Bayesian Decision Making

In this lab we will implement an OCR system. For simplicity we assume the texts contain only two letters: A and C, and that the letters are already well segmented into 10×10 pixel image chips. We will extract a simple feature from the image chips and use it in a bayesian decision framework to partition the image set. Furthermore, the relative frequency of occurrence of A and C in text is given. We will consider two cases: (1) the measurements are discrete, and (2) the measurements are continuous.

In the discrete probabilities case, we will find the optimal strategy, compute its Bayesian risk, experiment with the loss function and classify independent data samples by applying the theory from the lecture slides directly.

In the continuous measurements case, an assumption is made that the extracted feature for each letter class (A,C) is generated by a normal distribution. The 'zero-one' loss will be used to estimate risk, which simplifies the decision strategy.

The task is to design a classifier (Bayesian strategy) $q(x)$, which distinguishes between 10×10 images of two letters (classes) A and C using only a single measurement (feature):

$$ q: \mathcal{X} \rightarrow D $$ $$ \mathcal{X} = \{-10, \ldots, 10\}, \quad D = K = \{A, C\}$$

The measurement $x$ is obtained from an image by computing

mu = -563.9
sigma = 2001.6
x_unnormalized = (sum of pixel values in the left half of image)
                -(sum of pixel values in the right half of image)
x = (x_unnormalized - mu) / (2 * sigma) * 10
limit x to the interval <-10, 10>

Computation of this measurement is implemented in bayes.compute_measurement_lr_discrete.

The strategy $q(x)$ will be represented in numpy by a (21, ) vector containing 0 if the classification for that value of $x$ is supposed to be A, and 1 if the classification is supposed to be C. Thus given the vector $q$ and some $x$ we can decide for one of the classes.

As required in the formulation of Bayesian task, we are given all the necessary probabilities (specified in the bayes.ipynb):

1. Complete the function template bayes_risk_discrete so that it returns the Bayesian risk for a given strategy. The input parameters are the distributions, the strategy $q$ and arbitrary loss matrix $W$ (stored in (2, 2) array).
   
   

   q_discrete = np.array([0]*10 + [1] + [0]*10)
   W = np.array([[0, 1], [1, 0]])
   discreteA = {'Prior': 0.6153846153846154,
                'Prob': np.array([0.0125, 0., 0., 0.0125, 0.025, 0.0125, 0.025, 0.0375, 0.075, 0.1, 0.2125, 0.1375, 0.15, 0.1, 0.0875, 0.0125, 0., 0., 0., 0., 0.])}
   discreteC = {'Prior': 0.38461538461538464,
                'Prob': np.array([0., 0., 0., 0.02, 0.02, 0.22, 0.46, 0.16, 0.1, 0.02, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])}
   R_discrete = bayes_risk_discrete(discreteA, discreteC, W, q_discrete)  # -> 0.5153846153846154

2. Complete the function template find_strategy_discrete so that it returns the optimal Bayesian strategy. When the bayesian risk is the same for both classes for a particular $x$, prefer the “A” class to pass the automatic checks in the upload system.
   
   Hint: Use the second line of the derivation on slide 11 in the lecture slides.
   
   

   W = np.array([[0, 1], [1, 0]])
   q_discrete = find_strategy_discrete(discreteA, discreteC, W)
   # -> array([0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
    
   distribution1 = {}
   distribution2 = {}
   distribution1['Prior'] = 0.3
   distribution2['Prior'] = 0.7
   distribution1['Prob'] = np.array([0.2, 0.3, 0.4, 0.1])
   distribution2['Prob'] = np.array([0.5, 0.4, 0.1, 0.0])
   W = np.array([[0, 1], [1, 0]])
   q = find_strategy_discrete(distribution1, distribution2, W)
    
   # -> array([1, 1, 0, 0])

3. Compute the risk of the optimal strategy.
   
   
4. Plot the classification using the visualise_discrete function for two loss functions: W1=np.array(0, 1], [1, 0) and W2=np.array(0, 5], [1, 0) and save the figures as classif_W1.png and classif_W2.png.
   
   
5. Complete the classify_discrete function template which realizes the found strategy, so that it returns N labels given (10, 10, N) array of images.
   
   
6. Complete the classification_error_discrete function so that it returns the error on the whole image set images_test. The error is defined as number of incorrectly classified samples divided by total number of samples. To measure the error compare your classification with the ground truth in labels_test variable.
   
   

   W = np.array([[0, 1], [1, 0]])
   q_discrete = find_strategy_discrete(discreteA, discreteC, W)
   error_discrete = classification_error_discrete(images_test, labels_test, q_discrete)
    
   # -> 0.225

7. Show the images classified as A and images classified as C in a figure. Save it as decision_discrete.png.

Fill the correct answers to your answers.txt file.

question1 : gen_zero_matrix
question2 : 12.5467
question2 : [b,d,f]

In the second part of the assignment the probabilities $p_{X|k}(x|k)$ are given as continuous normal distributions (specified in the bayes.ipynb): $$p_{X|k}(x|A) = \frac{1}{\sigma_k\sqrt{2\pi}}e^{-\frac{(x-\mu_k)^2}{2\sigma_k^2}}$$

Example distribution:

contA = {'Mean': 124.2625,
         'Sigma': 1434.45420083,
         'Prior': 0.61538462}
contC = {'Mean': -2010.98,
         'Sigma': 558.42857106,
         'Prior': 0.38461538}

In this particular case, as explained in lecture notes, slides 21-22, the optimal strategy $q$ can be found by solving the following problem:

$$q(x) = \arg\max_k p_{K|X}(k|x) = \arg\max_k p_{XK}(x,k)/p_X(x) = \arg\max_k p_{XK}(x,k) = \arg\max_k p_K(k)p_{X|K}(x|k),$$

where $p_{K|X}(k|x)$ is called posterior probability of class $k$ given the measurement $x$ (i.e. it is the probability of the data being from class $k$ if we measure feature value $x$). Symbol $\arg\max_k$ denotes finding $k$ maximising the argument.

We will also work with an unnormalised measurement here instead of the discrete one used above

x = ((sum of pixel values in the left half of image)
   -(sum of pixel values in the right half of image))

At the beginning of the labs we will together show that in the case when $p_{X|k}(x|k)$ are normal distributions and when we consider zero-one loss function, the optimal Bayesian strategy $q(x)$ corresponds to a quadratic inequality (i.e. $ax^2+bx+c \ge 0$).

Hint: $$ \arg\max_{k\in\{A,C\}} p_K(k)p_{X|k}(x|k) \quad \mbox{translates into} \quad p_K(A)p_{X|k}(x|A) \ge p_K(C)p_{X|k}(x|C) $$

Optimal strategy representation: Solving the above quadratic inequality gives up to two thresholds $t_1$ and $t_2$ (zero crossing points) and a classification decision (A or C) in each of the intervals $\mathcal{X}_1 = (-\infty, t_1>$, $\mathcal{X}_2 = (t_1, t_2>$, $\mathcal{X}_3 = (t_2, \infty)$. So similarly to the discrete case we can represent the strategy in python as a dict with the following fields:

1. Fill in the formulas for quadratic inequality coefficients $a$, $b$ and $c$ into the find_strategy_2normal.
   
   
2. Complete the find_strategy_2normal template by filling in code for the determining the decision vectors. Make sure that you return proper strategy for all corner cases (same variances different means, prior = 1 for one class, same mean and one prior significantly higher, both distributions the same but maybe prior is different, high prior + high variance, p(x|A)p(A) = p(x|C)p(C) in exactly one point). All of these conditions can be tested by observing the quadratic inequality only, no pdf evaluation is needed!
   
   

   q_cont = find_strategy_2normal(contA, contC)
    
   # -> {'t2': -1248.7684903033442, 't1': -3535.997150276402, 'decision': array([0, 1, 0], dtype=int32)}

3. Using the visualize_2norm function in the template to display the probabilities $p_{Xk}(x,k)$ and the classification thresholds $t_1$ and $t_2$ and save the figure as thresholds.png.
   
   

   

4. Complete the template function bayes_risk_2normal. We can reformulate the equations from lecture slides, slide 21 as $$ R(q) = \int_X \sum_{k\in K}p_{XK}(x,k)W(k,q(x)) dx = \\ = \int_X p_X(x) \sum_{k\ne q(x)}p_{K|x}(k|x) dx = \\ = \int_X p_X(x)(1 - p_{K|x}(q(x)|x)) dx = \\ = 1 - \int_X p_X(x)p_{K|x}(q(x)|x) dx = \\ = 1 - \int_X p_K(q(x))p_{X|k}(x|q(x)) dx\,,$$ which in our special case simplifies further to $$R(q) = 1-\int_X p_K(q(x))p_{X|k}(x|q(x)) = \\ = 1 - \left( \int_{\mathcal{X}_1} p_K(\mbox{q.decision}(1))p_{X|k}(x|\mbox{q.decision}(1)) + \int_{\mathcal{X}_2} p_K(\mbox{q.decision}(2))p_{X|k}(x|\mbox{q.decision}(2)) + \\ \int_{\mathcal{X}_3} p_K(\mbox{q.decision}(3))p_{X|k}(x|\mbox{q.decision}(3))\right).$$ Thus we only need to compute integrals of the normal distribution corresponding to the strategy decision class over three intervals, multiply them by apriori probabilities and subtract their sum from one.
   
   Hint: To integrate exponential functions you can use the scipy.stats.norm.cdf function.
   
   

   R_cont = bayes_risk_2normal(contA, contC, q_cont)
    
   # -> 0.13519281686757106

5. Complete the classify_2normal function template so that it returns a label given a 10×10 image.
   
   
6. Complete the classification_error_2normal function so that it returns the error on the whole image set images. To measure the error compare your classification with the ground truth in labels variable.
   
   

   error_cont = classification_error_2normal(images_test, labels_test, q_cont)
    
   # -> 0.1

7. Show the images classified as A and images classified as C using the montage function. Save them as decision_2normal.png (see bayes.ipynb)

Fill the correct answers to your answers.txt file.

This task is not compulsory. Work on bonus tasks deepens your knowledge about the subject. Successful solution of a bonus task(s) will be rewarded by up to 4 points.

The data for this task is located in data_33rpz_bayes_bonus.npz. Use the following distributions D1, D2 and D3:

D1 = {'Sigma': np.array([1431.41915097, 694.63442162]),
      'Mean': np.array([151.61, 1154.01]),
      'Prior': 0.4166666666666667,
      'Cov': np.array([[2048960.78575758, 169829.10494949],
                       [ 169829.10494949, 482516.97969697]])}
 
D2 = {'Sigma': np.array([554.56287158, 696.83182077]),
      'Mean': np.array([-2055.76666667,    54.3]),
      'Prior': 0.25,
      'Cov': np.array([[307539.97853107,  -2813.69830508],
                       [ -2813.69830508, 485574.58644068]])}
 
D3 = {'Sigma': np.array([984.34517453, 637.97896862]),
      'Mean': np.array([  492.7125, -2353.1125]),
      'Prior': 0.3333333333333333,
      'Cov': np.array([[968935.42262658,  12248.03053797],
                       [ 12248.03053797, 407017.16439873]])}

y = (sum of pixel values in the upper half of image)
   -(sum of pixel values in the lower half of image)

The task is to classify the input images into one of the three classes. Use both measurements $x$ and $y$ here. The density function $p_{X}(\mathbf{x}) = p_K(A)p_{X|k}(\mathbf{x}|A)+p_K(C)p_{X|k}(\mathbf{x}|C)+p_K(T)p_{X|k}(\mathbf{x}|T)$, where $\mathbf{x} = [x, y]$ is visualised below.

Output:

• Classification error of the optimal Bayesian strategy on the provided image set,
• Classification of input images, i.e. which image was assigned to which class – draw the images per class.

Hints:

• Multivariate Gaussian distribution density is given by $$f_{\mathbf x}(x_1,\ldots,x_k) = \frac{1}{\sqrt{(2\pi)^k|\boldsymbol\Sigma|}} \exp\left(-\frac{1}{2}({\mathbf x}-{\boldsymbol\mu})^T{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu}) \right),$$ where $k$ is number of dimensions, $\boldsymbol\Sigma$ is $k \times k$ covariance matrix and $\boldsymbol\mu$ is $k$-dimensional mean vector.

• Calculate numerically the Bayesian risk for all decisions for every input image and select the optimal decision.

• [1] Duda R., Hart P., Stock D.: Pattern Classification, 2001
• [2] Michail I. Schlesinger, Vaclav Hlavac. Ten Lectures on Statistical and Structural Pattern Recognition. Kluwer Academic Publishers, 2002.
• [3] Slides from lecture by Vaclav Hlavac and Jiri Matas.
• [4] Old labs text.