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）
[1 points] A random variable $X \sim \mathcal{N}\left(\mu, \sigma^2\right)$ is Gaussian distributed with mean $\mu$ and variance $\sigma^2$. Given that for any $a, b \in \mathbb{R}$, we have that $Y=a X+b$ is also Gaussian, find $a, b$ such that $Y \sim \mathcal{N}(0,1)$.
Since $\$ Y=a X+b \$$, we have that $\$ E(Y)=E(a X+b)=a E(X)+b=a \backslash m u+b \$$ and $\$ \backslash$ operatorname ${\operatorname{Var}}(Y)=$ loperatorname ${\operatorname{Var}}(a X+b)=$ $a^{\wedge} 2 \backslash$ operatorname ${\operatorname{Var}}(X)=a^{\wedge} 2 \backslash$ sigma^ $2 \$$.
We want $\$ Y \$$ to be Gaussian distributed with mean $\$ 0 \$$ and variance $\$ 1 \$$. Therefore, we need to find $\$ a \$$ and $\$ b \$$ such that $\$ a \backslash m u+b=0 \$$ and $\$ a^{\wedge} 2 \backslash \operatorname{sigma}^{\wedge} 2=1 \$$.
Solving $\$ a \backslash m u+b=0 \$$ for $\$ b \$$, we get $\$ b=-a \backslash m u \$$. Substituting this into $\$ a^{\wedge} 2 \backslash \operatorname{sigma}^{\wedge} 2=1 \$$, we get $\$ a^{\wedge} 2=\backslash d f r a c{1} \backslash \mid$ sigma $\left.^{\wedge} 2\right} \$$. Taking the square root, we have $\$ a=\backslash d f r a c{1}{$ sigma $}$ or $\$ a=-\backslash d f r a c{1}{\backslash$ sigma $}$.
Therefore, the two possible choices for $\$ a \$$ and $\$ \mathrm{~b} \$$ are: $a=\frac{1}{\sigma}, \quad b=-\frac{\mu}{\sigma} \quad$ or $\quad a=-\frac{1}{\sigma}, \quad b=\frac{\mu}{\sigma}$
Both choices give $\$ Y \backslash \operatorname{sim} \backslash$ mathcal ${N}(0,1) \$$.
[2 points] For a random variable $Z$, its mean and variance are defined as $\mathbb{E}[Z]$ and $\mathbb{E}\left[(Z-\mathbb{E}[Z])^2\right]$, respectively. Let $X_1, \ldots, X_n$ be independent and identically distributed random variables, each with mean $\mu$ and variance $\sigma^2$. If we define $\widehat{\mu}n=\frac{1}{n} \sum{i=1}^n X_i$, what is the mean and variance of $\sqrt{n}\left(\widehat{\mu}_n-\mu\right)$ ?
We have:
\begin{align*} \mathbb{E}[\sqrt{n}(\widehat{\mu}_n – \mu)] &= \sqrt{n}\mathbb{E}[\widehat{\mu}_n] – \sqrt{n}\mu \ &= \sqrt{n}\mu – \sqrt{n}\mu \ &= 0. \end{align*}
Now, let’s compute the variance:
\begin{align*} \mathbb{E}[(\sqrt{n}(\widehat{\mu}_n – \mu))^2] &= \mathbb{E}[n(\widehat{\mu}_n – \mu)^2] \ &= n\mathbb{E}[(\widehat{\mu}_n – \mu)^2] \ &= n\operatorname{Var}(\widehat{\mu}n) \ &= n\operatorname{Var}\left(\frac{1}{n}\sum{i=1}^n X_i\right) \ &= n\frac{\sigma^2}{n^2} \ &= \frac{\sigma^2}{n}. \end{align*}
Therefore, the mean of $\sqrt{n}(\widehat{\mu}_n – \mu)$ is $0$ and its variance is $\dfrac{\sigma^2}{n}$.
MY-ASSIGNMENTEXPERT™可以为您提供UNIVERSITY OF ILLINOIS URBANA-CHAMPAIGN MATH2940 linear algebra线性代数课程的代写代考和辅导服务！ 请认准MY-ASSIGNMENTEXPERT™. MY-ASSIGNMENTEXPERT™为您的留学生涯保驾护航。
