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assignments/hwk01/HW1.md
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---
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.}
1. **(20 points)** \c{Derive the VC dimension of the following classifiers.}
2.
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.}
3. **(20 points)**
\c{Let $P (x|C)$ denote a Bernoulli density function for a class $C \in {C_1, C_2}$
- 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
$x = 1$.)}
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 $x = 1$.)}
For $x=0$, the posteriors $P(C_i | x)$ are given by $P(C_i | x = 0) = \frac{p(x = 0 | C_i) p(C_i)}{p(x = 0)}$.
- 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 $C_2$
- $p(x = 0 | C_i)$ is given to us as $p_1$
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
```

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assignments/hwk01/hw1solve.py
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from itertools import product
from sympy import symbols, log, diff, exp, Product, Pow
def calc_posterior(p_c1: float, D: int, p_ij: dict[tuple[int, int], float]):
@ -8,19 +9,11 @@ def calc_posterior(p_c1: float, D: int, p_ij: dict[tuple[int, int], float]):
}
def p_x_given_Ci(xs: list[int], i: int):
s = 0
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])
s *= pow(p_ij[i, j], 1.0 - xs[j]) * pow(1.0 - p_ij[i, j], xs[j])
return s
# print("===")
# for i in priors.keys():
# for xs in product([0, 1], repeat=2):
# xs = list(xs)
# print("xs", xs)
# print(i, xs, p_x_given_Ci(xs, i))
# print("===")
posteriors = {}
for i in [1, 2]:
for xs in product([0, 1], repeat=D):
@ -51,5 +44,19 @@ def prob_3c():
calc_posterior(0.8, D, p_ij)
# prob_2a()
def prob_2a():
p, x, theta, k, n = symbols("p x theta k n")
def get_mle(expr):
likelihood_func = Product(expr, (k, 1, n))
log_likelihood_func = log(likelihood_func)
print(diff(log_likelihood_func, x).simplify())
# print(diff(expr, x))
print(get_mle(Pow(p, x) * Pow(1 - p, 1 - x)))
print(get_mle((1 / theta) * exp(-x / theta)))
prob_2a()
prob_3c()