### 计算机代写|机器学习代写machine learning代考|Conditional Distributions

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 计算机代写|机器学习代写machine learning代考|Conditional Distributions

For the beam example illustrated in figure 4.5, our prior knowledge for the resistance $\left{X_{1}, X_{2}\right}$ of two adjacent beams is
and we know that the beam resistances are correlated with $\rho_{12}=$ 0.8. Such a correlation could arise because both beams were fabricated with the same process, in the same factory. This prior knowledge is described by the joint bivariate Normal PDF,
f_{X_{1} X_{2}}\left(x_{1}, x_{2}\right)=\mathcal{N}\left(\mathbf{x} ; \boldsymbol{\mu}{\mathbf{X}}, \mathbf{\Sigma}{\mathbf{X}}\right)\left{\begin{aligned} \boldsymbol{\mu}{\mathbf{X}} &=\left[\begin{array}{c} 500 \ 500 \end{array}\right] \ \boldsymbol{\Sigma}{\mathbf{X}} &=\left[\begin{array}{cc} 150^{2} & 0.8 \cdot 150^{2} \ 0.8 \cdot 150^{2} & 150^{2} \end{array}\right] \end{aligned}\right.
If we observe that the resistance of the second beam $x_{2}=700 \mathrm{kN} \cdot \mathrm{m}$, we can employ conditional probabilities to estimate the PDF of the strength $X_{1}$, given the observation $x_{2}$,
$$f_{X_{1} \mid x_{2}}\left(x_{1} \mid x_{2}\right)=\mathcal{N}\left(x_{1} ; \mu_{1 \mid 2}, \sigma_{1 \mid 2}^{2}\right),$$
where
\begin{aligned} &\mu_{1 \mid 2}=500+0.8 \times 150 \frac{\overbrace{700}^{\text {observation }}-500}{150}=660 \mathrm{kN} \cdot \mathrm{m} \ &\sigma_{1 \mid 2}=150 \sqrt{1-0.8^{2}}=90 \mathrm{kN} \cdot \mathrm{m} . \end{aligned}
Figure $4.6$ presents the joint and conditional PDFs corresponding to this example. For the joint PDF, the highlighted pink slice corresponding to $x_{2}=700$ is proportional to the conditional probability $f_{X_{1} \mid x_{2}}\left(x_{1} \mid x_{2}=700\right)$. If we want to obtain the conditional distribution from the joint PDF, we have to divide it by the marginal PDF $f_{X_{2}}\left(x_{2}=700\right)$. This ensures that the conditional PDF for $x_{1}$ integrates to 1. This example is trivial, yet it sets the foundations for the more advanced models that will be presented in the following chapters.

## 计算机代写|机器学习代写machine learning代考|Sum of Normal Random Variable

Figure $4.7$ presents steel cables where each one is made from dozens of individual wires. Let us consider a cable made of 50 steel wires, each having a resistance $x_{i}: X_{i} \sim \mathcal{N}\left(x_{i} ; 10,3^{2}\right) \mathrm{kN}$. We use equation $4.2$ to compare the cable resistance $X_{\text {cable }}={ }{i=1}^{50} X{i}$ depending on the correlation coefficient $\rho_{i j}$. With the hypothesis
$\sum$

that $X_{i} \Perp X_{j} \Leftrightarrow \rho_{i j}=0$, all nondiagonal terms of the covariance $\operatorname{matrix}[\boldsymbol{\Sigma} \mathbf{x}]{i j}=0, \forall i \neq j$, which leads to $$X{\text {cable }} \sim \mathcal{N}(x ; 50 \times 10 \mathrm{kN}, \underbrace{2}{\sigma{X_{\text {chble }}=3 \sqrt{50} \approx 21 \mathrm{kN}}^{50 \times(3 \mathrm{kN})^{2}}} .$$
With the hypothesis $\rho_{i j}=1$, all terms in $\left[\Sigma_{\mathbf{x}}\right]{i j}=(3 \mathrm{kN})^{2}, \forall i, j$, so that $$X{\text {cable }} \sim \mathcal{N}(x ; 50 \times 10 \mathrm{kN}, \underbrace{}{\sigma{\mathrm{X}{\text {cable }}=3 \mathrm{kN} \times 50=150 \mathrm{kN}}^{50^{2} \times(3 \mathrm{kN})^{2}}} \text {. }$$ Figure $4.8$ presents the resulting PDFs for the cable resistance, given each hypothesis. These results show that if the uncertainty in the resistance for each wire is independent, there will be some cancellation; some wires will have a resistance above the mean, and some will have a resistance below. The resulting coefficient of variation for $\rho=0$ is $\delta{\text {cable }}=\frac{31}{500}=0.11$, which is approximately three times smaller than $\delta_{\text {wire }}=\frac{3}{10}=0.3$, the variability associated with each wire. In the opposite case, if the resistance is linearly correlated $(\rho=1)$, the uncertainty adds up as you increase the number of wires, so $\delta_{\text {cable }}=\frac{150}{500}=\delta_{\text {wire }}$.

## 计算机代写|机器学习代写machine learning代考|Univariate Log-Normal

The random variable $X \sim \ln \mathcal{N}(x ; \lambda, \zeta)$ is $\log$-normal if $\ln X \sim$ $\mathcal{N}\left(\ln x ; \lambda, \zeta^{2}\right)$ is Normal. Given the transformation function $x^{\prime}=$ $\ln x$, the change of variable rule presented in $\S 3.4$ requires that
$$\begin{gathered} \overbrace{f_{X},\left(x^{\prime}\right)}^{N\left(x^{\prime} ; \lambda, \zeta^{2}\right)} d x^{\prime}=f_{X}(x) d x \ f_{X^{\prime}}\left(x^{\prime}\right)\left|\frac{d x^{\prime}}{d x}\right|=\underbrace{f_{X}(x)}_{\ln \mathcal{N}(x ; \lambda, \zeta)}, \end{gathered}$$
where the derivative of $\ln x$ with respect to $\mathrm{x}$ is
$$\frac{d x^{\prime}}{d x}=\frac{d \ln x}{d x}=\frac{1}{x} .$$
Therefore, the analytic formulation for the log-normal PDF is given by the product of the transformation’s derivative and the Normal

PDF evaluated for $x^{\prime}=\ln x$,
\begin{aligned} f_{X}(x) &=\frac{1}{x} \cdot \mathcal{N}\left(\ln x ; \lambda, \zeta^{2}\right) \ &=\frac{1}{x} \cdot \frac{1}{\sqrt{2 \pi} \zeta} \exp \left(-\frac{1}{2}\left(\frac{\ln x-\lambda}{\zeta}\right)^{2}\right), \quad x>0 \end{aligned}
The univariate log-normal PDF is parameterized by the mean $\left(\mu_{\ln x}=\lambda\right)$ and variance $\left(\sigma_{\ln x}^{2}=\zeta^{2}\right)$ defined in the log-transformed space $(\ln x)$. The mean $\mu_{X}$ and variance $\sigma_{X}^{2}$ of the log-normal random variable can be transformed in the log-space using the relations
\begin{aligned} &\lambda=\mu_{\mathrm{m} \mathrm{n}}=\ln \mu_{X}-\frac{\zeta^{2}}{2} \ &\zeta=\sigma_{\ln X}=\sqrt{\ln \left(1+\left(\frac{\sigma_{X}}{\mu_{X}}\right)^{2}\right)}=\sqrt{\ln \left(1+\delta_{X}^{2}\right)} \end{aligned}
Note that for $\delta_{X}<0.3$, the standard deviation in the log-space is approximately equal to the coefficient of variation in the original space, $\zeta \approx \delta x$. Figure $4.9$ presents an example of log-normal PDF plotted (a) in the original space and (b) in the log-transformed space. The mean and standard deviation are $\left{\mu_{X}=2, \sigma_{X}=1\right}$ in the original space and ${\lambda=0.58, \zeta=0.47}$ in the log-transformed space.

## 计算机代写|机器学习代写machine learning代考|Conditional Distributions

，我们知道梁电阻与ρ12=0.8。之所以会出现这种相关性，是因为两根梁都是在同一家工厂使用相同的工艺制造的。该先验知识由联合二元正态 PDF 描述，
$$f_{X_{1} X_{2}}\left(x_{1}, x_{2}\right)=\mathcal{N}\left(\ mathbf{x} ; \boldsymbol{\mu}{\mathbf{X}}, \mathbf{\Sigma}{\mathbf{X}}\right)\left{ μX=[500 500] ΣX=[15020.8⋅1502 0.8⋅15021502]\正确的。 我F在和○bs和r在和吨H一个吨吨H和r和s一世s吨一个nC和○F吨H和s和C○ndb和一个米X2=700ķñ⋅米,在和C一个n和米pl○是C○nd一世吨一世○n一个lpr○b一个b一世l一世吨一世和s吨○和s吨一世米一个吨和吨H和磷DF○F吨H和s吨r和nG吨HX1,G一世在和n吨H和○bs和r在一个吨一世○nX2, f_{X_{1} \mid x_{2}}\left(x_{1} \mid x_{2}\right)=\mathcal{N}\left(x_{1} ; \mu_{1 \mid 2 }, \sigma_{1 \mid 2}^{2}\right), 在H和r和 μ1∣2=500+0.8×150700⏞观察 −500150=660ķñ⋅米 σ1∣2=1501−0.82=90ķñ⋅米.$$

## 计算机代写|机器学习代写machine learning代考|Sum of Normal Random Variable

X电缆 ∼ñ(X;50×10ķñ,2⏟σXchble =350≈21ķñ50×(3ķñ)2.

X电缆 ∼ñ(X;50×10ķñ,⏟σX电缆 =3ķñ×50=150ķñ502×(3ķñ)2. 数字4.8给出了给定每个假设的电缆电阻的结果 PDF。这些结果表明，如果每根导线的电阻不确定性是独立的，就会有一些抵消；有些电线的电阻高于平均值，有些电线的电阻低于平均值。由此产生的变异系数ρ=0是d电缆 =31500=0.11, 大约比d金属丝 =310=0.3，与每根电线相关的可变性。在相反的情况下，如果电阻是线性相关的(ρ=1)，不确定性会随着电线数量的增加而增加，所以d电缆 =150500=d金属丝 .

## 计算机代写|机器学习代写machine learning代考|Univariate Log-Normal

FX,(X′)⏞ñ(X′;λ,G2)dX′=FX(X)dX FX′(X′)|dX′dX|=FX(X)⏟ln⁡ñ(X;λ,G),

dX′dX=dln⁡XdX=1X.

PDF 评估为X′=ln⁡X,

FX(X)=1X⋅ñ(ln⁡X;λ,G2) =1X⋅12圆周率G经验⁡(−12(ln⁡X−λG)2),X>0

λ=μ米n=ln⁡μX−G22 G=σln⁡X=ln⁡(1+(σXμX)2)=ln⁡(1+dX2)

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