From cb2b3db4938fa04ef615e5d0c4984a016d216856 Mon Sep 17 00:00:00 2001 From: Michael Zhang Date: Mon, 16 Oct 2023 16:50:38 -0500 Subject: [PATCH] d --- .gitignore | 3 +- assignments/hwk01/HW1.typ | 209 ++++++++++++++++++++++++++++++++++++++ 2 files changed, 211 insertions(+), 1 deletion(-) create mode 100644 assignments/hwk01/HW1.typ diff --git a/.gitignore b/.gitignore index 3d8c46a..036c568 100644 --- a/.gitignore +++ b/.gitignore @@ -1,4 +1,5 @@ *.asv .vscode *.pdf -*.zip \ No newline at end of file +*.zip +.DS_Store diff --git a/assignments/hwk01/HW1.typ b/assignments/hwk01/HW1.typ new file mode 100644 index 0000000..9365f8f --- /dev/null +++ b/assignments/hwk01/HW1.typ @@ -0,0 +1,209 @@ +#set document( + title: "Assignment 1", + author: "Michael Zhang ", +) + +#let c(body) = { + set text(gray) + body +} +#let boxed(body) = { + box(stroke: 0.5pt, inset: 2pt, outset: 2pt, baseline: 0pt, body) +} + +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 $RR$? The classification function + is specified by $f (x) = +1$ if $x > c$ and $f (x) = -1$ if $x lt.eq c$. 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 $RR$? 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 + + #table( + columns: (auto, auto, auto), + [2], [4], [interval], + [+], [+], [[1, 5]], + [+], [-], [[1, 3]], + [-], [+], [[3, 5]], + [-], [-], [[6, 8]], + ) + - 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: + $ frak(L) (theta|x) &= log( limits(Pi)_t frac(1,theta) e^(-frac(x^t, theta)) ) \ + &= sum_t ( log(frac(1, theta)) + log(e^(-frac(x^t,theta))) ) \ + &= sum_t ( log(frac(1,theta)) -frac(x^t,theta) ) $ + - Then take the partial with respect to $theta$ + $$\begin{split} + \frac{\partial\mathfrak{L}}{\partialtheta} &= \sum\limits_t \frac{\partial}{\partialtheta} \left( \log(\frac{1}{theta}) -\frac{x^t}{theta} \right) \\ + &=\sum\limits_t \left( -\frac{1}{theta} + \frac{x^t}{theta^2} \right) + \end{split}$$ + - 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[$f(x|theta) = 2theta x^(2theta - 1) , 0P(C_2|x=0)$ 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 $p_i j equiv p(x_j = 0|C_i)$ for i = 1, 2 and $j = 1, 2, \cdots , D$. Derive the classification rules for classifying a sample $x$ into $C_1$ and $C_2$. It is sufficient to give your rule as a function of $x$.} + + - The posteriors $P(C_i|x)$ can be found by expanding the Bayes' theorem equation: + - $P(C_i|x)=frac( p(bold(x)|C_i) P(C_i) , sum_k^{\{1,2\}} p(bold(x)|C_k) P(C_k) )$ + - Since $p_{i j}=p(x_j=0|C_i)$ , we can expand this into a general case for $p(bold(x)|C_i)$ by using the multivariate form of the Bernoulli: $p(bold(x)|C_i)= pi {j=1}^{D} p_{i j}^{(1-x_j)} (1-p_{i j})^{x_j}$ + - To determine the classification rules, pick the $C_i$ with the maximum posterior + - We use the discriminant function found in the slides $g_i(bold(x)) = p(bold(x) |C_i)P(C_i)$ to select the posterior + - If $g_1(bold(x)) > g_2(bold(x))$ , 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 $(x_1, x_2) \in \{(0, 0), (0, 1), (1, 0), (1, 1)\}$).} + + - 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 + ```