The maximum a posteriori probability of Bayesian inference (MAP)

This paper records the mathematical principle of Bayesian posterior probability distribution in detail , A binary classification problem based on Bayesian posterior probability , Let's talk about my understanding of Bayesian inference .

1. Dichotomous problem

Given N The data set of 2 samples , use $$X$$ To express , Each sample $$x_n$$ There are two properties , Finally, it belongs to a certain category $$t$$

$$t=\left\{0,1\right\}$$

$$\mathbf{x_n}=\begin{pmatrix}x_{n1} \\ x_{n2} \\ \end{pmatrix}$$, Suppose the model parameters $$w=\begin{pmatrix} w_1 \\ w_2\end{pmatrix}$$

$$\mathbf{X}=\begin{bmatrix} x_1^T \\ x_2^T \\. \\. \\ x_n^T\end{bmatrix}$$

Picture the sample set as follows ： According to Bayesian formula, there are ：

$p(w|t,X)=\frac {p(t|X,w)p(w)} {p(t|X)}$ ( The formula 1)

$$p(w | t,X)$$ Tell us ： In a given training sample set $$X$$ And a classification of these samples $$t$$ ( It's a supervised learning , Because we already have a sample set $$X$$、 And the classification of each sample in the sample set $$t$$), You need to solve the model parameters $$w$$ . therefore ,$$w$$ It is unknown. , It needs to be solved by Bayesian probability formula according to samples . We have found $$p(w|t,X)$$ The distribution of , So you know the model parameters $$w$$

When we get the optimal model parameters $$w^*$$ after , Given a sample to be predicted $$\mathbf{x_{new}}$$ According to the formula $P(T_{new}=1|x_{new}, w^*)$ To calculate the new sample $$\mathbf{x_{new}}$$ Classified as 1 What's the probability of , That's what the model predicts .

The formula 1 The medium size has three parts on the right ,$$p(t|X,w)$$ It's called likelihood probability (likelihood),$$p(w)$$ It's called a priori probability , These two parts are generally easier to solve . The hardest part to solve is the denominator : $$p(t|X)$$ It's called the boundary likelihood function (marginal likelihood). But fortunately , Boundary likelihood function and model parameters $$w$$ irrelevant , therefore , You can think of the denominator as about $$w$$ A constant of .

In Mathematics , If the prior probability is conjugate with the likelihood probability , So the posterior distribution probability $$p(w|t, X)$$ It follows the same distribution as the prior probability . for instance ： A priori probability obeys Beta Distribution , Likelihood probability follows binomial distribution , The prior probability distribution is conjugate with the likelihood probability distribution , Then the posterior probability also obeys Beta Distribution .

therefore , When using Bayes formula , If the selected prior probability distribution is conjugate with the likelihood probability distribution , The posterior probability distribution can be easily calculated （ Or to be able to accurately calculate a Concrete / The precise Model parameters $$w^*$$）, This is it. ：can compute posterior analytically. But the reality is , They are not conjugate , So there are three common approximate calculation methods ：

• Point estimation (Point Estimate--MAP Method )
• Laplace approximation method (Laplace approximation)
• Sampling method (Sampling--Metropolis-Hastings)

This paper only introduces the point estimation method .

Back to the formula 1, Let's start with a priori probability $$p(w)$$ , A priori probability is similar to when making a decision , Existing experience . Because we already have training samples $$X$$, Draw the contour lines corresponding to these samples , They are very close to Gaussian distribution , Then it can be considered that the prior probability obeys Gaussian distribution . That is to say $p(w)=N(0,\sigma^2I)$. among ,$$w$$ It's a vector ,$$I$$ It's a unit matrix .

And then there's likelihood probability $$p(t|X,w)$$ , Suppose that given the model parameters $$w$$ And sample sets $$X$$ Under the condition of , The classification results of each sample are independent of each other , therefore ：

$p(t|X,w)=\prod_{n=1}^N p(t_n|x_n, w)$ ( The formula 2)

for instance , When the model parameters are known $$w$$ when ,$$w$$ take $$x_1$$ The prediction is positive , take $$x_2$$ The prediction is negative …… take $$x_n$$ Predict the positive , The prediction results of each sample are independent of each other . namely ：$$w$$ Yes $$x_1$$ Forecast results of It won't affect its effect on $$x_2$$ Forecast results of .

because , It's a dichotomy problem ,$$t_n=\left\{0,1\right\}$$ , We can further extend the formula 2 It's written in ：$p(t|X,w)=\prod_{n=1}^N p(T_n=t_n|x_n, w)$ , among $$T_n$$ Representative sample $$x_n$$ Classified into a certain class A random variable ,$$t_n$$ Is the value of the random variable . such as $$t_n=0$$ Presentation sample $$x_n$$ Be classified as positive ,$$t_n=1$$ Means to be classified as negative .

2. sigmod function

Because the probability of a random variable taking a certain value is in [0,1] Between , So ask for the solution $$p(t|X,w)$$, Our goal is ： Find a function $$f(\mathbf{x_n};w)$$ This function produces a probability value . To simplify the discussion , choice $$sigmod(w^T*x)$$ , therefore ：

$P(T_n=1|x_n,w)=\frac{1}{1+exp(-w^T*x_n)}$

that ：

$P(T_n=0|x_n,w)=1-P(T_n=1|x_n,w)=\frac{exp(-w^T*x_n)}{1+exp(-w^T*x_n)}$

Combine the above two formulas into one ：

$P(T_n=t_n|x_n,w)=P(T_n=1|x_n,w)^{t_n}P(T_n=0|x_n,w)^{1-t_n}$

about N Samples , The formula 2 Can be written as ：

$p(t|X,w)=\prod_{n=1}^N (\frac{1}{1+exp(-w^T*x_n)})^{t_n}(\frac{exp(-w^T*x_n)}{1+exp(-w^T*x_n)})^{1-t_n}$ ( The formula 3)

thus , The prior probability follows Gaussian distribution , The likelihood probability is determined by the formula 3 give , Then we can solve the posterior probability formula mentioned above ：$p(w|X,t,\sigma^2)$

As long as we get the posterior probability , You can use the following formula to calculate the probability that the new sample will be divided into negative categories ：

$P(t_{new}=1|x_{new}, X, t)=E_{p(w|X,t,\sigma^2)}\left(\frac{1}{1+exp(-w^T*x_{new})}\right)$

Explain the formula ： Because we've got the posterior probability $$p(w|X,t,\sigma^2)$$ The expression of , It's about $$f(x_n;w)$$ Function of , Calculate the expected value of this function $$E$$, The expectation is Predict new samples $$x_{new}=1$$ Probability .

good , The next step is to solve the posterior probability .

3. Find the posterior probability

I've said that before , The prior probability follows Gaussian distribution $$N(0,\sigma^2I)$$, The likelihood distribution is determined by The formula 3 give , And the denominator -- The boundary likelihood function is a function of $$w$$ The constant , So define a function $$g(w;X,t,\sigma^2)=p(t|X,w)p(w|\sigma^2)$$ , function $$g$$ Obviously with posterior probability $$p(w|X,t,\sigma^2)$$ Is proportional to the . therefore , We get the function $$g$$ The maximum of , It is equivalent to finding the optimal parameter of posterior probability $$w^*$$.

The problem here is ： How can we maximize the function g Well ？$$g$$ yes $$w$$ Function of ,$$w$$ Which value does the function take $$g$$ To the maximum ？

Here we need to use a method called Newton's method (Newton-Raphson method). Newton's method can be used to Looking for zeros in a function . It works through the following formula ：

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

Continuous iteration , Finally, we find that the value of the function is 0 The point of .

In mathematics, when the judgment function takes the extreme value at a certain point , There are the following theorems ：

Let's take the derivative function of one variable $$h(x)$$ For example ,$$h(x)$$ Derivative is 0 The point of is the extreme point , But this extreme point is a minimum , Or the maximum value ？ In this case, we can judge $$h(x)$$ It is the second derivative to judge whether the extreme point is a minimum or a maximum . if $$h'(x_n)=0$$ And $$h''(x_n)<0$$, be , be $$h(x)$$ In $$x_n$$ Take the maximum .

therefore , If you can judge $$g(w;X,t,\sigma^2)$$ About $$w$$ The second derivative of is less than 0, Then you can use Newton's method to solve $$g(w;X,t,\sigma^2)$$ The first derivative of is about $$w$$ Zero point of , namely $$g'(w;X,t,\sigma^2)=0$$ when $$w$$ The values for $$w_0$$, This $$w_0$$ It's the best solution $$w^*$$ 了 .

good , Let's prove it $$g(w;X,t,\sigma^2)$$ About $$w$$ The second derivative of is less than 0 Of . because $$w$$ It's a vector , In multivariate functions , It's equivalent to proving that ：$$g(w;X,t,\sigma^2)$$ About $$w$$ The Hessian matrix of is negative definite .

Will function $$g$$ Take the logarithm , Maximize $$log (g(w;X,t,\sigma^2))$$

$$log (g(w;X,t,\sigma^2))=log({p(t|X,w)p(w|\sigma^2}))$$

$=log(p(t|X,w)+log(p(w|\sigma^2)$ To simplify the formula , Make the following agreement ： hypothesis $$w$$ It's a $$D$$ Dimension vector ： The first three terms are after the prior distribution obeys the Gaussian distribution , The result of simplification . According to the vector derivative formula ：$\frac{\partial w^Tw}{\partial w}=w$ By the chain derivation rule ： obtain ： therefore ：$$log (g(w;X,t,\sigma^2))$$ Yes $$w$$ The first partial number of is as follows ： The second partial derivative is as follows ： $$I$$ It's a unit matrix ,$$0=<P_n<=1$$ It's a probability value , The second partial derivative is the Hessian matrix , It's negative .

The proof is complete .

thus , You can safely use Newton's method to iterate , find $$g(w;X,t,\sigma^2)$$ When the maximum value is taken, the parameter $$w$$ The value of the , And this value is $$w^*$$

Now? ,$$w^*$$ Come on , You can use the following formula to predict new samples $$x_{new}$$ Predicted to be negative ($$T_{new}$$ The value is 1) The probability of

$P(T_{new}=1|x_{new},w^*)=\frac{1}{1+exp(-w^{*T}x_{new})}$

decision boundary

Because it's a binary problem , Let's look at classification using Bayesian posterior probability Decide the boundary What does it look like . Because the output is a probability value , It's obvious that $$P(T_{new}=1|x_{new},w^*)>0.5$$ The prediction is negative ,$$P(T_{new}=1|x_{new},w^*)<0.5$$ The prediction is positive . That's equal to 0.5 When ？

according to ：$P(T_{new}=1|x_{new},w^*)=\frac{1}{1+exp(-w^{*T}x_{new})}=0.5$ obtain ：

$-w^{*T}*x=0=w_1^*x_1+w_2^*x_2$

$x_2=\frac{w_1^*}{w_2^*}*(-x_1)$

That is to say, samples $$\mathbf{x}=\begin{pmatrix}x_{1} \\ x_{2} \\ \end{pmatrix}$$ Two properties of $$x_1$$ and $$x_2$$ It's linearly proportional . And this straight line is decision boundary summary

Bayesian method is a common method in machine learning , There are three parts in Bayesian formula , A priori probability distribution function 、 Likelihood probability distribution function 、 And the boundary likelihood probability distribution function （ The denominator of Bayesian formula ）. The three parts are worked out , Then we get the posterior probability distribution , And then for a new sample $$x_{new}$$ Calculate the expected value of the posterior probability distribution , This expectation is the prediction result of Bayesian model .

Because the calculation of posterior probability distribution depends on prior probability distribution function 、 Likelihood probability distribution function , When the two are conjugate , Posterior probability and prior probability obey the same distribution function , So we can deduce and calculate the posterior probability distribution (posterior could be computed analytically). however , When the two are not conjugate , It is an approximation of the posterior probability distribution . There are three ways to calculate approximations , Point estimation (point estimate --- MAP), Laplace approximation ,Metropolis-Hastings Sampling method . And this article mainly introduces It's the first way ： Point estimation (point estimate --- maximum a posteriori).

maximum a posteriori Where is the maximization of the quality of life ？ In fact, it is reflected in the maximization of the likelihood distribution function . The negative definiteness of Hesse matrix proves $$g(w;X,t,\sigma^2)$$ With maximum , And then we use Newton's method to find this function $$g$$ Take the optimal parameter solution of the maximum value $$w^*$$. And the optimal parameters are obtained $$w^*$$, The posterior probability distribution formula is obtained . For a new sample to be predicted $$x_{new}$$, Calculate the expected value of the posterior probability distribution of the sample , The expected value is the prediction result of Bayesian model for new samples .

Reference material

Newton method ：https://zh.wikipedia.org/wiki/ Newton method
Blog Garden Markdown The formula is messy ：http://www.cnblogs.com/cmt/p/markdown-latex.html

original text ：http://www.cnblogs.com/hapjin/p/8834794.html

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