EM Algorithm Derivation--Three Coin Model Derivation Process

This blog mainly introduces the case of the three-coin model involved in the EM algorithm explained in Li Hang's "Statistical Learning Methods (Second Edition)". The derivation process of the model is omitted in the original text. This blog is mainly about the specific derivation process of the model.

1 Three-coin model

  Suppose there are 3 coins, denoted as A, B, and C respectively. The probabilities of these coins appearing heads are $pp$ _$p$ and$q$ . Carry out the following coin tossing test: first toss coin A, choose coin B or coin C according to the result, choose coin B for heads, and choose coin C for tails;nnindependently$n$ trials (here$n=10$ ), the observation results are as follows:$$1,1,0,1,0,0,1,0,1,1$$ At present, only the result of the coin toss can be observed, but the process of the coin toss cannot be observed. The problem to be solved by the EM algorithm is how to estimate π \piin the absence of coin toss process information$pp$ _$p$ and$The\; value\; of\; q$ .

2 Derivation process

  Suppose the observed data is recorded as $Y=(y_1,y_2,\dots ,y_n{)}^T$ , where$y_i$Indicates the $The\; observation\; result\; of\; the\; i$ trial is 1 or 0; the hidden variable data is recorded as$Z=(z_1,z_2,\dots ,z_n{)}^T$ , among which$z_i$Indicates the $The\; outcome\; of\; i$ unobserved coin flips A. $i=(p,p,q)$ is recorded as a model parameter. For any observed variable yi y_{i}in an experiment$y_i$,the function $$\begin{array}{rl}P({and}_i|\theta )& =\sum _{z_i}P({and}_i,z_i|\theta )=\sum _{z_i}p(z_i|\theta )P({and}_i|z_i,i)\\ & =\pi p^{y_i}(1-p{)}^{1-y_i}+(1-\pi \left)q^{y_i}\right(1-q{)}^{1-y_i}\end{array}$$Therefore, the observed data YYThe functional function of$Y$$$\begin{array}{rl}P(Y|\theta )& =\sum _{Z}P(Z|\theta )P(Y|Z,i)\\ & =\prod _{i=1}^n[\pi p^{y_i}(1-p_i{)}^{1-y_i}+(1-\pi )q^{y_i}(1-q{)}^{1-y_i}]\end{array}$$If the traditional maximum likelihood method is used to solve the parameter $\theta$ ,indefinitely$$\hat{i}=arg\underset{i}{\max }logP(Y|\theta )$$ But this formula does not have an analytical solution, and this method does not work. At this time, the EM algorithm needs to be used. The EM algorithm is an iterative process, and ends when the termination condition is met. The following process mainly shows how to use the$The\; parameter\; result\; of\; the\; i$ time is derived from the$i+1$ parameter.

• First assign initial values ​​to the model parameters. Here the initial value is $i^{(0)}=(p^{(0)},p^{\left(0\right)},q^{(0)})$，
• Suppose we have now obtained the iiThe parameter value after$i$$i^{(i)}=(p^{(i)},p^{\left(i\right)},q^{\left(i\right)})$. Now what we have to do is:$i^{\left(i\right)}\to i^{(i+1)}$。
• root $i^{\left(i\right)}$ Estimate the value of hidden data and calculate$Q$ function (E step);
  what is to be estimated here is the hidden data$The\; probability\; of\; Z$ occurring, that is, the probability of picking coin B or coin C in each trial. After getting$i^{After\; (i)}$ , observation data$y_j$From Coin $B$的概率如下：$$u_j^{(i+1)}=P(z_j=B|y_j,i^{(i)})=\frac{{Pi}^{\left(i\right)}\left(p^{\left(i\right)}{)}^{y_j}\right(1-p^{\left(i\right)}{)}^{1-y_j}}{{Pi}^{(i)}(p^{\left(i\right)}{)}^{y_j}(1-p^{\left(i\right)}{)}^{1-y_j}+(1-{Pi}^{(i)})(q^{\left(i\right)}{)}^{y_j}(1-q^{\left(i\right)}{)}^{1-y_j}}$$Observation data $y_j$The probability from coin C is: $1-u_j^{(i+1)}$. Calculate $Q$ function (not talking aboutQQLet$Q$$$\begin{array}{rl}Q(\theta ,i^{(i)})& =E_Z[logP(Y,Z|\theta )|Y,i^{(i)}]\\ & =\sum _{Z}P(Z|Y,i^{(i)})logP(Y,Z|\theta )\end{array}$$Here to introduce $The\; meaning\; of\; the\; Q$ function: the logarithmic likelihood function of complete data$logP(Y,Z|\theta )$ in the given observation data$Y$ and the current parameter$i^{\left(i\right)}$ For unobserved dataZZThe conditional probability distribution of$Z$$P(Z|Y,i^{\left(i\right)})$Specify the current value$i^{(i+1)}=arg\underset{i}{\max }Q(\theta ,i^{\left(i\right)})$.
  Here is a little more knowledge about conditional probability expectations. at$X=$random variableYY$under\; x$The formula for calculating the expectation of$Y$$$E(Y|X=x)=\{\begin{array}{ll}\sum y_{j}P(Y)=y_j|X=x)& \text{(X,Y) is a two-dimensional discrete random variable}\\ {\int }_{-\infty }^{\infty }yp(y|x)d& \text{(X,Y) is a two-dimensional continuous random}\end{array}$$Next, in order to fully display $Q(\theta ,i^{\left(i\right)})$, we first adjust the number of trials of the three-coin model to 2.
  Assume that the observation data obtained under the process of 2 random experiments is$Y=(y_1,y_2)$。那么$Q(\theta ,i^{\left(i\right)})$represents every possible observation data set$The\; conditional\; probability\; of\; Z$ appearing and the likelihood function of complete data$logP(Y,Z|\theta )$ is the sum of the products.
  Now take$Z=(1,0)$ , when$P(Z|Y,i^{(i)})=u_1^{(i+1)}(1-u_2^{(i+1)})$ ,then$$\begin{array}{rl}logP(Y,Z|\theta )& =log[\pi p^{y_1}(1-p{)}^{1-y_1}(1-\pi )q^{y_2}(1-q{)}^{1-y_2}]\\ & =log\pi +y_{1}logp+(1-y_1)log(1-p)+log(1-p)+y_{2}I\text{'}msorry\_+(1-y_2)log(1-q)\end{array}$$Add $Z=(1,1)$、$Z=(0,1)$、$Z=(0,0)$ P ( Z ∣ Y , θ ( i ) ) P(Z|Y,\theta^{(i)})$P(Z|Y,i^{\left(i\right)})$和$logP(Y,Z|\theta )$ to get the corresponding$Q(\theta ,i^{\left(i\right)})$. This information is$Q_2(i,i^{\left(i\right)})$, then$$\begin{array}{rl}{Q}_2(i,i^{(i)})& =u_1^{i+1}(1-u_2^{(i+1)})(log\pi +log(1-p)+y_{1}logp+(1-y_1)log(1-p)+y_{2}I\text{'}msorry\_+(1-y_2\left)log\right(1-q))\\ & +u_1^{(i+1)}{u}_2^{(i+1)}(log\pi +log\pi +y_{1}logp+(1-y_1\left)log\right(1-p)+y_{2}logp+(1-y_2\left)log\right(1-p))\\ & +(1-u_1^{(i+1)})u_2^{(i+1)}(log(1-p)+log\pi +y_{1}I\text{'}msorry\_+(1-y_1)log(1-q)+y_{2}logp+(1-y_2\left)log\right(1-p))\\ & +(1-u_1^{(i+1)})(1-u_2^{(i+1)})(log(1-p)+log(1-p)+y_{1}I\text{'}msorry\_+(1-y_1\left)log\right(1-q)+y_{2}I\text{'}msorry\_+(1-y_2\left)log\right(1-q))\\ & =(u_1^{(i+1)}+u_2^{(i+1)})log\pi +(2-u_1^{(i+1)}-u_2^{(i+1)})log(1-p)\\ & +u_1^{(i+1)}[y_{1}logp+(1-y_1\left)log\right(1-p)]+u_2^{(i+1)}[y_{2}logp+(1-y_2)log(1-p)]\\ & +(1-u_1^{(i+1)})[y_{1}I\text{'}msorry\_+(1-y_1\left)log\right(1-q)+(1-u_2^{(i+1)}\left)\right[y_{2}I\text{'}msorry\_+(1-y_2)log(1-q)]\end{array}$$Analyzing the above results, it can be inferred that $When\; n$ times of random trials, itsQQDetermine the value of$Q$$$Q(\theta ,i^{(i)})=\sum _{j=1}^n[u_j^{(i+1)}[log\pi +y_{j}logp+(1-y_j)log(1-p)]+(1-u_j^{(i+1)})[log(1-p)+y_{j}I\text{'}msorry\_+(1-y_j)log(1-q)]]$$
• Next, in order to get a new round of parameters $i^{(i+1)}$ , demand usage$Q(\theta ,i^{\left(i\right)})$function in turn for$\pi$ ,$p$ and$q$ seeks the partial derivative. The details are as follows:
  For$\pi$求偏密 get the following formula：$$\frac{\partial Q}{\partial \pi }=\sum _{j=1}^n[u_j^{(i+1)}\cdot \frac{1}{Pi}-\frac{1-u_j^{(i+1)}}{1-Pi}]$$ Let the calculation formula be 0, then get:$${Pi}^{(i+1)}=\frac{1}{n}\cdot \sum _{j=1}^n{u}_j^{(i+1)}$$to ppCalculate the partial derivative$of\; p$$$\frac{\partial Q}{\partial p}=\sum _{j=1}^n{u}_j^{(i+1)}(\frac{y_j}{p}+\frac{y_j-1}{1-p})$$ Let the calculation formula be 0, get:$$p^{(i+1)}=\frac{{\sum }_{j=1}^n{u}_j^{(i+1)}\cdot y_j}{{\sum }_{j=1}^n{u}_j^{(i+1)}}$$to $Calculate\; the\; partial\; derivative\; of\; q$ to get the following formula:$$\frac{\partial Q}{\partial q}=\sum _{j=1}^n(1-u_j^{(i+1)})(\frac{y_j}{q}+\frac{y_j-1}{1-q})$$ the same, possible and attainable:$$q^{(i+1)}=\frac{{\sum }_{j=1}^n(1-u_j^{(i+1)})y_j}{{\sum }_{j=1}^n(1-u_j^{(i+1)})}$$

So far, the derivation of the EM algorithm of the three-coin model is completed.

References

1. https://blog.csdn.net/weixin_41566471/article/details/106219019
2. "Statistical Learning Methods"