math> So either P=0 or P=1 <math>\rightarrow N_1(1-P) </math> | ||
+ | |||
+ | <math>\Leftrightarrow P=\frac{N_1}{N}</math> | ||
+ | |||
+ | ---- | ||
+ | |||
+ | <br> | ||
+ | **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> | ||
+ | |||
+ | |||
+ | |||
+ | ---- | ||
+ | |||
+ | <br> | ||
+ | **Example three: I.D. Gaussian, both <math>\mu</math> and <math>\sigma^2</math> unknown | ||
+ | |||
+ | <math>\theta = (\theta_1, \theta_2) = (\mu, \sigma^2)</math> | ||
+ | |||
+ | We have | ||
+ | <math>ln\rho(x_k|\mu,\sigma^2) = </math> | ||
+ | <math>ln(\frac{1}{\sqrt{2\pi}\sigma}\cdot e^{-\frac{(x-\mu)^2}{2\sigma^2}})</math> | ||
+ | |||
+ | =<math>-1/2ln(2\pi\sigma^2)-1/(2\sigma^2)(x_k-\mu)^2</math> | ||
+ | <math>ln\rho_D(D|\mu, \sigma^2)</math> | ||
+ | |||
+ | =<math>ln\prod\limits_{k=1}{N}\rho(x_k|\mu,\theta^2)</math> | ||
+ | |||
+ | =<math> \sum\limits_{k=1}^{N}(-\frac{1}{2}ln(2\pi\sigma^2)</math> | ||
+ | <math>-\frac{1}{2\sigma^2}(x_k-\mu)^2)</math> | ||
+ | <math>\nabla_{\mu,\sigma^2}ln_D(D|\mu,\sigma^2)</math> | ||
+ | |||
+ | =<math>\begin{bmatrix} | ||
+ | \frac{\partial}{\partial \mu}ln\rho_D(D|\mu,\sigma^2)\\ | ||
+ | \frac{\partial}{\partial \sigma^2}ln\rho_D(D|\mu,\sigma^2)\\ | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | =<math>\begin{bmatrix} | ||
+ | \frac{\partial}{\partial \mu}(-\frac{N}{2}ln(2\pi\sigma^2) | ||
+ | -\frac{1}{2\sigma^2}\sum\limits_{k=1}^{N}(x_k-\mu)^2)\\ | ||
+ | \frac{\partial}{\partial \sigma^2}(-\frac{N}{2}ln(2\pi\sigma^2) | ||
+ | -\frac{1}{2\sigma^2}\sum\limits_{k=1}^{N}(x_k-\mu)^2)\\ | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | =<math>\begin{bmatrix} | ||
+ | \frac{1}{\sigma^2}\sum\limits_{k=1}^{N} | ||
+ | (x_k-\mu)\\ | ||
+ | -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}- | ||
+ | \frac{-1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2 | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | =<math>\begin{bmatrix} | ||
+ | \frac{1}{\sigma^2}\sum\limits_{k=1}^{N} | ||
+ | (x_k-\mu)\\ | ||
+ | -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}+ | ||
+ | \frac{1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2 | ||
+ | \end{bmatrix}</math> set to be 0 | ||
+ | |||
+ | from <math>\frac{1}{\sigma^2}\sum\limits_{k=1}^{N} (x_k-\mu)=0</math> | ||
+ | <math>\Leftrightarrow \mu=</math> | ||
+ | <math>\sum\limits_{k=1}^{N}x_k-N\mu=0</math> | ||
+ | |||
+ | <math>\Leftrightarrow \mu=\frac{1}{N}\sum\limits_{k=1}^{N}x_k</math> which is sample mean. | ||
+ | |||
+ | From <math>-\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}-</math> | ||
+ | <math>\frac{-1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2=0</math> and | ||
+ | <math>\hat{\mu}=\frac{1}{N}\sum\limits_{k=1}^{N}x_k \Rightarrow</math> | ||
+ | |||
+ | <math>-\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}+</math> | ||
+ | <math>\frac{1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2=0 \Leftrightarrow</math> | ||
+ | |||
+ | <math>-\frac{N}{2}+\frac{1}{2\sigma^2}</math> | ||
+ | <math>\sum\limits_{k=1}^{N}(x_k-\mu)^2=0 \Leftrightarrow</math> | ||
+ | |||
+ | <math>\frac{1}{2\sigma^2}=</math> | ||
+ | <math>\frac{N}{2}\cdot \frac{1}{\sum\limits_{k=1}^{N}(x_k-\mu)^2} </math> | ||
+ | <math>\Leftrightarrow</math> | ||
+ | |||
+ | <math>\sigma^2 = \frac{1}{N}\cdot \sum\limits_{k=1}^{N}(x_k-\mu)^2</math> | ||
+ | =<math>\hat{\sigma^2}</math> which is the MLE of <math>\sigma</math> | ||
+ | |||
+ | In general, when <math>x\sim N(\vec{\mu}, \Sigma), </math> | ||
+ | <math>x\in \mathbb{R}^n, \vec{\mu}, \Sigma</math> unknown, | ||
+ | the MLE for <math>\vec{\mu}</math> and <math>\Sigma </math>are: | ||
+ | <math>\hat{\mu}=\frac{1}{N}\sum\limits_{k=1}^{N}x_k</math> | ||
+ | <math>, \hat{\Sigma} = \frac{1}{N}\sum\limits_{k=1}^{N}</math> | ||
+ | <math>(x_k-\mu)(x_k-\mu)^T</math> | ||
+ | |||
+ | <math>\Sigma</math> is non singular, but <math>\hat{\Sigma}</math> can be singular | ||
+ | <math>\Rightarrow</math> no inverse <math>\rightarrow</math> this happens when number | ||
+ | of points N<n: feature space down. | ||
+ | |||
+ | What happens when repeat sampling and estimating? | ||
+ | |||
+ | Sample: <math>(x_1^i, x_2^i,...,x_N^i) \Rightarrow</math> | ||
+ | <math>\hat{\mu}^i = \frac{1}{N}\sum\limits_{k=1}^{N}x_k^i</math> | ||
+ | |||
+ | |||
+ | **<math>E(\hat{u})=?</math> | ||
+ | |||
+ | We have <math>E(\hat{u})= E(\frac{1}{N}\sum\limits_{k=1}^{N}(x_k))</math> | ||
+ | <math>\frac{1}{N}E(x_k)=\frac{1}{N}\sum\limits_{k=1}^{N}E(x)=</math> | ||
+ | <math>\frac{1}{N}\sum\limits_{k=1}^{N}u = \mu</math> | ||
+ | |||
+ | But how far do we expect to derivate from the mean? | ||
+ | |||
+ | <math>E(|\hat{\mu}-\mu|^2) = E((\hat{\mu}-\mu)(\hat{\mu}-\mu))</math> | ||
+ | <math>=E(\hat{\mu}\cdot\hat{\mu}-\hat{\mu}\cdot{\mu}</math> | ||
+ | <math>-{\mu}\cdot\hat{\mu}+{\mu}\cdot{\mu})</math> | ||
+ | |||
+ | <math>=E(\hat{\mu}\cdot\hat{\mu})-2\cdot \mu E(\hat{u})+\mu \cdot \mu</math> | ||
+ | |||
+ | <math>=E(\hat{\mu}\cdot\hat{\mu})-\mu\cdot\mu</math> | ||
+ | |||
+ | <math>=E(\frac{1}{N}\sum\limits_{k=1}^{N}x_k \cdot \frac{1}{N}\sum\limits_{j=1}^{N}x_j)-\mu\cdot\mu</math> | ||
+ | |||
+ | <math>=\frac{1}{N^2}\sum\limits_{k=1}^{N}E(x_k \cdot x_j)-\mu\cdot\mu</math> | ||
+ | |||
+ | <math>=\frac{1}{N^2}[\sum\limits_{k,j=1,k\neq j}^{N}</math> | ||
+ | <math>E(x_k )\cdot E(x_j)+\sum\limits_{k,j=1,k\neq j}^{N}</math> | ||
+ | <math>E(x_k )\cdot E(x_k)]-\mu\cdot\mu</math> | ||
+ | |||
+ | <math>=\frac{1}{N^2}[N\cdot (N-1)\mu\cdot \mu+</math> | ||
+ | <math>\sum\limits_{k=1}^{N}E(x^2)]-\mu\cdot\mu</math> | ||
+ | |||
+ | <math>-\frac{1}{N}\mu\cdot\mu+\frac{1}{N^2}\sum\limits_{k=1}^{N}E(x^2)</math> | ||
+ | |||
+ | by <math>E[(x-\mu)(x-\mu)] = \sigma^2 \Rightarrow</math> | ||
+ | <math>E(x \cdot x)-\mu^2 = \sigma^2 \rightarrow </math> | ||
+ | <math>E(x \cdot x) = \sigma^2+\mu^2</math> | ||
+ | |||
+ | |||
+ | So: <math>E(|\hat{\mu}-\mu|^2) = -\frac{1}{N}\mu \cdot \mu +</math> | ||
+ | <math> \frac{1}{N}(\sigma^2+\mu \cdot \mu) = \frac{1}{N}\sigma^2</math> | ||
+ | |||
+ | ---- | ||
+ | |||
+ | <br> | ||
+ | *Bias: The maximum likelihood for the variance $\sigma^2$ is biased means | ||
+ | the expected value over all data sets of size n of the sample variance is not equal to the true variance: | ||
+ | |||
+ | <math>E[\frac{1}{n}\sum\limits_{k=1}^{N}(x_k-\bar{x})] = \frac{n-1}{n}</math> | ||
+ | <math>\sigma^2 \neq \sigma^2</math> | ||
+ | |||
+ | But we can tell that as n <math>\rightarrow \infty</math>, the MLE of <math>\sigma</math> is closing to <math>\sigma^2</math> | ||
---- | ---- | ||
Latest revision as of 10:51, 22 January 2015
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 $
- Example three: I.D. Gaussian, both $ \mu $ and $ \sigma^2 $ unknown
$ \theta = (\theta_1, \theta_2) = (\mu, \sigma^2) $
We have $ ln\rho(x_k|\mu,\sigma^2) = $ $ ln(\frac{1}{\sqrt{2\pi}\sigma}\cdot e^{-\frac{(x-\mu)^2}{2\sigma^2}}) $
=$ -1/2ln(2\pi\sigma^2)-1/(2\sigma^2)(x_k-\mu)^2 $ $ ln\rho_D(D|\mu, \sigma^2) $
=$ ln\prod\limits_{k=1}{N}\rho(x_k|\mu,\theta^2) $
=$ \sum\limits_{k=1}^{N}(-\frac{1}{2}ln(2\pi\sigma^2) $ $ -\frac{1}{2\sigma^2}(x_k-\mu)^2) $ $ \nabla_{\mu,\sigma^2}ln_D(D|\mu,\sigma^2) $
=$ \begin{bmatrix} \frac{\partial}{\partial \mu}ln\rho_D(D|\mu,\sigma^2)\\ \frac{\partial}{\partial \sigma^2}ln\rho_D(D|\mu,\sigma^2)\\ \end{bmatrix} $
=$ \begin{bmatrix} \frac{\partial}{\partial \mu}(-\frac{N}{2}ln(2\pi\sigma^2) -\frac{1}{2\sigma^2}\sum\limits_{k=1}^{N}(x_k-\mu)^2)\\ \frac{\partial}{\partial \sigma^2}(-\frac{N}{2}ln(2\pi\sigma^2) -\frac{1}{2\sigma^2}\sum\limits_{k=1}^{N}(x_k-\mu)^2)\\ \end{bmatrix} $
=$ \begin{bmatrix} \frac{1}{\sigma^2}\sum\limits_{k=1}^{N} (x_k-\mu)\\ -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}- \frac{-1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2 \end{bmatrix} $
=$ \begin{bmatrix} \frac{1}{\sigma^2}\sum\limits_{k=1}^{N} (x_k-\mu)\\ -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}+ \frac{1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2 \end{bmatrix} $ set to be 0
from $ \frac{1}{\sigma^2}\sum\limits_{k=1}^{N} (x_k-\mu)=0 $ $ \Leftrightarrow \mu= $ $ \sum\limits_{k=1}^{N}x_k-N\mu=0 $
$ \Leftrightarrow \mu=\frac{1}{N}\sum\limits_{k=1}^{N}x_k $ which is sample mean.
From $ -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}- $ $ \frac{-1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2=0 $ and $ \hat{\mu}=\frac{1}{N}\sum\limits_{k=1}^{N}x_k \Rightarrow $
$ -\frac{N}{2}\cdot\frac{2\pi}{2\pi\sigma^2}+ $ $ \frac{1}{2\sigma^4}\sum\limits_{k=1}^{N}(x_k-\mu)^2=0 \Leftrightarrow $
$ -\frac{N}{2}+\frac{1}{2\sigma^2} $ $ \sum\limits_{k=1}^{N}(x_k-\mu)^2=0 \Leftrightarrow $
$ \frac{1}{2\sigma^2}= $ $ \frac{N}{2}\cdot \frac{1}{\sum\limits_{k=1}^{N}(x_k-\mu)^2} $ $ \Leftrightarrow $
$ \sigma^2 = \frac{1}{N}\cdot \sum\limits_{k=1}^{N}(x_k-\mu)^2 $ =$ \hat{\sigma^2} $ which is the MLE of $ \sigma $
In general, when $ x\sim N(\vec{\mu}, \Sigma), $ $ x\in \mathbb{R}^n, \vec{\mu}, \Sigma $ unknown, the MLE for $ \vec{\mu} $ and $ \Sigma $are: $ \hat{\mu}=\frac{1}{N}\sum\limits_{k=1}^{N}x_k $ $ , \hat{\Sigma} = \frac{1}{N}\sum\limits_{k=1}^{N} $ $ (x_k-\mu)(x_k-\mu)^T $
$ \Sigma $ is non singular, but $ \hat{\Sigma} $ can be singular $ \Rightarrow $ no inverse $ \rightarrow $ this happens when number of points N<n: feature space down.
What happens when repeat sampling and estimating?
Sample: $ (x_1^i, x_2^i,...,x_N^i) \Rightarrow $ $ \hat{\mu}^i = \frac{1}{N}\sum\limits_{k=1}^{N}x_k^i $
- $ E(\hat{u})=? $
We have $ E(\hat{u})= E(\frac{1}{N}\sum\limits_{k=1}^{N}(x_k)) $ $ \frac{1}{N}E(x_k)=\frac{1}{N}\sum\limits_{k=1}^{N}E(x)= $ $ \frac{1}{N}\sum\limits_{k=1}^{N}u = \mu $
But how far do we expect to derivate from the mean?
$ E(|\hat{\mu}-\mu|^2) = E((\hat{\mu}-\mu)(\hat{\mu}-\mu)) $ $ =E(\hat{\mu}\cdot\hat{\mu}-\hat{\mu}\cdot{\mu} $ $ -{\mu}\cdot\hat{\mu}+{\mu}\cdot{\mu}) $
$ =E(\hat{\mu}\cdot\hat{\mu})-2\cdot \mu E(\hat{u})+\mu \cdot \mu $
$ =E(\hat{\mu}\cdot\hat{\mu})-\mu\cdot\mu $
$ =E(\frac{1}{N}\sum\limits_{k=1}^{N}x_k \cdot \frac{1}{N}\sum\limits_{j=1}^{N}x_j)-\mu\cdot\mu $
$ =\frac{1}{N^2}\sum\limits_{k=1}^{N}E(x_k \cdot x_j)-\mu\cdot\mu $
$ =\frac{1}{N^2}[\sum\limits_{k,j=1,k\neq j}^{N} $ $ E(x_k )\cdot E(x_j)+\sum\limits_{k,j=1,k\neq j}^{N} $ $ E(x_k )\cdot E(x_k)]-\mu\cdot\mu $
$ =\frac{1}{N^2}[N\cdot (N-1)\mu\cdot \mu+ $ $ \sum\limits_{k=1}^{N}E(x^2)]-\mu\cdot\mu $
$ -\frac{1}{N}\mu\cdot\mu+\frac{1}{N^2}\sum\limits_{k=1}^{N}E(x^2) $
by $ E[(x-\mu)(x-\mu)] = \sigma^2 \Rightarrow $ $ E(x \cdot x)-\mu^2 = \sigma^2 \rightarrow $ $ E(x \cdot x) = \sigma^2+\mu^2 $
So: $ E(|\hat{\mu}-\mu|^2) = -\frac{1}{N}\mu \cdot \mu + $
$ \frac{1}{N}(\sigma^2+\mu \cdot \mu) = \frac{1}{N}\sigma^2 $
- Bias: The maximum likelihood for the variance $\sigma^2$ is biased means
the expected value over all data sets of size n of the sample variance is not equal to the true variance:
$ E[\frac{1}{n}\sum\limits_{k=1}^{N}(x_k-\bar{x})] = \frac{n-1}{n} $ $ \sigma^2 \neq \sigma^2 $
But we can tell that as n $ \rightarrow \infty $, the MLE of $ \sigma $ is closing to $ \sigma^2 $
(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.