# 随机过程 Stochastic Processes ST227-10

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Let $\left(\mathbf{u}{t}\right){0 \leq t \leq T}$ be an $m$-dimensional process and
\begin{aligned} &\mathbf{a}:[0, T] \times \Omega \rightarrow \mathbb{R}^{m}, \mathbf{a} \in \mathcal{C}{1 \mathbf{w}}([0, T]) \ &b:[0, T] \times \Omega \rightarrow \mathbb{R}^{m n}, b \in \mathcal{C}{1 \mathbf{w}}([0, T]) \end{aligned}
The stochastic differential $d \mathbf{u}(t)$ of $\mathbf{u}(t)$ is given by
$$d \mathbf{u}(t)=\mathbf{a}(t) d t+b(t) d \mathbf{W}(t)$$
if, for all $0 \leq t_{1}<t_{2} \leq T$
$$\mathbf{u}\left(t_{2}\right)-\mathbf{u}\left(t_{1}\right)=\int_{t_{1}}^{t_{2}} \mathbf{a}(t) d t+\int_{t_{1}}^{t_{2}} b(t) d \mathbf{W}(t)$$

## ST227-10 COURSE NOTES ：

Let $A$ be a nonempty subset of a metric space $(X, d)$. Define
$$d_{A}(x):=\inf {d(x, a): a \in A}, \quad x \in X .$$
Then $d_{A}$ is continuous. (Geometrically, we think of $d_{A}(x)$ as the distance of $x$ to $A$.)

We give a proof even though it is easy, because of the importance of this result. Let $x, y \in X$ and $a \in A$ be arbitrary. We have, from the triangle inequality $d(a, x) \leq d(a, y)+d(y, x)$,
\begin{aligned} &d(a, y) \geq d(a, x)-d(y, x) \ &d(a, y) \geq d_{A}(x)-d(y, x) \end{aligned}
The inequality says that $d_{A}(x)-d(y, x)$ is a lower bound for the set ${d(a, y): a \in A}$. Hence the greatest lower bound of this set, namely, $\inf {d(a, y)=a \in \backslash A} n d$ is greater than or equal to this lower bound, that is,
$$d_{A}(y) \geq d_{A}(x)-d(y, x)$$

# 随机过程 Stochastic Processes STATS4024_1 /STATS5026_1

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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
$$E[Y \mid X=x] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }$$

## STATS4024_1 /STATS5026_1COURSE 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}

# 随机过程IStochastic Processes MATH2012W1-01

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with measurement equation
$$z(k)=\left[\begin{array}{ll} 1 & 0 \end{array}\right] x^{i}(k)+w(k)$$
The models differ in the control gain parameter $b^{i}$. The process and measurement noises are mutually uncorrelated with zero mean and variances given by
$$E[v(k) v(j)]=0.16 \delta_{k j}$$
and
$$E[w(k) w(j)]=\delta_{k j}$$

The control gain parameters were chosen to be $b^{1}=2$ and $b^{2}=0.5$.
The Markov transition matrix was selected to be
$$\left[\begin{array}{ll} 0.8 & 0.2 \ 0.1 & 0.9 \end{array}\right]$$
For this example $N=7$, and the cost parameters $R(k)$ and $Q(k)$, were selected as
$$R(k)=5.0 \quad k=1,2, \ldots, N-1$$

## MATH2012W1-01COURSE NOTES ：

Proof: Since $f(d)$ is the minimal polynomial of $\boldsymbol{F}, p(d)$ can be factored as
$$p(d)=g(d) . f(d)$$
for some polynomial $g(d)=\sum_{i=0}^{s} b_{i} d^{s-i}$
Let $z_{k}$ and $\bar{z}{k}$ be linear combinations of $y{k}$ defined as in by using polynomials $f(d)$ and $p(d)$, respectively.
$\bar{z}{k} \quad$ can be expressed in terms of $z{k}$ as
$$\bar{z}{k}=\sum{i=0}^{s} b_{i} z_{k-i}$$

# 随机过程 Stochastic Processes  MATH97113

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Proof:
First note that $e\left(f_{1}\right), \ldots, e\left(f_{n}\right)$ are linearly independent whenever $f_{1}, \ldots, f_{n}$ are distinct, from which it is clear that $\sum_{i=1}^{n} x_{i} \otimes e\left(f_{i}\right)=0$ implies $x_{i}=0$ for all $i$, whenever $f_{i}$ ‘s are distinct. This will establish that the processes are well defined. The second part of the lemma will follow from Lemma 4.2.10 with the choice of the dense set $\mathcal{E}$ to be $\mathcal{E}(k)$ and $\mathcal{H}=\Gamma(k)$ and by noting the fact that $\mathcal{L}, \delta, \delta^{\prime}$ and $\sigma$ have appropriate ranges. For example,
$$\left\langle e(g), a_{g}^{\dagger}(\Delta)(x \otimes e(f))\right\rangle=\langle e(g), e(f)\rangle \int_{\Delta}\left\langle g(s), \delta^{\prime}(x)\right\rangle d s,$$

which belongs to $\mathcal{A}$, so the range of $a_{g^{\prime}}^{\dagger}(\Delta)$ is contained in $\mathcal{A} \otimes \Gamma(k)$. Similarly, one verifies that
$$\left\langle e(g), \Lambda_{a}(\Delta)(x \otimes e(f))\right\rangle=\langle e(g), e(f)\rangle \int_{\Delta}\left\langle g(s), \sigma(x){f(s)}\right\rangle d s,$$ which belongs to $\mathcal{A}$ since $\sigma(x) \in \mathcal{A} \otimes \mathcal{B}\left(k{0}\right)$.

## MATH97113 COURSE NOTES ：

For fixed $x \in \mathcal{A}, u \in h$ and $f \in L_{\mathrm{bc}}^{4}$, we define the integral $\int_{0}^{t} Y(s) \circ\left(a_{\delta}+\mathcal{I}{\mathcal{L}}\right)(d s)(x \otimes e(f)) u$ by setting it to be equal to $$\int{0}^{t} Y(s)\left(\left(\mathcal{L}(x)+\left\langle\delta\left(x^{}\right), f(s)\right\rangle\right) \otimes e(f)\right) u d s$$ This integral exists and is finite since $s \mapsto Y(s)\left(\left(\mathcal{L}(x)+\left\langle\delta\left(x^{}\right), f(s)\right\rangle\right) \otimes\right.$ $e(f)) u$ is strongly integrable over $[0, t]$. We define the integral involving the other two processes, that is, $\int_{0}^{t} Y(s) \circ\left(\Lambda_{a}+a_{g}^{\dagger}\right)(d s)(x \otimes e(f)) u$ by setting it to be equal to
$$\left(\int_{0}^{t} \Lambda_{T_{x}}(d s)+a_{S_{x}}^{\dagger}(d s)\right) u e(f),$$
which is well-defined by Corollary

# 随机过程|Stochastic Processes MATH0060

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$$\phi_{k}(x)=\sum_{m=1}^{N^{\prime}} L_{k}^{(m)^{*}} x L_{k}^{(m)}, \text { for all } x \in \mathcal{A} .$$
Here the Linbladian $\mathcal{L}^{\phi}$ corresponding to the partial state $\phi_{0}$ is formally given by
$$\mathcal{L}^{\phi}(x)=\sum_{k \in \mathbb{Z}^{d}} \mathcal{L}_{k}^{\phi}(x),$$

where
$$\mathcal{L}{k}^{\phi}(x)=\phi{k}(x)-x=\frac{1}{2} \sum_{m=1}^{N^{\prime}}\left[L_{k}^{(m)^{}}, x\right] L_{k}^{(m)}+L_{k}^{(m)^{}}\left[x, L_{k}^{(m)}\right]$$

## MATH0060 COURSE NOTES ：

for all $\alpha \in I$, whenever $v \notin J_{0}$. Fixing this $J_{0}$, we choose $I_{0}$ to be the union of $I_{v, n}, v \in J_{0}, n=1,2, \cdots, \infty$, such that
$\left\langle e\left(g^{t}+\frac{1}{n}\left(H_{I} P R_{\Delta}\right){e{\nu}, v e e\left(g_{t}\right)}\right), k_{\alpha}\right\rangle=0=\left\langle k_{\alpha}, e\left(f^{t}+\frac{1}{n}\left(H_{t}^{\prime} P S_{\Delta^{\prime}}\right){e{\nu}, u e\left(f_{t}\right)}\right)\right\rangle$ for all $\alpha \notin I_{v, n}$ when $n<\infty$, and
$$\left\langle e\left(g^{t}\right), k_{\alpha}\right\rangle=0=\left\langle k_{\alpha}, e\left(f^{t}\right)\right\rangle \text { for } \alpha \notin I_{v, \infty}$$
We now have
$$\left\langle H_{l} a_{R}^{\dagger}(\Delta)(v e(g)), H_{t}^{\prime} a_{S}^{\dagger}\left(\Delta^{\prime}\right)(u e(f))\right\rangle$$

# 随机过程作业代写Stochastic process代考

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## 泊松过程Poisson point process代写

• Skorokhod空间Skorokhod space
• 马尔可夫链Markov chain

## 随机过程的相关

A stochastic or random process can be defined as a collection of random variables that is indexed by some mathematical set, meaning that each random variable of the stochastic process is uniquely associated with an element in the set.The set used to index the random variables is called the index set. Historically, the index set was some subset of the real line, such as the natural numbers, giving the index set the interpretation of time.

## 随机过程课后作业代写

If we further assume that $v_{L}$ is a function of $v$, so that all granules of a given size have the same liquid content, then $\beta$ is no longer an explicit function of $v_{L}$, and we can write:
\begin{aligned} \frac{\partial n(v, t)}{\partial t}=\frac{1}{2} \int_{0}^{\infty} d \varepsilon \beta(v-\varepsilon, \varepsilon) n(v-\varepsilon, t) n(\varepsilon, t) \ &-n(v, t) \int_{0}^{\infty} d \varepsilon \beta(v, \varepsilon) n(\varepsilon, t) \end{aligned}
This implies an equation for the mass distribution of liquid:
\begin{aligned} \frac{\partial M(v, t)}{\partial t}=\frac{1}{2} \int_{0}^{\infty} d \varepsilon \beta(v-\varepsilon, \varepsilon) M(v-\varepsilon, t) n(\varepsilon, t) \ &-M(v, t) \int_{0}^{\infty} d \varepsilon \beta(v, \varepsilon) n(\varepsilon, t) \end{aligned}
Here we recognize that for physical reasons the integrands are zero if $v_{L}$ exceeds $v$.