Tag Archives: 格拉斯哥大学代写

高级衍射几何学与拓扑学 5M:Adv Diff Geom & Topology MATHS5039_1 5M

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高级衍射几何学与拓扑学 5M:Adv Diff Geom & Topology MATHS5039_1 5M

Proof of (b) is easy:
$$
d\left(y_{n}, x\right) \leq d\left(y_{n}, x_{n}\right)+d\left(x_{n}, x\right) \rightarrow 0 .
$$
(c) is also easy. Look at
$$
d\left(x_{n}, y_{n}\right) \leq d\left(x_{n}, x_{n}^{\prime}\right)+d\left(x_{n}^{\prime}, y_{n}^{\prime}\right)+d\left(y_{n}^{\prime}, y_{n}\right) .
$$
Taking limits, we get $\lim {n} d\left(x{n}, y_{n}\right) \leq \lim {n} d\left(x{n}^{\prime}, y_{n}^{\prime}\right)$. Similar argument shows the other way inequality and hence the proof.


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MATHS5039_1 COURSE NOTES ：

For, if $\varepsilon>0$ is given, by the Cauchy nature of $\left(x_{n}\right)$, there exists $N \in \mathbb{N}$ such that $d\left(x_{m}, x_{n}\right)<\varepsilon$ for $m, n \geq N$. We have, for $n \geq N$,
$$
d\left(\tilde{x}{n}, \xi\right):=\lim {m \rightarrow \infty} d\left(x_{n}, x_{m}\right)<\varepsilon .
$$
It is clear that the map $\varphi: X \rightarrow \tilde{X}$ given by $\varphi(x)=\tilde{x}$ is an isometry. $d(\varphi(x), \varphi(y)):=\lim _{n} d(x, y)$, the limit of a constant sequence.
We claim that the image $\varphi(X)$ of $X$ under the map $\varphi: x \mapsto \tilde{x}$ is dense









数学 MATH MATHS2025_1/MATHS3016_1

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数学 MATH MATHS2025_1/MATHS3016_1

Let $f$ be a function from a nonempty open subset $E$ of $\mathbb{R}$ to $\mathbb{R}$. The function $f$ is said to be differentiable at $c \in E$ if
$$
\lim {x \rightarrow c} \frac{f(x)-f(c)}{x-c} $$ or, equivalently, $$ \lim {h \rightarrow 0} \frac{f(c+h)-f(c)}{h}
$$
exists. This limit (if it exists) is called the derivative of $f$ at $c$. If the derivative of $f$ exists at every $c \in E$, then $f$ is said to be differentiable on $E$ (or just differentiable). The derivative of $f$ as a function from $E$ to $\mathbb{R}$ is denoted by
$$
f^{\prime} \text { or } \frac{d f}{d x}
$$
Note that the limit in Eq. (7.1) is understood as the limit of the function
$$
g(x)=\frac{f(x)-f(c)}{x-c}, \quad x \in E \backslash{c}
$$


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MATHS2025_1/MATHS3016_1COURSE NOTES ：

From the definition of $f^{\prime}(c)$, it follows that for every $\varepsilon>0$, there exists $\delta>0$ such that $x \in E,|x-c|<\delta$, and $x \neq c$ imply
$$
\left|\frac{f(x)-f(c)}{x-c}-f^{\prime}(c)\right|<\varepsilon .
$$
Thus, for every $x \in E$ with $|x-c|<\delta$,
$$
|f(x)-\varphi(x)| \leq \varepsilon|x-c|,
$$
where $\varphi$ is the linear function defined by
$$
\varphi(x)=f(c)+f^{\prime}(c)(x-c), \quad x \in \mathbb{R}
$$









复杂分析的方法 Methods in Complex Analysis MATHS4076_1

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复杂分析的方法 Methods in Complex Analysis MATHS4076_1

To find the best probability assignment subject to the constraint of given mean, we use the MaxEnt method. I.e. maximse
$$
S[p]=-k_{\mathrm{B}} \sum_{k \geq 0} p(k) \ln p(k)
$$
subject to the constraint
$$
\sum_{k \geq 0} k p(k)=\mu .
$$
using the method of Lagrangian multipliers to take constraints into account.
To simplify notation we can assume $k_{\mathrm{B}}=1$; this amounts to redefining Lagrangian multipliers in units of $k_{\mathrm{B}}$ (can you see why?). Thus define
$$
\mathcal{L}[p]=-\sum_{k \geq 0} p(k) \ln p(k)+\lambda_{0}\left(\sum_{k \geq 0} p(k)-1\right)+\lambda_{1}\left(\sum_{k \geq 0} k p(k)-\mu\right)
$$


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MATHS4076_1 COURSE NOTES ：

$$
\mu=\sum_{k \geq 0} k p(k)=\left(1-\mathrm{e}^{\lambda_{1}}\right) \sum_{k=1}^{\infty} k \mathrm{e}^{\lambda_{1} k}=\frac{\mathrm{e}^{\lambda_{1}}}{1-\mathrm{e}^{\lambda_{1}}}
$$
required by the constraint of the given mean. Hence
$$
\mathrm{e}^{\lambda_{1}}=\frac{\mu}{1+\mu} \quad \Longrightarrow \quad p(k)=\frac{1}{1+\mu}\left(\frac{\mu}{1+\mu}\right)^{k}
$$
1









随机过程 Stochastic Processes STATS4024_1 /STATS5026_1

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随机过程 Stochastic Processes STATS4024_1 /STATS5026_1

from which it follows that, by monotonicity,
$$
\forall n \in \mathbb{N}: \quad E\left[Y_{n+1} \mid X=x\right] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }
$$
Moreover,
$$
\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y_{n} d P=\int_{B} E\left[Y_{n} \mid X=x\right] d P_{X}(x)
$$
and
$$
\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y d P \geq \int_{[X \in B]} Y_{n} d P
$$
Thus
$$
E[Y \mid X=x] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }
$$


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STATS4024_1 /STATS5026_1 COURSE NOTES ：

Therefore,
$$
P([Y \in A] \mid \cdot)=P([Y \in A]), \quad P_{X} \text {-a.s. }
$$
and if $Y$ is a real-valued integrable random variable, then
$$
E[Y \mid \cdot]=E[Y], \quad P_{X}-a . s .
$$
Proof: Independence of $X$ and $Y$ is equivalent to
$$
P([X \in B] \cap[Y \in A])=P([X \in B]) P([Y \in A]) \quad \forall A \in \mathcal{B}{1}, B \in \mathcal{B} $$ or $$ \begin{aligned} \int{[X \in B]} I_{[Y \in A]}(\omega) P(d \omega) &=P([Y \in A]) \int I_{B}(x) d P_{X}(x) \
&=\int_{D} P([Y \in A]) d P_{X}(x)
\end{aligned}
$$









进一步的复杂分析 Further Complex Analysis MATHS5070_1 /MATHS4104_1

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进一步的复杂分析 Further Complex Analysis MATHS5070_1 /MATHS4104_1

$$
l_{\mathrm{p}}(\psi)=c-\frac{n}{2} \log (\mathrm{SSB}+\lambda \mathrm{SSW})+\frac{n-m}{2} \log \lambda,
$$
where $\lambda=1+k \psi$. The maximum of $l_{\mathrm{p}}$ is given by
$$
\hat{\lambda}=\left(1-\frac{1}{m}\right) \frac{\mathrm{MSB}}{\mathrm{MSW}},
$$
where MSB $=\operatorname{SSB} /(m-1)$ and MSW $=\operatorname{SSW} /(n-m)$, or
$$

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MATHS5070_1 /MATHS4104_1 COURSE NOTES ：

$$
\hat{F}{r}(x)=\frac{1}{m{r}} \sum_{u=1}^{m_{r}} 1_{\left(\hat{\alpha}{r}, u \leq x\right)} \stackrel{P}{\longrightarrow} F{r}(x), \quad x \in C\left(F_{r}\right),
$$
where $\hat{\alpha}{r, u}$ is the $u$ th component of $\hat{\alpha}{r}, 1 \leq r \leq s$, and
$$
\hat{F}{0}(x)=\frac{1}{n} \sum{u=1}^{n} 1_{\left(\tilde{\epsilon}{u} \leq x\right)} \stackrel{P}{\longrightarrow} F{0}(x), \quad x \in C\left(F_{0}\right),
$$









线性混合模型 Linear Mixed Models STATS5054_1/STATS4045_1

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线性混合模型 Linear Mixed Models STATS5054_1/STATS4045_1

The methods discussed above for the full parameter vector now directly carry over to calculate local influences on the geometric surface defined by $\mathrm{LD}{1}(\omega)$. We now partition $\ddot{L}$ as $$ \ddot{L}=\left(\begin{array}{ll} \ddot{L}{11} & \ddot{L}{12} \ \ddot{L}{21} & \ddot{L}{22} \end{array}\right), $$ according to the dimensions of $\theta{1}$ and $\theta_{2}$. Cook (1986) has then shown that the local influence on the estimation of $\theta_{1}$, of perturbing the model in the direction of a normalized vector $\boldsymbol{h}$, is given by
$$
C_{h}\left(\boldsymbol{\theta}{1}\right)=2\left|\boldsymbol{h}^{\prime} \Delta^{\prime}\left[\ddot{L}^{-1}-\left(\begin{array}{cc} 0 & 0 \ 0 & \ddot{L}{22}^{-1}
\end{array}\right)\right] \Delta \boldsymbol{h}\right|
$$
Because all eigenvalues of the malrix
$$
\left(\begin{array}{ll}
\ddot{L}{11} & \ddot{L}{12} \
\ddot{L}{21} & \ddot{L}{22}
\end{array}\right)\left(\begin{array}{cc}
0 & 0 \
0 & \ddot{L}{22}^{-1} \end{array}\right)=\left(\begin{array}{cc} 0 & \ddot{L}{12} \ddot{L}_{22}^{-1} \
0 & I
\end{array}\right)
$$

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MATHS5077_1 COURSE NOTES ：

A measure of influence, proposed , is then
$$
\rho_{i}=-\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)^{\prime} \ddot{L}\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)
$$$$
\boldsymbol{\Delta}{i}=-\sum{j \neq i} \boldsymbol{\Delta}{j}=\ddot{L}{(i)}(\widehat{\boldsymbol{\theta}})\left(\hat{\boldsymbol{\theta}}{(i)}^{1}-\widehat{\boldsymbol{\theta}}\right) $$ such that expression (11.4) becomes $$ C{i}=-2\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)^{\prime} \ddot{L}{(i)} \ddot{L}^{-1} \ddot{L}{(i)}\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)
$$









代数专题 : Topics in Algebra MATHS5077_1

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代数专题 : Topics in Algebra MATHS5077_1

Although the inverse of a homomorphism may not exist, a homomorphism does share some properties in common with isomorphisms. For example, if $\theta: G \rightarrow$ $G^{\prime}$ is a homomorphism, and if $e$ and $e^{\prime}$ are the identities in $G$ and $G^{\prime}$, respectively, then $\theta(e)=\theta\left(e^{2}\right)=\theta(e) \theta(e)$, so that $\theta(e)=e^{\prime}$. Thus a homomorphism maps the identity in $G$ to the identity in $G^{\prime}$. Similarly, as $e^{\prime}=\theta(e)=\theta\left(g g^{-1}\right)=$ $\theta(g) \theta\left(g^{-1}\right)$, we see that for all $g$ in $G$
$$
\theta(g)^{-1}=\theta\left(g^{-1}\right) .
$$
The most obvious example of a homomorphism is a linear map between vector spaces. Indeed, a vector space is a group with respect to addition, and as any linear map $\alpha: V \rightarrow W$ satisfies
$$
\alpha(u+v)=\alpha(u)+\alpha(v)
$$

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MATHS5077_1 COURSE NOTES ：

Proof First, every element of $x_{0} K$ is a solution to the given equation, for the general element of $x_{0} K$ is $x_{0} k$, where $k \in K$, and $\theta\left(x_{0} k\right)=\theta\left(x_{0}\right) \theta(k)=y e^{\prime}=$ $y$. Next suppose that $\theta(x)=y$, and consider the element $x_{0}^{-1} x$ in $G$. Then
$$
\theta\left(x_{0}^{-1} x\right)=\theta\left(x_{0}^{-1}\right) \theta(x)=\left(\theta\left(x_{0}\right)\right)^{-1} y=y^{-1} y=e^{\prime},
$$
so that $x_{0}^{-1} x \in K$. This means that $x \in x_{0} K$ and so the set of solutions of $\theta(x)=y$ is the coset $x_{0} K$.









通用线性模型 Generalised Linear Models STATS4043_1

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通用线性模型 Generalised Linear Models STATS4043_1

warning message, Clearly missing value cannot be allowed in certain contexts and wil1 be faulted, for instance an array cannot be given a shape containing a wissing value. In order to allow the user to detect missing values and replace them, if required, three special functions are supplied;
$$
\begin{aligned}
\text { \&EQMN }(X) &=1 \text { (true) if } x=* \
&=0 \text { (false) otherwise } \
\text { xMYV }(X ; Y) &=* \text { if } y=1 \quad \text { (true) } \
&=x \text { if } y=0 \quad \text { (false) } \
\not{\gamma V M}(X ; Y) &=x \text { if } x \neq * \
&=y \text { if } x=*
\end{aligned}
$$

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STATS4043_1 COURSE NOTES ：

The likelihood function can be taken to be
$$
L(u)=\exp \left(-\frac{1}{2} \frac{\Sigma\left(y_{i}-\mu\right)^{2}}{a^{2}}\right)
$$
with $10 g-1$ ikelihood function
$$
\ell(\mu)=-\frac{1}{2} \frac{\varepsilon\left(\gamma_{i}-\mu\right)^{2}}{\sigma^{2}}
$$
Following through the usual maximam 1 ikelihood calculations, we have
$$
\begin{aligned}
\mathbb{E}^{\prime}(u) &=E\left(y_{i}-\mu\right) / d^{2} \
Q^{*}(\mu) &=n / o^{2}
\end{aligned}
$$









数学物理 Mathematical Physics MATHS5073_1 /MATHS4107_1

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数学物理 Mathematical Physics MATHS5073_1 /MATHS4107_1

Now approaching this problem using vector notation, we note that the torque about the fulcrum due to $F_{1}$ is $\tau_{1}=r_{1} \times F_{1}$. Here $r_{1}$ is a vector from the fulcrum to the point of application of $\mathbf{F}{1}$, and $$ \tau{1}=\mathbf{r}{1} \times \mathbf{F}{1}=r_{1} F_{1} \sin \theta_{1} \hat{\mathbf{n}}=d_{1} F_{1} \hat{\mathbf{n}}
$$
where $\hat{\mathbf{n}}$ is a unit vector normal to, and out of the plane of the figure. A similar analysis of the torque $\tau_{2}$ reveals that it is
$$
\tau_{2}=\mathbf{r}{2} \times \mathbf{F}{2}=-r_{2} F_{2} \sin \theta_{2} \hat{\mathbf{n}}=-d_{2} F_{2} \hat{\mathbf{n}}
$$
the minus sign occurring because $\hat{\mathbf{n}}$ has the same meaning as before but this torque is directed into the plane of the figure. We now observe that $\theta_{2}=\pi-\theta_{1}$, and therefore $\sin \theta_{2}=\sin \theta_{1}$. At equilibrium the sum of these torques is zero.

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MATHS5073_1 /MATHS4107_1 COURSE NOTES ：

Sometimes the determination of the gradient is straightforward. Given
$$
f(x, y, z)=\frac{x y}{z}
$$
we easily find
$$
\frac{\partial f}{\partial x}=\frac{y}{z}, \quad \frac{\partial f}{\partial y}=\frac{x}{z}, \quad \frac{\partial f}{\partial z}=-\frac{x y}{z^{2}}
$$
leading to $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$









弹性回归 Flexible Regression STATS5052_1/STATS4040_1

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弹性回归 Flexible Regression STATS5052_1/STATS4040_1

where $G_{0}$ is a standard exponential distribution. The distribution of $\epsilon_{t}$ is therefore
$$
f_{v, \xi}\left(\epsilon_{t}\right)=\sum_{j=1}^{\infty} w_{j} v\left(-\xi_{j} e^{-\lambda}, \xi_{j} e^{\lambda}\right)
$$
and the conditional return distribution is
$$
f_{G, \lambda}\left(\gamma_{t} \mid \sigma_{t}\right)=\sum_{j=1}^{\infty} w_{j} v\left(\gamma_{t} \mid-\xi_{j} \sigma_{t} e^{-\lambda}, \xi_{j} \sigma_{t} e^{\lambda}\right)
$$

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STATS5052_1/STATS4040_1 COURSE NOTES ：

Utilising the latent variable, the copula yields
$$
P\left(Y_{1}=0, Y_{2} \leq Y_{2}\right)=P\left(Y_{1}^{} \leq 0, Y_{2} \leq Y_{2}\right)=C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)
$$
and
$$
P\left(Y_{1}=1, Y_{2} \leq \gamma_{2}\right)=P\left(Y_{1}^{}>0, Y_{2} \leq \gamma_{2}\right)=F_{2}\left(\gamma_{2}\right)-C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right) \text {, }
$$
leading to the mixed binary-continuous density
$$
p\left(\gamma_{1}, \gamma_{2}\right)=\left(\frac{\partial C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)}{\partial F_{2}\left(\gamma_{2}\right)}\right)^{1-\gamma_{1}} \cdot\left(1-\frac{\partial C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)}{\partial F_{2}\left(\gamma_{2}\right)}\right)^{\gamma_{1}} \cdot p_{2}\left(\gamma_{2}\right),
$$