广义线性模型 | Generalized Linear Models代写 STAT 504代考

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这是一份psu宾夕法尼亚州立大学 STAT 504作业代写的成功案

(广义) 线性楛梨 | (Generalized) Linear Models代写 STAT 504代考
问题 1.

$$
f(y ; \theta)=\exp [y \theta-b(\theta)+c(y)]
$$
the conjugate distribution for the random parameter is
$$
p(\theta ; \zeta, \gamma)=\exp [\zeta \theta-\gamma b(\theta)+s(\zeta, \gamma)]
$$


证明 .

where $s(\zeta, \gamma)$ is a term not involving $\theta$. This conjugate is also a member of the exponential family. The resulting compound distribution, for $n$ observations, is
$$
f(y ; \zeta, \gamma)=\exp [s(\zeta, \gamma)+c(y)-s(\zeta+y, \gamma+n)]
$$
This is not generally a member of the exponential family.

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STAT 504 COURSE NOTES :


$$
y=K \frac{\alpha \mathrm{e}^{\beta t}}{1+\alpha \mathrm{e}^{\beta t}}
$$
where $K$ is the asymptotic maximum value of the response.
We can transform this to a linear structure by using a logit link:
$$
\log \left(\frac{y}{K-y}\right)=\log (\alpha)+\beta t
$$




统计计算| Statistical Computing代写 MATH6173代考

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这是一份southampton南安普敦大学 MATH6173作业代写的成功案

统计计算| Statistical Computing代写 MATH6173代考
问题 1.

We estimate $\beta$ by minimizing the penalized least squares criterion
$$
H(\beta)=(\mathbf{y}-\mathbf{H} \beta)^{T}(\mathbf{y}-\mathbf{H} \beta)+\lambda|\beta|^{2} .
$$
The solution is
$$
\hat{\mathbf{y}}=\mathbf{H} \hat{\beta}
$$


证明 .

with $\hat{\beta}$ determined by
$$
-\mathbf{H}^{T}(\mathbf{y}-\mathbf{H} \hat{\beta})+\lambda \hat{\beta}=0
$$
From this it appears that we need to evaluate the $M \times M$ matrix of inner products in the transformed space. However, we can premultiply by $\mathbf{H}$ to give
$$
\mathbf{H} \hat{\beta}=\left(\mathbf{H H}^{T}+\lambda \mathbf{I}\right)^{-1} \mathbf{H H}^{T} \mathbf{y}
$$

.

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MATH6173 COURSE NOTES :


$$
P(G, X)=\sum_{r=1}^{R} \pi_{r} P_{r}(G, X)
$$
a mixture of joint densities. Furthermore we assume
$$
P_{r}(G, X)=P_{r}(G) \phi\left(X ; \mu_{r}, \Sigma\right) .
$$




统计推断| Statistical Inference代写 MATH10028代考

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这是一份drps爱丁堡大学MATH10028作业代写的成功案

统计推断| Statistical Inference代写 MATH10028代考
问题 1.

\begin{aligned}
&Y_{1}=\frac{1}{\sqrt{k}} Z_{1}+\frac{1}{\sqrt{k}} Z_{2}+\cdots+\frac{1}{\sqrt{k}} Z_{k} \
&Y_{2}=\frac{1}{\sqrt{2 \times 1}} Z_{1}-\frac{1}{\sqrt{2 \times 1}} Z_{2}
\end{aligned}


证明 .

The variance of a r.v. $X$ is denoted by $\operatorname{Var}(X)$ and is defined by:
$$
\operatorname{Var}(X)=E(X-E X)^{2} \text {. }
$$
The explicit expression of the right-hand side in (8) is taken from (6) for $g(x)=(x-E X)^{2}$. The alternative notations $\sigma^{2}(X)$ and $\sigma_{X}^{2}$ are also often used for the $\operatorname{Var}(X)$.

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MATH10028 COURSE NOTES :


F_{n}(x)=\left{\begin{array}{ll}
0, & \text { if } x<1-\frac{1}{n} \
\frac{1}{2}, & \text { if } 1-\frac{1}{n} \leq x<1+\frac{1}{n}, \
1, & \text { if } x \geq 1+\frac{1}{n}
\end{array} \quad G(x)= \begin{cases}0, & \text { if } x<1 \
1, & \text { if } x \geq 1\end{cases}\right.




多元统计分析| Multivariate Statistical Analysis 代写 STAT 505代考

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这是一份PSU宾夕法尼亚州立大学STAT 505作业代写的成功案

多元统计分析| Multivariate Statistical Analysis 代写 STAT 505代考
问题 1.

\begin{aligned}
&\text { Wilks’ lambda }=\prod_{i=1}^{f} \frac{1}{1+\tau_{i}}=\frac{|\mathbf{E}|}{|\mathbf{E}+\mathbf{H}|} \
&\text { Piliai’stracc }=\sum_{i=1}^{s} \frac{\eta_{1}}{1+\tau_{i}}=\operatorname{tr}\left[\mathbf{H}(\mathbf{H}+\mathbf{E})^{-1}\right]
\end{aligned}


证明 .

\begin{gathered}
\text { Hotelling-Lawley trace }=\sum_{i=1}^{\pi} \eta_{i}=\mathrm{tr}{\left[\mathrm{HE}^{-1}\right]} \ \text { Roy’s greatest root }=\frac{\eta{1}}{1+\eta_{1}}
\end{gathered}

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STAT 505COURSE NOTES :


\begin{array}{ll}
\lambda_{1}=5.83, & \mathbf{e}{1}^{\prime}=[383,-924,0] \ \lambda{2}=2.00, & \mathbf{e}{2}^{\prime}=[0,0,1] \ \lambda{3}=0.17, & \mathbf{e}_{3}^{\prime}=[.924, .383,0]
\end{array}




方差分析和实验设计| Analysis of Variance and Design of Experiments代写 STAT 502代考

0

这是一份PSU宾夕法尼亚州立大学STAT 502作业代写的成功案

方差分析和实验设计| Analysis of Variance and Design of Experiments代写 STAT 502代考
问题 1.

The estimates use the $c_{i k}$ and $C_{k}$ values in Table 10.4:
$$
\begin{aligned}
&\hat{a}{1}^{}=75 /\left[(-5)^{2}+(-3)^{2}+\cdots+(5)^{2}\right]=75 / 70=1.071 \ &\hat{a}{3}^{}=-31 /\left[(-5)^{2}+(7)^{2}+\cdots+(5)^{2}\right]=-31 / 180=-.172
\end{aligned}
$$


证明 .

The estimate of $\mu$ (from Table $10.1$ ) is $\bar{X}{. .}=14.3$, and the estimation formula is $$ \hat{\mu}{i}=14.3+1.071 c_{i 1}-.172 c_{i 3} .
$$
From this equation, we estimate $\mu_{1}$ by inserting $c_{11}=-5$ and $c_{13}=-5$ to obtain
$$
\hat{\mu}_{1}=14.3+1.071(-5)-.172(-5)=9.8
$$

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STAT 502 COURSE NOTES :


$$
S S_{a \text { linear }(b)}=3,698.0+6,160.5+5,304.5-15,000=163.0 \text {. }
$$
We divide this by its degrees of freedom to obtain
$$
M S_{a \text { linear }(b)}=163 / 2=81.5 .
$$
Finally, we divide this by $M S_{w}$ to obtain the $F$ ratio
$$
F_{(2,108)}=81.5 / 40=2.04
$$