--- geometry: margin=2cm output: pdf_document title: Assignment 1 subtitle: CSCI 5521 date: \today author: | | Michael Zhang | zhan4854@umn.edu $\cdot$ ID: 5289259 --- \renewcommand{\c}[1]{\textcolor{gray}{#1}} 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{$f(x|\theta) = 2\theta x^{2\theta - 1} , 0P(C_2|x=0)$ 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 ```