Tag Archives: MATH4061代考

度量空间|MATH4061 Metric Spaces代写 Sydney代写

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这是一份Sydney悉尼大学MATH4061的成功案例

度量空间|MATH4061 Metric Spaces代写 Sydney代写


问题 1.

set $U_{2}$ to get $x_{2} \in B\left(x_{1}, r_{1}\right) \cap U_{2}$. Again, we can find $r_{2}$ such that $0<r_{2}<2^{-2}$ and $B\left[x_{2}, r_{2}\right] \subset B\left(x_{1}, r_{1}\right) \cap U_{2}$. Proceeding this way, for each $n \in \mathbb{N}$, we get $x_{n} \in X$ and an $r_{n}$ with the properties
$$
B\left[x_{n}, r_{n}\right] \subset B\left(x_{n-1}, r_{n-1}\right) \cap U_{n} \text { and } 0<r_{n}<2^{-n}
$$

证明 .

Clearly, the sequence $\left(x_{n}\right)$ is Cauchy: if $m \leq n$,
$$
d\left(x_{m}, x_{n}\right) \leq d\left(x_{n}, x_{n-1}\right)+\cdots+d\left(x_{m+1}, x_{m}\right) \leq \sum_{k=m}^{n} 2^{-k}
$$
Since $\sum_{k} 2^{-k}$ is convergent, it follows that $\left(x_{n}\right)$ is Cauchy.





英国论文代写Viking Essay为您提供作业代写代考服务

MATH4061 COURSE NOTES ：


We shall sketch the argument. Let $m, M$ be such that
$$
m \leq \frac{\partial f}{\partial y}(x, y) \leq M, \text { for }(x, y) \in D
$$
Let $X:=\left(C[a, b], |_{\infty}\right)$. Consider
$$
T(\varphi)(x):=\varphi(x)-\frac{1}{M} f(x, \varphi(x)), \text { for } x \in[a, b]
$$
One shows that $T$ is a contraction by an obvious use of the mean value theorem.


















高级数学 Adv. fin Mathematics MATH4061

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这是一份nottingham诺丁汉大学MATH4061作业代写的成功案例

高级数学 Adv. fin Mathematics MATH4061

Let $D_{n}:=E\left(X_{n} \mid \mathcal{B}{n-1}\right)-X{n-1}$ for $n=-1,-2, \ldots$. By assumption, $D_{n} \geq 0$ a.s. Now $\sum_{n \leq-1} E D_{n}=\sum_{n \leq-1} E\left(X_{n}-X_{n-1}\right)=E X_{-1}-K$; note that $E X_{n} \downarrow K$ as $n \rightarrow-\infty$. Since the sum is finite, the series $\sum D_{n}$ converges by monotone convergence to an integrable function.

Let $Z_{n}:=\sum_{k \leq n} D_{k}$ for each $n$. Then $Z_{n}$ is nondecreasing as $n$ increases almost surely, and $Z_{n} \downarrow 0$ a.s. as $n \rightarrow-\infty$. Let $Y_{n}:=X_{n}-Z_{n}$. To show that $Y_{n}$ is a (reversed) martingale, we have
$$
\begin{aligned}
E\left(Y_{n} \mid \mathcal{B}{n-1}\right) &=E\left(X{n}-Z_{n} \mid \mathcal{B}{n-1}\right)=D{n}+X_{n-1}-\sum_{k \leq n} D_{k} \
&=X_{n-1}-\sum_{k<n} D_{k}=X_{n-1}-Z_{n-1}=Y_{n-1}
\end{aligned}
$$
proving the decomposition as desired. Now $Z_{n}$ converges in $\mathcal{L}^{1}$ by monotone or dominated convergence, and $Y_{n}$ converges a.s. and in $\mathcal{L}^{1}$ by Theorem 10.6.1, so $X_{n}$ converges likewise.

英国论文代写Viking Essay为您提供作业代写代考服务

MATH4061 COURSE NOTES ：

If $M, P$, and $Q$ are laws on $S, \rho(M, P)<x$ and $\rho(P, Q)<y$, then for any Borel set $A$,
$$
M(A) \leq P\left(A^{x}\right)+x \leq Q\left(\left(A^{x}\right)^{y}\right)+y+x \leq Q\left(A^{x+y}\right)+x+y,
$$
so $\rho(M, Q) \leq x+y$. Letting $x \downarrow \rho(M, P)$ and $y \downarrow \rho(P, Q)$ gives $\rho(M, Q) \leq$ $\rho(M, P)+\rho(P, Q)$.

The metric $\rho$ is called the Prohorov metric, or sometimes the LévyProhorov metric. Now for any laws $P$ and $Q$ on $S$, let $\int f d(P-Q):=$ $\int f d P-\int f d Q$ and