统计学模型|MA50260 Applied stochastic differential equations代写

0

Write the relevant mathematical arguments in a premake inferences on the basis of a fitted model and recognise the assumptions underlying these inferences and possible limitations to their accuracy.

这是一份Bath巴斯大学MA50260作业代写的成功案

统计学模型|MA50260 Applied stochastic differential equations代写

$$
Q(p)=\lambda+(\eta / 2)[(1+\delta) S(p)-(1-\delta) S(1-p)],-1 \leq \delta \leq 1
$$
The distribution lies in the range of $(\lambda-a(\eta / 2)(1-\delta), \lambda+a(\eta / 2)(1+\delta))$, which for $a=\infty$ is $(-\infty, \infty)$. The quantile density, using $s(p)$ for the quantile density of $S(p)$, is
$$
q(p)=(\eta / 2)[(1+\delta) s(p)+(1-\delta) s(1-p)]
$$
A study of the quantile statistics of this distribution will show why this form has been adopted. Notice that using $R$ to denote the reflected distribution we have that $M_{R}=-M_{S}, U Q_{R}=-L Q_{S}$ and $L Q_{R}=-U Q_{S}$. Substituting $p=0.5$ gives
$$
\begin{aligned}
M=\lambda+(\eta / 2) & {\left[(1+\delta) M_{S}+(1-\delta) M_{R}\right] } \
=& \lambda+\eta \delta M_{\delta}
\end{aligned}
$$


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MA50260 COURSE NOTES :

value, $x$, which for each one is $p$. This probability is, by the multiplication law of probability, $p^{n}$. Hence
$$
p_{(n)}=p^{n} \text { so } p=p_{(n)}^{1 / n} \text { and } F(x)=p_{(n)}^{1 / n}
$$
Finally, therefore, inverting the CDF, $F(x)$, to get the quantile function, we have the equivalent statements that the $p_{(n)}$ quantile of the largest value is given by both $Q_{(n)}\left(p_{(n)}\right)$ and $Q\left(p_{(n)}^{1 / n}\right)$.
Hence
$$
Q_{(n)}\left(p_{(n)}\right)=Q\left(p_{(n)}^{1 / n}\right) .
$$
The quantile function of the largest observation is thus found from the original quantile function in the simplest of calculations.



椭圆的数值解法|MA50251 Applied stochastic differential equations代写

0

To perform simple cWrite the relevant mathematical arguments in a precise and lucid fashion.

这是一份Bath巴斯大学MA50251作业代写的成功案

椭圆的数值解法|MA50170 Numerical solution of elliptic PDEs代写

For a nonnegative integer $m$, we denote by $C^{m}(\mathcal{O})$ the set of all $m$-times continuously differentiable real-valued functions in $\mathcal{O}$, and by $C_{0}^{m}(\mathcal{O})$ the subspace of $C^{m}(\mathcal{O})$ consisting of those functions which have compact supports in O. For $u \in C^{m}(\mathcal{O})$ and $1 \leq p<\infty$, we define
$$
|u|_{m, p}=\left(\int_{\mathcal{O}} \sum_{|\alpha| \leq m}\left|D^{\alpha} u(x)\right|^{p} d x\right)^{1 / p}
$$
and for $p=2, u, v \in C^{m}(\mathcal{O})$,
$$
\langle u, v\rangle_{m, 2}=\int_{\mathcal{O}} \sum_{|\alpha| \leq m} D^{\alpha} u(x) \cdot D^{\alpha} v(x) d x .
$$


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MA50251 COURSE NOTES :

We start our discussion by considering the following linear system on the $n$-dimensional Euclidean space $\mathbf{R}^{n}$ :
$$
\frac{d X_{t}\left(x_{0}\right)}{d t}=A X_{t}\left(x_{0}\right), \quad X_{0}\left(x_{0}\right)=x_{0} \in \mathbf{R}^{n},
$$
where $A$ is some $n \times n$ constant matrix. Clearly, the equation has a unique solution which is given by
$$
X_{t}\left(x_{0}\right)=e^{A t} x_{0}
$$



椭圆的数值解法|MA50170 Numerical solution of elliptic PDEs代写

0

To perform simple calculations to compute certain quantities relating to Brownian motion, and to understand how these quantities can be important in pricing financial derivatives

这是一份Bath巴斯大学MA50170作业代写的成功案

椭圆的数值解法|MA50170 Numerical solution of elliptic PDEs代写

by $\S$ 4. Apply successive point-overrelaxation (point SOR) to, with the particular (over)relaxation factor
$$
\text { (12) } \omega_{b}=2 /\left(1+\sqrt{1-\mu^{2}}\right)=1+\left[\mu /\left(1+\sqrt{1-\mu^{2}}\right)\right]^{2} \text {. }
$$
Kahan has proved that for this $\omega_{b}$,
$$
\omega_{b}-1 \leqq \rho\left(L_{\omega_{b}}\right) \leqq \sqrt{\omega_{b}-1}
$$
hence this $\omega_{b}$ is a good relaxation factor, since $\rho\left(L_{\omega}\right) \geqq \omega_{b}-1$ for any relaxation factor. For $\rho(B)=1-\varepsilon$, where $\varepsilon$ is small, the asymptotic convergence rate $\gamma=-\log \rho\left(L_{\omega_{\Delta}}\right)$ therefore satisfies
$$
\sqrt{2 \varepsilon}=-\frac{1}{2} \log \left(\omega_{b}-1\right) \leqq \gamma \leqq \log \left(\omega_{b}-1\right)=2 \sqrt{2 \varepsilon}
$$


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MA50170 COURSE NOTES :

has the desired property. This was shown by L. F. Richardson, Lanczos , and Stiefel. (Here $C_{m}(t)=\cos \left(m \cos ^{-1} t\right)$ on $(-1,1)$.) Moreover, from classic recursion formulas for the Chebyshev polynomials, it follows that where the $r$ th relaxation factor is
$$
\omega_{r}=1+C_{r-2}(1 / \rho) / C_{r}(1 / \rho), \quad \rho=\rho(B)
$$
Furthermore, as was first observed by Golub and Varga
$$
\lim {r \rightarrow \infty} \omega{r}=\omega_{b}=2 /\left[1+\left(1-\rho^{2}(B)\right)^{1 / 2}\right]
$$



随机过程与金融|MA50089 Stochastic processes & finance代写

0

To perform simple calculations to compute certain quantities relating to Brownian motion, and to understand how these quantities can be important in pricing financial derivatives

这是一份Bath巴斯大学MA50089作业代写的成功案

随机过程与金融|MA50089 Stochastic processes & finance代写

In the univariate two-way model, we measure one dependent variable $y$ on each experimental unit. The balanced two-way fixed-effects model with factors $A$ and $B$ is
$$
\begin{aligned}
y_{i j k} &=\mu+\alpha_{i}+\beta_{j}+\gamma_{i j}+\varepsilon_{i j k} \
&=\mu_{i j}+\varepsilon_{i j k}, \
i &=1,2, \ldots, a, \quad j=1,2, \ldots, b, \quad k=1,2, \ldots, n
\end{aligned}
$$
where $\alpha_{i}$ is the effect (on $y_{i j k}$ ) of the $i$ th level of $A, \beta_{j}$ is the effect of the $j$ th level of $B, \gamma_{i j}$ is the corresponding interaction effect, and $\mu_{i j}$ is the population mean for the $i$ th level of $A$ and the $j$ th level of $B$. In order to obtain $F$-tests, we further assume that the $\varepsilon_{i j k}$ ‘s are independently distributed as $N\left(0, \sigma^{2}\right)$.

Let $\bar{\mu}{i .}=\sum{j} \mu_{i j} / b$ be the mean at the $i$ th level of $A$ and define $\bar{\mu}{. j}$ and $\bar{\mu}{. .}$ similarly. Then if we use side conditions $\sum_{i} \alpha_{i}=\sum_{j} \beta_{j}=\sum_{i} \gamma_{i j}=\sum_{j} \gamma_{j}=0$,


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MA50089 COURSE NOTES :

0 can be tested by $T^{2}$ or any of the four MANOVA test statistics. To test $H_{0}: \sum_{i} c_{i} \overline{\boldsymbol{\mu}}{i}=\mathbf{0}$, for example, we can use $$ T^{2}=\frac{n b}{\sum{i=1}^{a} c_{i}^{2}}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)^{\prime}\left(\frac{\mathbf{E}}{v{E}}\right)^{-1}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right) $$ which is distributed as $T{p, v_{E}}^{2}$ when $H_{0}$ is true. Alternatively, the hypothesis matrix
$$
\mathbf{H}{1}=\frac{n b}{\sum{i=1}^{a} c_{i}^{2}}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)\left(\sum{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)^{\prime} $$ can be used in $$ \Lambda=\frac{|\mathbf{E}|}{\left|\mathbf{E}+\mathbf{H}{1}\right|}
$$



希尔伯特空间的分析|MA40256 Analysis in Hilbert spaces代写

0

Review of Lagrangian and Eulerian descriptions: Jacobian, Euler’s identity and Reynold’s transport theorem. The continuity equation and incompressibility condition.

这是一份Bath巴斯大学MA40256作业代写的成功案

粘性流体动力学 |MA40255 Viscous fluid dynamics代写


Let $\tilde{Q}{j}$ denote a cube strictly contained in $Q{j}$ such that $\left|Q_{j}\right| \leq\left|\tilde{Q}{j}\right|+$ $\epsilon / 2^{j}$, where $\epsilon$ is arbitrary but fixed. Then, for every $N$, the cubes $\tilde{Q}{1}, \tilde{Q}{2}, \ldots, \tilde{Q}{N}$ are disjoint, hence at a finite distance from one another, and repeated applications of Observation 4 imply
$$
m_{}\left(\bigcup_{j=1}^{N} \tilde{Q}{j}\right)=\sum{j=1}^{N}\left|\tilde{Q}{j}\right| \geq \sum{j=1}^{N}\left(\left|Q_{j}\right|-\epsilon / 2^{j}\right) .
$$
Since $\bigcup_{j=1}^{N} \tilde{Q}{j} \subset E$, we conclude that for every integer $N$, $$ m{}(E) \geq \sum_{j=1}^{N}\left|Q_{j}\right|-\epsilon
$$
In the limit as $N$ tends to infinity we deduce $\sum_{j=1}^{\infty}\left|Q_{j}\right| \leq m_{}(E)+\epsilon$ for every $\epsilon>0$, hence $\sum_{j=1}^{\infty}\left|Q_{j}\right| \leq m_{}(E)$. Therefore, combined with Observation 2, our result proves that we have equality.


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MA40256 COURSE NOTES :

So, suppose $F$ is compact (so that in particular $m_{}(F)<\infty$ ), and let $\epsilon>0$. By Observation 3 we can select an open set $\mathcal{O}$ with $F \subset \mathcal{O}$ and $m_{}(\mathcal{O}) \leq m_{}(F)+\epsilon$. Since $F$ is closed, the difference $\mathcal{O}-F$ is open, and by Theorem $1.4$ we may write this difference as a countable union of almost disjoint cubes $$ \mathcal{O}-F=\bigcup_{j=1}^{\infty} Q_{j} $$ For a fixed $N$, the finite union $K=\bigcup_{j=1}^{N} Q_{j}$ is compact; therefore $d(K, F)>0$ (we isolate this little fact in a lemma below). Since $(K \cup$ $F) \subset \mathcal{O}$, Observations 1,4 , and 5 of the exterior measure imply $$ \begin{aligned} m_{}(\mathcal{O}) & \geq m_{}(F)+m_{}(K) \
&=m_{}(F)+\sum_{j=1}^{N} m_{}\left(Q_{j}\right)
\end{aligned}
$$



粘性流体动力学 |MA40255 Viscous fluid dynamics代写

0

Review of Lagrangian and Eulerian descriptions: Jacobian, Euler’s identity and Reynold’s transport theorem. The continuity equation and incompressibility condition.

这是一份Bath巴斯大学MA40255作业代写的成功案

粘性流体动力学 |MA40255 Viscous fluid dynamics代写


to $\sigma^{2}$ as $d_{i j}$ tends to zero. In some cases, there will further baseline variability and the

Solve the Jeffery-Hamel wedge-flow relation, Eq. (3-195), for creeping flow, $R e=0$ but $\alpha \neq 0$. Show that the proper solution is
$$
f(\eta)=1+\frac{1}{2} \csc ^{2} \alpha\left[\sin \left(\frac{\pi}{2}-2 \alpha \eta\right)-1\right]
$$
Show also that the constant $C=4 \alpha^{2} \cot ^{2} \alpha$ and sketch a few profiles. Show that backflow always occurs for $\alpha>90^{\circ}$.
In spherical polar coordinates, when the variations $\partial / \partial \phi$ vanish, an incompressible stream function $\psi(r, \theta)$ can be defined such that
$$
u_{r}=\frac{\partial \psi / \partial \theta}{r^{2} \sin \theta} \quad u_{\theta}=-\frac{\partial \psi / \partial r}{r \sin \theta}
$$
The particular stream function
$$
\psi(r, \theta)=\frac{2 v r \sin ^{2} \theta}{1+a-\cos \theta} \quad a=\text { const }
$$


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MA40255 COURSE NOTES :

So let us prove that the entries increase until the middle (then they begin to decrease by the symmetry of the table). We want to compare two consecutive entries:
$$
\left(\begin{array}{l}
n \
k
\end{array}\right) ?\left(\begin{array}{c}
n \
k+1
\end{array}\right) .
$$
If we use the formula in , we can write this as
$$
\frac{n(n-1) \cdots(n-k+1)}{k(k-1) \cdots 1} ? \frac{n(n-1) \cdots(n-k)}{(k+1) k \cdots 1} .
$$
There are many common factors on both sides that are positive, and so we can simplify. We get the really simple comparison
$$
1 ? \frac{n-k}{k+1} \text {. }
$$
Rearranging, we get
$$
k ? \frac{n-1}{2} .
$$



贝叶斯统计学的主题 |MA40189 Topics in Bayesian statistics代写

0

Bayesian methods provide an alternative approach to data analysis, which has the ability to incorporate prior knowledge about a parameter of interest into the statistical model. 

这是一份Bath巴斯大学MA40189作业代写的成功案

多变量数据分析 |MA40090 Multivariate data analysis代写


to $\sigma^{2}$ as $d_{i j}$ tends to zero. In some cases, there will further baseline variability and the covariance may be defined as
$$
\Sigma=\tau^{2} \quad \sigma^{2} R(d)
$$
with the limiting variance as $d_{i j}$ tends to zero being $\tau^{2} \quad \sigma^{2}$ instead of $\sigma^{2}$. Another formulation in this case involves a discontinuity when $d_{i i}=0$, so that
$$
\begin{aligned}
&\Sigma_{i j}=\sigma^{2} r_{i j}\left(d_{i j}\right) \quad i \neq j \
&\Sigma_{i i j}=\tau^{2}
\end{aligned}
$$
Whatever the model adopted for the correlation between errors, the problem reduces to simultaneously estimating the regression coefficients and the parameters of the distance decay function.


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MA40189 COURSE NOTES :

A technique often used to estimate the covariance structure focuses on functions of dissimilarity between errors or observations. Let $\Sigma(d)$ denote the covariance matrix at distance $d$ between points. Whereas
$$
\Sigma(d)=\sigma^{2} R(d)
$$
diminishes to zero for widely separated points and attains its maximum as $d_{i j}$ tends to zero, the variogram function
$$
\gamma(d)=\sigma^{2}-\Sigma(d)=\sigma^{2}(I-R(d))
$$
has value zero when $d_{i j}=0$, and reaches its maximum at $\sigma^{2}$ as spatial covariation in $\Sigma(d)$ disappears; hence $\sigma^{2}$ is known as the sill in geostatistics. For instance, the variogram for the exponential model is
$$
\gamma\left(d_{i j}\right)=\sigma^{2}\left(1-e^{-3 d_{i} / h}\right)
$$
As above, an extra variance parameter to allow for measurement error at $d=0$ (the ‘nugget’ error) may be added to give
$$
\gamma(d)=\nu^{2} \quad \sigma^{2}[\mathrm{I}-R(d)]
$$
and the sill is now $v^{2} \quad \sigma^{2}$. An alternative version of a nugget error model is
$$
\gamma(d)=v^{2} \quad\left(\sigma^{2}-v^{2}\right)[I-R(d)]
$$



多变量数据分析 |MA40090 Multivariate data analysis代写

0

select and apply an appropriate technique for the analysis of multivariate data to look for structure in such data or to achieve dimensionality reduction

这是一份Bath巴斯大学MA40090作业代写的成功案

多变量数据分析 |MA40090 Multivariate data analysis代写


When the units within missingness pattern $s$ are crossclassified only by their observed variables, the result is a table with counts that we shall denote by
$$
z_{O_{x}(y)}^{(s)}=\sum_{M_{s}(y) \in M_{x}} x_{y}^{(s)} \text { for all } O_{s}(y) \in O_{s} .
$$
The marginal probability that an observation falls within cell $O_{s}(y)$ of this table will be called
$$
\beta_{O_{x}(y)}=\sum_{M_{x}(y) \in M_{x}} \theta_{y}
$$


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MA40090 COURSE NOTES :


that is,
$$
x_{O_{x}(y)}^{(s)} \mid z_{O_{x}(y)}^{(s)}, \theta \sim M\left(z_{O_{x}(y)}^{(s)}, \gamma_{O_{x}(y)}\right)
$$
Notice that (7.34) is simply the portion of $\theta$ corresponding to $x_{O_{s}(y)}^{(s)}$, rescaled so that its elements sum to one. It follows that the conditional expectation of an element of $x^{(s)}$ is
$$
E\left(x_{y}^{(s)} \mid z^{(s)}, \theta\right)=z_{O_{x}(y)}^{(s)} \theta_{y} / \beta_{O_{x}(y)}
$$
The E-step consists of calculating for every $s=1, \ldots, S$ and summing the results,



数学方法|MA40059 Mathematical methods 2代写

0

To stimul should be able to obtain solutions to certain important PDEs using a variety of techniques e.g. Green’s functions, separation of variables, and in cases interpret these in physical terms. They should also be familiar with important analytic properties of the solution.

这是一份Bath巴斯大学MA40059作业代写的成功案

数学方法|MA40059 Mathematical methods 2代写

$$
\begin{gathered}
Y(x, \epsilon)=y(x)=\epsilon \eta(x), \
Y^{\prime}(x, \epsilon)=y^{\prime}(x)+\epsilon \eta^{\prime}(x) .
\end{gathered}
$$
Then the meaning of $\delta y$ is
$$
\delta y=\left(\frac{\partial Y}{\partial \epsilon}\right){\epsilon=0} d \epsilon=\eta(x) d \epsilon ; $$ this is just like a differential $d Y$ if $\epsilon$ is the variable. The meaning of $\delta y^{\prime}$ is $$ \delta y^{\prime}=\left(\frac{\partial Y^{\prime}}{\partial \epsilon}\right){\epsilon=0} d \epsilon=\eta^{\prime}(x) d \epsilon .
$$
This is identical with
$$
\frac{d}{d x}(\delta y)=\frac{d}{d x}[\eta(x) d \epsilon]=\eta^{\prime}(x) d \epsilon
$$


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MA40059 COURSE NOTES :

We can show that these nine quantities are the components of a second-rank tensor which we shall denote by UV. Note that this is not a dot product or a cross product; it is called the direct product of $\mathrm{U}$ and $\mathrm{V}$ (or outer product or tensor product). Since $\mathbf{U}$ and $\mathbf{V}$ are vectors, their components in a rotated coordinate system are:
$$
U_{k}^{\prime}=\sum_{i=1}^{3} a_{k i} U_{i}, \quad V_{l}^{\prime}=\sum_{j=1}^{3} a_{l j} V_{j} .
$$
Hence the components of the second-rank tensor UV are
$$
U_{k}^{\prime} V_{l}^{\prime}=\sum_{i=1}^{3} a_{k i} U_{i} \sum_{j=1}^{3} a_{l j} V_{j}=\sum_{i, j=1}^{3} a_{k i} a_{l j} U_{i} V_{j}
$$
which is just with $T_{i j}=U_{i} V_{j}$ and $T_{k l}^{\prime}=U_{k}^{\prime} V_{l}^{\prime}$.



有马太效应的概率论|MA40058 Probability with martingales代写

0

To stimulate through theory and especially examples, an interest and appreciation of the power and elegance of martingales in analysis and probability. 

这是一份Bath巴斯大学MA40058作业代写的成功案

有马太效应的概率论|MA40058 Probability with martingales代写

Let $\mathcal{K}$ be a vector subspace of $\mathcal{L}^{2}$ which is complete in that whenever $\left(V_{n}\right)$ is a sequence in $\mathcal{K}$ which has the Cauchy property that
$$
\sup {r, s \geq k}\left|V{r}-V_{s}\right| \rightarrow 0 \quad(k \rightarrow \infty)
$$
then there exists a $V$ in $\mathcal{K}$ such that
$$
\left|V_{n}-V\right| \rightarrow 0 \quad(n \rightarrow \infty)
$$
Then given $X$ in $\mathcal{L}^{2}$, there exists $Y$ in $\mathcal{K}$ such that
$$
|X-Y|=\Delta:=\inf {|X-W|: W \in \mathcal{K}}
$$
(ii)
$$
X-Y \perp Z, \quad \forall Z \in \mathcal{K}
$$
Properties (i) and (ii) of $Y$ in $\mathcal{K}$ are equivalent and if $\tilde{Y}$ shares either property (i) or (ii) with $Y$, then
$$
|\tilde{Y}-Y|=0 \quad \text { (equivalently, } Y=\tilde{Y}, \text { a.s. })
$$


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MA40058 COURSE NOTES :

Thus
$$
\mathrm{E}\left(S_{n}^{4}\right) \leq n K+3 n(n-1) K \leq 3 K n^{2}
$$
and (see Section 6.5)
$$
\mathrm{E} \sum\left(S_{n} / n\right)^{4} \leq 3 K \sum n^{-2}<\infty
$$
so that $\sum\left(S_{n} / n\right)^{4}<\infty$, a.s., and
$$
S_{n} / n \rightarrow 0, \quad \text { a,s. }
$$