csci5521/assignments/hwk01/HW1.md

geometry	output	title	subtitle	date	author
margin=2cm	pdf_document	Assignment 1	CSCI 5521	\today	| Michael Zhang | zhan4854@umn.edu $\cdot$ ID: 5289259

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1. (20 points) \c{Derive the VC dimension of the following classifiers.}

   a. \c{What is the VC dimension, d_c, of a threshold c in \mathbb{R}? The classification function is specified by f (x) = +1 if x > c and f (x) = -1 if x \le c. Prove your answer.}

   - VC dimension is \boxed{2}
   - Given c, pick one point below $c$ and another point above $c$
   - For ex: Choose points $\{2, 4\}$ . For any arrangement of + / - labels, you can always distinguish them by putting a threshold at 3
   - Cannot shatter 3 points since if there's something in the middle then it's not shatterable
   - Choose any points $\{a, b, c\}$ in increasing order. The labeling a=+, b=-, c=+ cannot be achieved with any threshold
   - The trivial case of any 2 equaling each other also doesn't work since the case where those 2 are labeled differently cannot be distinguished
   

   b. \c{What is the VC dimension, d_I , of intervals in \mathbb{R}? The classification function specified by an interval [a,b] labels any example positive iff it lies inside the interval [a,b]. Prove your answer.}

    - VC dimension is \boxed{2}
        - Given the interval, pick one point in the interval and one outside
        - For ex: Choose points $\{2, 4\}$
        - 2=+, 4=+ => interval (1, 5)
        - 2=+, 4=- => interval (1, 3)
        - 2=-, 4=+ => interval (3, 5)
        - 2=-, 4=- => interval (6, 8)
    - Cannot shatter 3 points with the (positive, negative, positive) pattern, since the inside of the interval must be interpreted as positive.
        - Same as above, choose any points $\{a, b, c\}$ in increasing order. The labeling a=+, b=-, c=+ cannot be achieved with any interval since the positives are separated by a negative in between
   

2. (20 points) \c{Find the Maximum Likelihood Estimation (MLE) for the following pdf. In each case, consider a random sample of size n. Show your calculation}

   a. \c{$f(x|\theta) = \frac{1}{\theta} e^{-\frac{x}{\theta}} , x>0 , \theta>0$}

   - To find MLE, first find the log likelihood function:
       $$\begin{split}
       \mathfrak{L} (\theta|x) &=\log( \prod\limits_t \frac{1}{\theta} e^{-\frac{x^t}{\theta}} ) \\
       &=\sum\limits_t \left( \log(\frac{1}{\theta}) + \log(e^{-\frac{x^t}{\theta}}) \right) \\
       &=\sum\limits_t \left( \log(\frac{1}{\theta}) -\frac{x^t}{\theta} \right)
       \end{split}$$
   - Then take the partial with respect to $\theta$
       $$\begin{split}
       \frac{\partial\mathfrak{L}}{\partial\theta} &= \sum\limits_t \frac{\partial}{\partial\theta} \left( \log(\frac{1}{\theta}) -\frac{x^t}{\theta} \right) \\
       &=\sum\limits_t \left( -\frac{1}{\theta} + \frac{x^t}{\theta^2} \right)
       \end{split}$$
   - Now set it to 0 to find a local maximum
       $$\begin{split}
       0&=\sum\limits_t \left( -\frac{1}{\theta} + \frac{x^t}{\theta^2} \right) \\
       \sum\limits_t \frac{1}{\theta} &= \sum\limits_t \frac{x^t}{\theta^2} \\
       \sum\limits_t 1 &= \sum\limits_t \frac{x^t}{\theta} \\
       \sum\limits_t 1 &= \frac{1}{\theta} \sum\limits_t x^t \\
       N &= \frac{1}{\theta} \sum\limits_t x^t \\
       \theta &= \boxed{\frac{\sum\limits_t x^t}{N}}
       \end{split}$$
   

   b. \c{$f(x|\theta) = 2\theta x^{2\theta - 1} , 0<x\le 1 , 0<\theta<\infty$}

   - Find the log likelihood function:
       $$\begin{split}
       \mathfrak{L}(\theta|x) &= \log \left( \prod\limits_t 2\theta {x^t}^{2\theta - 1} \right) \\
       &= \sum\limits_t \left( \log(2\theta) + \log({x^t}^{2\theta-1}) \right) \\
       &= \sum\limits_t \left( \log(2\theta) + (2\theta - 1)\log(x^t) \right)
       \end{split}$$
   - Take the partial with respect to $\theta$
       $$\begin{split}
       \frac{\partial\mathfrak{L}}{\partial\theta} &= \sum\limits_t \frac{\partial}{\partial\theta} \left( \log(2\theta) + (2\theta - 1)\log(x^t) \right) \\
       &= \sum\limits_t \left( \frac{1}{\theta} + 2\log(x^t) \right)
       \end{split}$$
   - Set to 0
       $$\begin{split}
       0 &= \sum\limits_t \left( \frac{1}{\theta} + 2\log(x^t) \right) \\
       -\sum\limits_t 2\log(x^t) &= \sum\limits_t \frac{1}{\theta} \\
       -\theta \sum\limits_t 2\log(x^t) &= \sum\limits_t 1 \\
       -\theta \sum\limits_t 2\log(x^t) &= N \\
       \theta &= \boxed{-\frac{N}{\sum\limits_t 2\log(x^t)}}
       \end{split}$$
   

3. (20 points) \c{Let P (x|C) denote a Bernoulli density function for a class C \in {C_1, C_2} and P (C) denote the prior}

   a. \c{Given the priors P (C_1) and P (C_2), and the Bernoulli densities specified by p_1 \equiv p(x = 0|C_1) and p_2 \equiv p(x = 0|C_2), derive the classification rules for classifying a sample x into C_1 and C_2 based on the posteriors P (C_1|x) and P (C_2|x). (Hint: give rules for classifying x = 0 and x = 1.)}

   - The posteriors $P(C_i | x)$ can be found by expanding the Bayes' theorem equation:
       - $P(C_i|x) = \frac{p(x|C_i) P(C_i)}{ \sum\limits_k^{\{1,2\}} p(x|C_k) P(C_k) }$
       - Since $p_1=p(x=0|C_1)$ , we can expand this into a general case for $p(x|C_1)$ by using the Bernoulli density formula: $p(x|C_1)=p_1^{(1-x)} (1-p_1)^x$
       - Since $p_2$ is defined in an analogous way, I'll write $p(x|C_i)=p_i^{(1-x)} (1-p_i)^x$
   - Expanded form: $P(C_i|x)=\frac{ p_i^{(1-x)} (1-p_i)^x P(C_i) }{ \sum\limits_k^{\{1, 2\}} p_k^{(1-x)} (1-p_k)^x P(C_k) }$
   - To determine the classification rules, pick the $C_i$ with the maximum posterior:
       - For $x=0$ , pick $C_1$ if $P(C_1|x=0)>P(C_2|x=0)$ else $C_2$
       - For $x=1$ , pick $C_1$ if $P(C_1|x=1)>P(C_2|x=1)$ else $C_2$
   

   b. \c{Consider D-dimensional independent Bernoulli densities}

   $$
   \c{
       P (x|C) = P (x_1, x_2, \cdots , x_D|C) = \prod\limits_j P (x_j |C)
   }
   $$
   
   \c{specified by $p_ij \equiv p(x_j = 0|C_i)$ for i = 1, 2 and $j = 1, 2, \cdots , D$. Derive the classification rules for classifying a sample $\mathbf{x}$ into $C_1$ and $C_2$. It is sufficient to give your rule as a function of $\mathbf{x}$.}
   
   - The posteriors $P(C_i|x)$ can be found by expanding the Bayes' theorem equation:
       - $P(C_i|x)=\frac{ p(\mathbf{x}|C_i) P(C_i) }{ \sum\limits_k^{\{1,2\}} p(\mathbf{x}|C_k) P(C_k) }$
       - Since $p_{ij}=p(x_j=0|C_i)$ , we can expand this into a general case for $p(\mathbf{x}|C_i)$ by using the multivariate form of the Bernoulli: $p(\mathbf{x}|C_i)= \prod\limits_{j=1}^{D} p_{ij}^{(1-x_j)} (1-p_{ij})^{x_j}$
   - To determine the classification rules, pick the $C_i$ with the maximum posterior
   - We use the discriminant function found in the slides $g_i(\mathbf{x}) = p(\mathbf{x} |C_i)P(C_i)$ to select the posterior
       - If $g_1(\mathbf{x}) > g_2(\mathbf{x})$ , then choose $C_1$ else choose $C_2$
   

   c. \c{Follow the definition in 3(b) and assume D = 2, p_{11} = 0.6, p_{12} = 0.1, p_{21} = 0.6, and p_{22} = 0.9. For two different priors (P (C_1) = 0.2 or 0.8 and P (C_2) = 1 - P (C_1)), calculate the posterior probabilities P (C_1|x) and P (C_2|x). (Hint: Calcu- late the probabilities for all possible samples (x1, x2) \in \{(0, 0), (0, 1), (1, 0), (1, 1)\}).}

   - I wrote the following Python program to compute these values:
   
     ```py
     def calc_posterior(p_c1: float, D: int, p_ij: dict[tuple[int, int], float]):
         priors = {
             1: p_c1,
             2: 1 - p_c1,
         }
   
         def p_x_given_Ci(xs: list[int], i: int):
             s = 1.0
             for j in range(len(xs)):
                 s *= pow(p_ij[i, j], 1.0 - xs[j]) * pow(1.0 - p_ij[i, j], xs[j])
             return s
   
         posteriors = {}
         for i in [1, 2]:
             for xs in product([0, 1], repeat=D):
                 numer = p_x_given_Ci(xs, i) * priors[i]
   
                 def each_denom(k): return p_x_given_Ci(xs, k) * priors[k]
                 denom = sum(map(each_denom, priors.keys()))
                 posteriors[*xs, i] = numer / denom
   
         print("Priors:", priors)
         for xs in product([0, 1], repeat=D):
             print(f"{xs = }")
             for i in [1, 2]:
                 prob = posteriors[*xs, i]
                 print(f" * C{i}: {prob:0.3f}")
             print()
   
   
     def prob_3c():
         D = 2
         p_ij = {}
         p_ij[1, 0] = 0.6
         p_ij[1, 1] = 0.1
         p_ij[2, 0] = 0.6
         p_ij[2, 1] = 0.9
   
         calc_posterior(0.2, D, p_ij)
         calc_posterior(0.8, D, p_ij)
     ```
   
   - The values that it output are:
   
     ```
     Priors: {1: 0.2, 2: 0.8}
     xs = (0, 0)
     * C1: 0.027
     * C2: 0.973
   
     xs = (0, 1)
     * C1: 0.692
     * C2: 0.308
   
     xs = (1, 0)
     * C1: 0.027
     * C2: 0.973
   
     xs = (1, 1)
     * C1: 0.692
     * C2: 0.308
   
     Priors: {1: 0.8, 2: 0.19999999999999996}
     xs = (0, 0)
     * C1: 0.308
     * C2: 0.692
   
     xs = (0, 1)
     * C1: 0.973
     * C2: 0.027
   
     xs = (1, 0)
     * C1: 0.308
     * C2: 0.692
   
     xs = (1, 1)
     * C1: 0.973
     * C2: 0.027
     ```