Pattern Recognition - Exam Review

Pattern Recognition - Exam Review

Bayesian estimation

The use of parzen window in meanshift

Initialization:
Choose an initial point as the center of the Parzen window.

Choose a window function $K\left(u\right)$ , usually using a Gaussian distribution as the window function. let$h$ is the bandwidth parameter of the window function.

Kernel density estimation:
for each sample point $x_i$, calculate its relationship with the target point xxThe distance between$x$$d(x,x_i)$ .

The distance $d(x,x_i)$ divided by the bandwidth parameter$h$ , get the standardized distance${in}_i=\frac{d(x,x_i}{h}$.

General standardization distance ${in}_i$Substitute into the window function $In\; K\left(u\right)$ , achievable weight${In}_i=K\left(u_i\right)$ .

Sample point xxThe value of the kernel density estimation function of$x$$f_h(x)=\frac{}{n{h}^d}{\sum }_{i=1}^n{In}_i$, where $n$ is the sample size,$d$ is the data dimension.

If Gaussian distribution is used as the window function, then the window function is $K\left(u\right)=\frac{}{(2p{)}^{d/2}}{It\; is}^{-\frac{}{2}|u{|}^2}$ , where$|u|$ represents the normalized distanceuuThe modulus length of$u\; .$

Calculate the mean shift vector:
multiply each sample point according to its weight in the window (that is, the value of the kernel density estimation function) by the vector from the point to the center of the window to obtain a vector set, and then add these vectors and average Get the mean shift vector.

Shift window:
Move the center of the Parzen window to the position indicated by the mean shift vector calculated in the previous step. Repeat the above steps until the center of the Parzen window is stable or reaches the preset number of iterations, judging that the algorithm has converged. Output the final Parzen window center as the clustering result of the algorithm.

middle distance method

The teacher said that the derivation of the formula should be tested, and the derivation is as follows:

在Δ \DeltaIn$\Delta$$\angle$ HJI immediately$\theta$ carries out the law of cosines as follows:

$$cos\theta =\frac{H{J}^2+J{K}^2-H{K}^{}}{2HJ\_\_xJK}=\frac{H{J}^2+\frac{}{4}J{I}^2-H{K}^{}}{HJxJI}\left(1\right)$$

在$In\; \Delta$ HJI, the same for$\angle$ HJI immediately$\theta$ carries out the law of cosines as follows:

$$cos\theta =\frac{H{J}^2+J{I}^2-H{I}^{}}{2HJ\_\_xJI}\left(2\right)$$

Obviously, formula 1 and formula 2 are the same:

$$\frac{H{J}^2+\frac{}{4}J{I}^2-H{K}^{}}{HJxJI}=\frac{H{J}^2+J{I}^2-H{I}^{}}{2HJ\_\_xJI}$$

解得：
$$HK=\sqrt{\frac{}{2}H{I}^2+\frac{}{2}H{J}^2-\frac{}{4}I{J}^2}$$

Dual transformation in SVM

This is our final exam question for this semester, and the teacher revealed it in advance. It is also similar to the above derivation, but it is simpler.

That is, if the objective function is not ${In}^T{x}_i+b$ is instead${In}^T{x}_i$, please convert the dual problem. In fact, it has become simpler, and there is no need to perform derivation and back substitution on b. The solution is as follows: