# 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
$$\left\langle H_{l} a_{R}^{\dagger}(\Delta)(v e(g)), H_{t}^{\prime} a_{S}^{\dagger}\left(\Delta^{\prime}\right)(u e(f))\right\rangle$$