MY-ASSIGNMENTEXPERT™可以为您提供washington.edu CSE446 Machine Learning机器学习课程的代写代考和辅导服务！
CSE446课程简介
Programming in Python
We will use Python for the programming portions of the assignments. Python is a powerful general-purpose programming language with excellent libraries for statistical computations and visualizations. During the first week of the quarter, we will provide a tutorial to jump-start your transition into working in Python.
Here are some Python related resources:
www.learnpython.org “Whether you are an experienced programmer or not, this website is intended for everyone who wishes to learn the Python programming language.”
Prerequisites
Catalog Description: Design of efficient algorithms that learn from data. Representative topics include supervised learning, unsupervised learning, regression and classification, deep learning, kernel methods, and optimization. Emphasis on algorithmic principles and how to use these tools in practice. Prerequisite: CSE 332; MATH 208 or MATH 136; and either STAT 390, STAT 391, or CSE 312.Prerequisites: CSE 332; MATH 208 or MATH 136; and either STAT 390, STAT 391, or CSE 312.
Credits: 4.0
CSE446 Machine Learning HELP（EXAM HELP， ONLINE TUTOR）
If $f(x)$ is a PDF, the cumulative distribution function (CDF) is defined as $F(x)=\int_{-\infty}^x f(y) d y$. For any function $g: \mathbb{R} \rightarrow \mathbb{R}$ and random variable $X$ with PDF $f(x)$, recall that the expected value of $g(X)$ is defined as $\mathbb{E}[g(X)]=\int_{-\infty}^{\infty} g(y) f(y) d y$. For a boolean event $A$, define $\mathbf{1}{A}$ as 1 if $A$ is true, and 0 otherwise. Thus, $\mathbf{1}{x \leq a}$ is 1 whenever $x \leq a$ and 0 whenever $x>a$. Note that $F(x)=\mathbb{E}[\mathbf{1}{X \leq x}]$. Let $X_1, \ldots, X_n$ be independent and identically distributed random variables with $\operatorname{CDF} F(x)$. Define $\widehat{F}n(x)=\frac{1}{n} \sum{i=1}^n \mathbf{1}\left{X_i \leq x\right}$. Note, for every $x$, that $\widehat{F}_n(x)$ is an empirical estimate of $F(x)$. You may use your answers to the previous problem.
a. [1 points] For any $x$, what is $\mathbb{E}\left[\widehat{F}_n(x)\right]$ ?
a. Since $\widehat{F}_n(x)$ is the average of $n$ independent Bernoulli random variables with parameter $F(x)$, we have $\mathbb{E}[\widehat{F}_n(x)] = F(x)$.
b. [1 points] For any $x$, the variance of $\widehat{F}_n(x)$ is $\mathbb{E}\left[\left(\widehat{F}_n(x)-F(x)\right)^2\right]$. Show that $\operatorname{Variance}\left(\widehat{F}_n(x)\right)=$ $\frac{F(x)(1-F(x))}{n}$.
b. We have
\begin{align*} \operatorname{Variance}\left(\widehat{F}n(x)\right) &= \mathbb{E}\left[\left(\widehat{F}n(x)-F(x)\right)^2\right] \ &= \mathbb{E}\left[\left(\frac{1}{n} \sum{i=1}^n \mathbf{1}\left{X_i \leq x\right}-F(x)\right)^2\right] \ &= \mathbb{E}\left[\left(\frac{1}{n} \sum{i=1}^n \left(\mathbf{1}\left{X_i \leq x\right}-F(x)\right)\right)^2\right] \ &= \frac{1}{n^2} \mathbb{E}\left[\left(\sum_{i=1}^n \left(\mathbf{1}\left{X_i \leq x\right}-F(x)\right)\right)^2\right] \ &= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \mathbb{E}\left[\left(\mathbf{1}\left{X_i \leq x\right}-F(x)\right)\left(\mathbf{1}\left{X_j \leq x\right}-F(x)\right)\right]. \end{align*}
If $i = j$, then $\mathbb{E}\left[\left(\mathbf{1}\left{X_i \leq x\right}-F(x)\right)\left(\mathbf{1}\left{X_j \leq x\right}-F(x)\right)\right] = \mathbb{E}\left[\left(\mathbf{1}\left{X_i \leq x\right}-F(x)\right)^2\right] = F(x)(1-F(x))$. If $i \neq j$, then $X_i$ and $X_j$ are independent, so
Therefore,
$$
\text { Variance }\left(\widehat{F}_n(x)\right)=\frac{1}{n} F(x)(1-F(x)) \text {. }
$$
c. $\left[1\right.$ points] Using your answer to b, show that for all $x \in \mathbb{R}$, we have $\mathbb{E}\left[\left(\widehat{F}_n(x)-F(x)\right)^2\right] \leq \frac{1}{4 n}$.
c. Using the result from part b, we have
\begin{align*} \mathbb{E}\left[\left(\widehat{F}_n(x)-F(x)\right)^2\right] &= \operatorname{Variance}\left(\widehat{F}_n(x)\right) \ &= \frac{1}{
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