Tag Archives: UCL代写

数论  Theory of Numbers MATH3701

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这是一份UCL伦敦大学 MATH003701作业代写的成功案例

数论  Theory of Numbers MATH3701
问题 1.

Proof. Assume, by way of contradiction, that there are only a finite number of prime numbers, say:
$$
p_{1}, p_{2}, \ldots, p_{n}
$$
Define
$$
N=p_{1} p_{2} \cdots p_{n}+1
$$


证明 .

Since $p_{1} \geq 2$, clearly $N \geq 3$. So by Lemma $10.2 N$ has a prime divisor $p$. By assumption $p=p_{i}$ for some $i=1, \ldots, n$. Let $a=p_{1} \cdots p_{n}$. Note that
$$
a=p_{i}\left(p_{1} p_{2} \cdots p_{i-1} p_{i+1} \cdots p_{n}\right),
$$
so $p_{i} \mid a$. Now $N=a+1$ and by assumption $p_{i} \mid a+1$. So by Exercise $3.2$ $p_{i} \mid(a+1)-a$, that is $p_{i} \mid$

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MATH003701 COURSE NOTES ：

$$
a, a+1, a+2, \ldots, a+(n-1)
$$
are all composite.
Proof. Given $n \geq 1$ let $a=(n+1) !+2$. We claim that all the numbers
$$
a+i, \quad 0 \leq i \leq n-1
$$
composite. Since $(n+1) \geq 2$ clearly $2 \mid(n+1)$ ! and $2 \mid 2$. Hence $2 \mid(n+1) !+2$. Since $(n+1) !+2>2,(n+1) !+2$ is composite. Consider
$$
a+i=(n+1) !+i+2
$$







动态系统 Dynamical Systems MATH3509

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这是一份UCL伦敦大学 MATH003509作业代写的成功案例

动态系统 Dynamical Systems MATH3509
问题 1.

show that
$$
\dot{\mathbf{x}}=\mathbf{A}(t) \mathbf{x}+\mathbf{B}(t)
$$
has solution
$$
x(t)=\varphi\left(t, t_{0}\right)\left(x_{0}+\int_{C_{0}}^{t} \varphi\left(t_{0}, t\right) B(\tau) d \tau\right)
$$


证明 .

when $\mathbf{x}=\mathrm{x}{0}$ at $t=t{0}$.
Find the solution of
$$
\dot{x}=\left(\begin{array}{rr}
0 & -1 \
1 & 0
\end{array}\right) x+\left(\begin{array}{l}
\sin t \
\cos t
\end{array}\right)
$$
when $\mathbf{x}(0)=\mathbf{x}_{0}$.

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MATH003509 COURSE NOTES ：

$$
\mathbf{f}(\pi(\mathbf{x}))=\pi(\mathrm{f}(\mathbf{x})) \text {, }
$$
for each $x \in \mathbb{R}^{n}$, where $\pi: \mathbb{R}^{n} \rightarrow T^{n}$ is given by
$$
\begin{aligned}
\pi(\mathbf{x})=\pi\left(\left(x_{1}, \ldots, x_{n}\right)^{\mathrm{T}}\right) &=\left(x_{1} \bmod 1, \ldots, x_{n} \bmod 1\right)^{\mathrm{T}} \
&=\left(\theta_{1}, \ldots, \theta_{n}\right)^{\mathrm{T}}=\theta
\end{aligned}
$$
(see Figure 3.10). If $k \in \mathbb{Z}^{\mathrm{n}}$, then (3.4.1) implies
$$
\pi(\bar{f}(\mathbf{x}+\mathbf{k}))=\mathbf{f}(\pi(\mathbf{x}+\mathbf{k}))=\mathbf{f}(\pi(\mathbf{x}))=\pi(\bar{f}(\mathbf{x})),
$$
for all $\mathbf{x} \in \mathbb{R}^{n}$. Continuity of $\overline{\mathbf{f}}$ then gives,
$$
\overline{\mathbf{f}}(\mathbf{x}+\mathbf{k})=\overline{\mathbf{f}}(\mathbf{x})+\mathbf{1}(\mathbf{k})
$$







离散数学 Discrete Mathematics MATH00103

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这是一份UCL伦敦大学 MATH00103作业代写的成功案例

离散数学 Discrete Mathematics MATH00·03
问题 1.

translates into
$$
c \cdot q^{n+1}=c \cdot q^{n}+c \cdot q^{n-1}
$$
which after simplification becomes
$$
q^{2}=q+1
$$


证明 .

So both numbers $c$ and $n$ disappear. ${ }^{1}$
So we have a quadratic equation for $q$, which we can solve and get
$$
q_{1}=\frac{1+\sqrt{5}}{2} \approx 1.618034, \quad q_{2}=\frac{1-\sqrt{5}}{2} \approx-0.618034 .
$$
This gives us two kinds of geometric progressions that satisfy the same recurrence as the Fibonacci numbers:
$$
G_{n}=c\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad G_{n}^{\prime}=c\left(\frac{1-\sqrt{5}}{2}\right)^{n}
$$

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MATH00103 COURSE NOTES ：

(a) if $a \mid b$ and $b \mid c$ then $a \mid c$;
(b) if $a \mid b$ and $a \mid c$ then $a \mid b+c$ and $a \mid b-c$;
(c) if $a, b>0$ and $a \mid b$ then $a \leq b$;
(d) if $a \mid b$ and $b \mid a$ then either $a=b$ or $a=-b$.







统计方法和数据分析Statistical Methods and Data Analysis MATH0099

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这是一份UCL伦敦大学 MATH0099作业代写的成功案例

统计方法和数据分析Statistical Methods and Data AnalysisMATH0099
问题 1.

$100(1-\alpha) \%$ confidence interval for $\mu_{1}-\mu_{2}$, independent samples; $y_{1}$ and $y_{2}$ approximately normal; $\sigma_{1}^{2}=\sigma_{2}^{2}$
$$
\bar{y}{1}-\bar{y}{2} \pm t_{\alpha / 2} s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}
$$
where
$$
s_{p}=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}} \quad \text { and } \quad \mathrm{df}=n_{1}+n_{2}-2
$$


证明 .

$t$ test for $\mu_{1}-\mu_{2}$, independent samples; $y_{1}$ and $y_{2}$ approximately normal; $\sigma_{1}^{2}=\sigma_{2}^{2}$
T.S.: $t=\frac{\bar{y}{1}-\bar{y}{2}-D_{0}}{s_{p} \sqrt{1 / n_{1}+1 / n_{2}}} \quad \mathrm{df}=n_{1}+n_{2}-2$

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MATH0099 COURSE NOTES ：

The $99 \%$ confidence interval for $\sigma$ is then
$$
\sqrt{\frac{29(3.433)^{2}}{52.34}}<\sigma<\sqrt{\frac{29(3.433)^{2}}{13.12}}
$$
or
$$
2.56<\sigma<5.10
$$
Thus, we are $99 \%$ confident that the standard deviation in the weights of coffee cans lies between $2.56$ and $5.10$ grams. The designed value for $\sigma, 4$ grams, falls within our confidence interval. Using our results from , a $99 \%$ confidence interval for $\mu$ is
$$
500.453 \pm 2.756 \frac{3.433}{\sqrt{30}} \quad 500.453 \pm 1.73
$$
or
$$
498.7<\mu<502.2
$$







金融和保险数学专题Topics in Financial and Insurance MathematicsMATH0095

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这是一份UCL伦敦大学 MATH0095作业代写的成功案例

金融和保险数学专题Topics in Financial and Insurance MathematicsMATH0095
问题 1.

$$
\mathbb{E}[d S]=r S d t=\mu S d t+\mathbb{E}[J-1] S \lambda(t) d t
$$
It follows that the risk-neutral drift is given by $\mu=r+\mu_{J}$ with
$$
\mu_{J}=-\lambda(t) \mathbb{E}[J-1]
$$


证明 .

$k:=\log (K / F)$ defined by
$$
\hat{G}(u, \tau)=\int_{-\infty}^{\infty} e^{\mathrm{i} u k} G(k, \tau) d x
$$

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MATH0095 COURSE NOTES ：

Because the Black-Scholes formula $C$ for a call option is linearly homogenous in the stock price $S$ and the strike price $K$, we have the relation
$$
C=S \frac{\partial C}{\partial S}+K \frac{\partial C}{\partial K}
$$
It follows that
$$
K^{2} \frac{\partial^{2} C}{\partial K^{2}}=S^{2} \frac{\partial^{2} C}{\partial S^{2}}
$$
Also, in the jump-to-ruin case with zero interest rates and dividends, we have
$$
\frac{\partial C}{\partial T}=\frac{1}{2} \sigma^{2} S^{2} \frac{\partial^{2} C}{\partial S^{2}}+\lambda S \frac{\partial C}{\partial S}-\lambda C
$$







市场风险、措施和投资组合理论Market Risk, Measures and Portfolio TheoryMATH0094

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这是一份UCL伦敦大学 MATH0094作业代写的成功案例

市场风险、措施和投资组合理论Market Risk, Measures and Portfolio TheoryMATH0094
问题 1.

have $\left(R_{1} R_{2}\right)$ followed by every possible result for draws $(3,4 \ldots k)$. In other words, the probability of $R_{j} R_{k}$ is the same as that of
$$
R_{1} R_{2}\left(R_{3}+W_{3}\right) \cdots\left(R_{k}+W_{k}\right)=R_{1} R_{2}
$$
and we have
$$
P\left(R_{j} R_{k} \mid B\right)=P\left(R_{1} R_{2} \mid B\right)=\frac{M(M-1)}{N(N-1)}
$$


证明 .

and likewise
$$
P\left(W_{j} R_{k} \mid B\right)=P\left(W_{1} R_{2} \mid B\right)=\frac{(N-M) M}{N(N-1)}
$$
Therefore by the product rule
$$
P\left(R_{k} \mid R_{j} B\right)=\frac{P\left(R_{j} R_{k} \mid B\right)}{P\left(R_{j} \mid B\right)}=\frac{M-1}{N-1}
$$
and
$$
P\left(R_{k} \mid W_{j} B\right)=\frac{P\left(W_{j} R_{k} \mid B\right)}{P\left(W_{j} \mid B\right)}=\frac{M}{N-1}
$$

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MATH0094 COURSE NOTES ：

$$
p^{r}(1-p)^{n-r}
$$
as in the binomial distribution $(3-79)$. But as these numbers increase, we can use relations of the form
$$
\left(1+\frac{\epsilon}{p}\right)^{c} \simeq \exp \left(\frac{\epsilon c}{p}\right)
$$
and $(3-81)$ goes into







金融和数字Finance and Numerics MATH0093

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这是一份UCL伦敦大学 MATH0093作业代写的成功案例

金融和数字Finance and Numerics MATH0093
问题 1.

Assume we have $n$ assets which have the following dynamics under a measure $Q$ :
$$
d z_{i}=-\frac{1}{2} \sigma_{i}^{2} d t+\sigma_{i} d w_{i}
$$
where $d w_{i} d w_{j}=\rho_{i j} d t$. Assume further we have a contingent claim with payoff of the form
$$
Y_{T}=e^{z_{k}(T)} X_{k}\left(e^{z_{i}(T)}\right)
$$


证明 .

where $X_{k}$ is a multilinear form. Then for the following changes of measure
$$
\frac{d Q_{p}}{d Q}=e^{z_{p}(T)-z_{p}}
$$

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MATH0093 COURSE NOTES ：

Consider the following valuation problem:
$$
E_{t}^{Q}\left(e^{z_{T}}-K\right)^{+}
$$
In terms of indicator functions, can be written as
$$
E_{t}^{Q}\left(e^{z_{T}}-K\right)^{+}=E_{t}^{Q} e^{z_{T}} 1\left(z_{T}>\log K\right)-K \cdot E_{t}^{Q} 1\left(z_{T}>\log K\right)
$$
Consider the second expectation in :
$$
E_{t}^{Q} 1\left(z_{T}>\log K\right)=\int_{-\infty}^{\infty} d z_{T} \operatorname{Pr}\left(z_{T} \mid z\right) 1\left(z_{T}>\log K\right)
$$







椭圆偏微分方程Elliptic Partial Differential Equations MATH0090

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这是一份UCL伦敦大学 MATH0090作业代写的成功案例

椭圆偏微分方程Elliptic Partial Differential Equations MATH0090
问题 1.

Symmetry: $a^{i j}(x)=a^{j i}(x)$ for all $i, j$ and $x \in \Omega$ (this is no serious restriction).
Ellipticity: There exists a constant $\lambda>0$ with
$$
\lambda|\xi|^{2} \leq \sum_{i, j=1}^{d} a^{i j}(x) \xi^{i} \xi^{j} \quad \text { for all } x \in \Omega, \xi \in \mathbb{R}^{d}
$$


证明 .

In particular, the matrix $\left(a^{y}(x)\right)_{i, j=1}, \ldots, d$ is positive definite for all $x$, and the smallest eigenvalue is greater than or equal to $\lambda$.
Boundedness of the coefficients: There exists a constant $K$ with
$$
\left|a^{i j}(x)\right|,\left|b^{i}(x)\right|,|c(x)| \leq K \text { for all } i, j \text { and } x \in \Omega
$$

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MATH0090 COURSE NOTES ：

Since at an interior maximum $x_{0}$ of $u$, we must have
$$
u_{x^{i}}\left(x_{0}\right)=0 \quad \text { for } i=1, \ldots, d
$$
and
$$
\left(u_{x^{i} x^{j}}\left(x_{0}\right)\right){i, j=1, \ldots, d} \quad \text { negative semidefinite, } $$ and thus by the ellipticity condition also $$ L u\left(x{0}\right)=\sum_{i, j=1}^{d} a^{i j}\left(x_{0}\right) u_{x^{i} x^{j}}\left(x_{0}\right) \leq 0
$$
such an interior maximum cannot occur.
Returning to the general case $L u \geq 0$, we now consider the auxiliary function
$$
v(x)=e^{\alpha x^{1}}
$$
for $\alpha>0$. Then







量化和计算Quantitative and ComputationalMATH0088

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这是一份UCL伦敦大学 MATH0088作业代写的成功案例

量化和计算Quantitative and ComputationalMATH0088
问题 1.

$$
\frac{\partial V}{\partial S}=\mathcal{H}(S-E) .
$$
What about the slope of the slope in theS direction? Well, if the slope is zero or one, the slope of the slope is zero. So you know the second term in the equation.
$$
\frac{\partial^{2} V}{\partial S^{2}}=0 .
$$
To recap, we’ve got
$$
\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^{2} S^{2} \times 0+r S \mathcal{H}(S-E)-r \max (S-E, 0)=0 .
$$


证明 .

This is an equation for $\frac{\partial V}{\partial t}$. For example, if $SE$ we have
$$
\frac{\partial V}{\partial t}=-r S+r S-R E=-r E
$$

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MATH0088 COURSE NOTES ：



Set up the portfolio of one long forward contract and short $\Delta$ of the underlying asset:
$$
\Pi=V(S, t)-\Delta S .
$$
This changes by an amount
$$
d \Pi=\frac{\partial V}{\partial t} d t+\frac{\partial V}{\partial S} d S+\frac{1}{2} \sigma^{2} S^{2} \frac{\partial^{2} V}{\partial S^{2}} d t-\Delta d S
$$
from $t$ to $t+d t$. Choose
$$
\Delta=\frac{\partial V}{\partial S}
$$






计算和模拟Asset Pricing in Continuous Time MATH0086

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这是一份UCL伦敦大学 MATH0086作业代写的成功案例

计算和模拟Asset Pricing in Continuous Time MATH0086
问题 1.

Solving for $x_{4}$ in the last equation yields
$$
x_{4}=\frac{6}{3}=2
$$
Using $x_{1}=2$ in the third equation, we obtain
$$
x_{3}=\frac{4-5(2)}{6}=-1
$$


证明 .

Now $x_{3}=-1$ and $x_{4}=2$ are used to find $x_{2}$ in the second equation:
$$
x_{2}=\frac{-7-7(-1)+4(2)}{-2}=-4 .
$$
Finally, $x_{1}$ is obtained using the first equation:
$$
x_{1}=\frac{20+1(-4)-2(-1)-3(2)}{4}=3 .
$$

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MATH0086 COURSE NOTES ：



Modify Program so that it will compute $A^{-1}$ by repeatedly solving $N$ linear systems
$$
A C_{J}=E_{J} \quad \text { for } J=1,2, \ldots, N
$$
Then and
$$
\begin{gathered}
A\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right]=\left[\begin{array}{llll}
E_{1} & \boldsymbol{E}{2} & \ldots & \boldsymbol{E}{N}
\end{array}\right] \
A^{-l}=\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right] .
\end{gathered}
$$
Make sure that you compute the $\boldsymbol{L} \boldsymbol{I}$ factorization only once!