0

Alternatively, the net capital stock is replaced by a linear time trend $t$ that should account for the technological change.
Formally, the cost function used here is given by
\begin{aligned} C\left(w, x^{f}, y^{r}, s, t\right)=& y_{s, t}^{r} \times\left[\sum_{i=1}^{n} \sum_{j=1}^{w} \beta_{i j}\left(w_{i} w\right){s, t}^{1 / 2}+\sum{i=1}^{n} \sum_{j=1}^{w} \gamma_{i j}\left(w_{i} f\right){s, t}\right.\ &+\sum{i=1}^{n} \delta_{i}(w){s, t} \times t+\sum{i=1}^{n} \delta_{i 2}\left(w w_{s, t} \times t^{2}\right] \end{aligned}
where $C$ is the real total variable cost, $s$ is the industry ( $s=1$ : banking, $s=2$ : insurance), and $\beta_{i j}, \gamma_{i j}, \delta_{i j}$ are unknown parameters.

Applying Shephard’s lemma to the above cost function, we derive the conditional (with respect to output and the quasi-fixed factors) demand equations for the variable labour inputs $x_{f}^{v}, i \in{U S, M S, H S}:\left(x_{i}^{v}\right){s t}=\left(\partial \mathrm{C}(\cdot) / \partial w{i}\right){s, t^{-}}$The division of these factor demands by real value added yields a system of input-output coefficients of the following form: $$(\pi){i, t} \equiv\left(\frac{x_{i}^{v}}{y^{r}}\right){s, t}=\sum{j=1}^{n} \beta_{i j}\left(w / w_{i}\right){s, t}^{1 / 2}+\sum{j=1}^{m} \gamma_{i j}\left(x_{j}^{f}\right){s, t}+\delta{f 1} \times t+\delta_{i 2} \times t^{2}$$

## ACFI304 COURSE NOTES ：

Each bank $i$ faces the decision whether or not to adopt technology $g$, by maximizing its utility. Since the amount of customers per bank is fixed, this is equivalent to maximizing its utility per customer: $U i\left(s_{f}, s_{g}, t_{p}\right)$. This utility is a function of three parameters: (1) the market share of bank $i$, denoted $s_{i} ;(2)$ the combined share $s_{g}$ of other banks using $g$; and (3) $t_{e}$, which represents the technology used by bank $i$, where $t=g$ if the bank adopts $g$ and $t=f$ otherwise. Now:
$$\begin{array}{ll} U_{l}\left(s_{p}, s_{g}, f\right)=0 & \text { if the bank does not adopt } g \ U_{l}\left(s_{p}, s_{g}, g\right)=\left(s_{i}+s_{g}\right) b-c & \text { if it does adopt } g . \end{array}$$
Since by definition $b>c, s_{g}=1$ (all banks adopt $g$ ) is always a Nash-equilibrium: unilateral deviation (dropping or not adopting $g$ ) from this equilibrium will lower a bank’s utility from a positive number to zero. Now let $s_{1}$ be the market share of the largest bank. Then there is a second equilibrium where $s_{g}=0$ (no bank adopts $g$ ) if:
$$s_{1} b-c<0 \Leftrightarrow s_{1}<c / b$$

# 计算博弈论入门 Introduction to Computational Game Theory COMP323

0

Given a set $S \subseteq N$, let $\chi_{S}: N \rightarrow{0,1}$ be the indicator function of the set $S$, i.e., let $\chi_{S}(i)=1$ if $i \in S$ and $\chi_{S}(i)=0$ if $i \in N \backslash S$. A collection of sets $\mathcal{S} \subseteq 2^{N} \backslash{\emptyset}$ is said to be balanced if there exists a vector $\left(\delta_{S}\right){S \in \mathcal{S}}$ of positive numbers such that $$\sum{S_{\hookrightarrow} \mathcal{S}} \delta_{S} \chi_{S}=\chi_{N}$$
The vector $\left(\delta_{S}\right){S \in \mathcal{S}}$ is called the balancing weight system for $\mathcal{S}$. Observe that any coalition structure $C S$ over $N$ is a balanced set system with $\delta{S}=1$ for any $S \in C S$. Indeed, balanced set systems can be viewed as generalized partitions of $N$. We are now ready to state the theorem.

## COMP323 COURSE NOTES ：

ranges from 0 to $n-1$, and $W$ ranges from 0 to $w(N)$. For $s=0, j=1, \ldots, n-1$, we have
$$X[j, W, 0]= \begin{cases}1 & \text { if } W=0 \ 0 & \text { otherwise. }\end{cases}$$
Further, for $j=1, s=1, \ldots, n-1$ we have
$$X[1, W, s]= \begin{cases}1 & \text { if } W=w_{1} \text { and } s=1 \ 0 & \text { otherwise. }\end{cases}$$
Now, having computed the values $X\left[j^{\prime}, W^{\prime}, s^{\prime}\right]$ for all $j^{\prime}<j$, all $W^{\prime}=0, \ldots, w(N)$, and all $s^{\prime}=$ $0, \ldots, n-1$, we can compute $X[j, W, s]$ for $W=0 \ldots, w(N)$ and $s=1, \ldots, n-1$ as follows:
$$X[j, W, s]=X[j-1, W, s]+X\left[j-1, W-w_{j}, s-1\right]$$
In the equation above, the first term counts the number of subsets that have weight $W$ and size $s$ and do not contain $j$, whereas the second term counts the number of subsets of this weight and size that do contain $j$.

Thus, we can inductively compute $X[n-1, W, s]$ for all $W=0, \ldots, w(N)$ and all $s=$ $0, \ldots, n-1$. Now, $N_{s}, s=0, \ldots, n-1$, can be computed as
$$N_{s}=X\left[n-1, q-w_{n}, s\right]+\cdots+X[n-1, q-1, s]$$

# 全球经济中的商业 Business in the Global Economy MKIB225

0

Masquerading: MA poses as another agent to gain access to services or data at a host. False Identity: Host assumes false identity in order to lure MAs, or MAs use false identity to defraud the host and infrastructure services.

Denial of Service: MAs may attempt to consume or corrupt resources of a host to preclude other MAs from accessing the host’s services.
Lack of Capacity and Credentials: Host can ignore an MA’s request for services or access to resources due to overload or inappropriate supply of credential information.

## MKIB225 COURSE NOTES ：

Regardless of the end-user application, all kinds of MAs raise a number of security issues but specifically the security problems concerning:
1) The protection of the platform or host that runs MAs against these kinds of attacks which can harm or use their resources without protection, guards, access rights or permission.
2) The essential protection necessary to guard MAs against malicious hosts that might alter or forge information it carries when it visits hosts in its applications’ mobility itineraries.

# 会计学理论 Accounting Theory ACFI202

0

Given two vectors $d$ and $c$ with $n$ components each, let $\min (d, c)$ be the vector whose $i$-th component is the minimum of the $i$-th components of $d$ and $c$ for $i=1, \ldots$, n. For example:
$$\min ((3,7,2),(0,5,3))=(0,5,2)$$
If $d$ and $c$ are nonnegative, then $d-\min (d, c)$ and $c-\min (d, c)$ are both nonnegative, and:
$$[d / / c]=[d-\min (d, c) / / c-\min (d, c)]$$
For instance:
\begin{aligned} {[(3,7,2) / /(0,5,3)] } &=[(3,7,2)-(0,5,2) / /(0,5,3)-(0,5,2)] \ &=[(3,2,0) / /(0,0,1)] . \end{aligned}

## ACFI202 COURSE NOTES ：

where one might note the difference between the $\mathrm{T}$-account $A s s e t s(T+1)$ and the vector $A S S E T S(T+1)$. The liabilities are:
\begin{aligned} \text { Liabilities }(T+1) &=[(0, \ldots, 0) / /(0, \ldots, 0, D)] \ &=[(0, \ldots, 0) / / \operatorname{DEBTS}(T+1)] \end{aligned}
Hence the end-of-the-period resultant equation zero-term is:
Assets
$$[(\ldots, 0, C A S H(T+1), F G(T+1), R M(T+1), 1,0,0) / /(0, \ldots)]$$
Liabilities
$$+[(0, \ldots, 0) / /(0, \ldots, 0, D)]$$
Total Assets and Liabilities
$$\begin{gathered} +[(0, \ldots, 0, D) / /(\ldots, 0, C A S H(T+1), F G(T+1) \ R M(T+1), 1,0,0)] \end{gathered}$$

# 金融数学的数值分析 Numerical Analysis for Financial Mathematics MATH371

0

In the CIR model, the state dynamics under the physical measure are given by a square-root diffusion process
$$\mathrm{d} x_{i t}=\kappa_{i}\left(\theta_{i}-x_{i t}\right) \mathrm{d} t+\sigma_{i} \sqrt{x_{i t}} \mathrm{~d} z_{i t}$$
The equivalent martingale dynamics determine bond prices. Assume that the market price of risk of each factor $\lambda_{i}$ is proportional to the factor. The dynamics under the equivalent martingale measure are given by
$$\mathrm{d} x_{i t}=\left(\kappa_{i} \theta_{i}-\left(\kappa_{i}+\lambda_{i}\right) x_{i t}\right) \mathrm{d} t+\sigma_{i} \sqrt{x_{i t}} \mathrm{~d} \tilde{z}{i t} .$$ Zero-coupon bonds maturing at time $t+\tau$ have prices and yields given by \begin{aligned} &P\left(x{t}, \tau\right)=\exp \left[A(\tau)-B(\tau)^{\prime} x_{t}\right] \ &Y\left(x_{t}, \tau\right)=(1 / \tau)\left[-A(\tau)+B(\tau)^{\prime} x_{t}\right] \end{aligned}

## MATH371 COURSE NOTES ：

The log-likelihood function is derived from the prediction error decomposition of the KF and is given by
$$\ln L(\Psi)=-\frac{N m}{2} \ln 2 \pi-\frac{1}{2} \sum_{t=0}^{N} \ln \left|P_{y y, t}\right|-\frac{1}{2} \sum_{t=0}^{N} u_{t} P_{y y, t}^{-1} u_{t}^{\prime},$$
where $N$ is the number of time steps with $t={h, 2 h, \ldots, T-h, T}, N=$ $T / h, m$ is the number of different maturity bonds used, and $\Psi$ is the parameter vector.

To apply the KF in the one-factor CIR models, initialize the algorithm with the unconditional mean and variance of the state
$$\hat{x}{0}=\theta, \quad P{0}=\theta \frac{\sigma^{2}}{2 \kappa}$$

# 理论与实践中的网络 Networks in Theory and Practice MATH367

0

Let $X^{}$ be a solution to the optimization problem: Minimize $f(X)$ subject to: $$X \in \mathcal{K} \text {, }$$ where $f$ is continuously differentiable and $\mathcal{K}$ is closed and convex. Then $X^{}$ is a solution of the variational inequality problem: determine $X^{} \in \mathcal{K}$, such that $$\left\langle\nabla f\left(X^{}\right), X-X^{*}\right\rangle \geq 0, \quad \forall X \in \mathcal{K},$$
where $\nabla f(X)$ is the gradient vector of $f$ with respect to $X$.

## MATH367 COURSE NOTES ：

A Nash equilibrium is a strategy vector
$$X^{}=\left(X_{1}^{}, \ldots, X_{m}^{}\right) \in \mathcal{K},$$ such that $$U_{i}\left(X_{i}^{}, \hat{X}{i}^{}\right) \geq U{i}\left(X_{i}, \hat{X}{i}^{}\right), \quad \forall X{i} \in \mathcal{K}^{i}, \forall i$$
where $\hat{X}{i}^{}=\left(X{1}^{}, \ldots, X_{i-1}^{}, X_{i+1}^{}, \ldots, X_{m}^{*}\right)$.
In other words, under Nash equilibrium, no unilateral deviation in strategy by any single player is profitable for that player.

# 数学风险理论 Mathematical Risk Theory MATH366

0

$$p_{n, n}(t, t+h)=e^{-\alpha h},$$
the function $h \mapsto p_{n, n}(t, t+h)$ is continuous, which proves (ii). Finally, we have
\begin{aligned} \lim {h \rightarrow 0} \frac{1}{h}\left(1-p{n, n}(t, t+h)\right) &=\lim {h \rightarrow 0} \frac{1}{h}\left(1-e^{-\alpha h}\right) \ &=\alpha \end{aligned} as well as \begin{aligned} \lim {h \rightarrow 0} \frac{1}{h} p_{n, n+1}(t, t+h) &=\lim _{h \rightarrow 0} \frac{1}{h} e^{-\alpha h} \alpha h \ &=\alpha . \end{aligned}

## MATH366 COURSE NOTES ：

(1) We have already noticed in the preceding part of the proof that the differential equation
$$\frac{d}{d t} p_{k, k}(r, t)=-p_{k, k}(r, t) \lambda_{k+1}(t)$$
with initial condition $p_{k, k}(r, r)=1$ has the unique solution
$$p_{k, k}(r, t)=e^{-\int_{r}^{t} \lambda_{k+1}(s) d s} .$$
(2) Assume now that $k<n$. Then the function $t \mapsto 0$ is the unique solution of the homogeneous differential equation
$$\frac{d}{d t} p_{k, n}(r, t)=-p_{k, n}(r, t) \lambda_{n+1}(t)$$
with initial condition $p_{k, n}(r, r)=0$. This implies that the inhomogeneous differential equation
$$\frac{d}{d t} p_{k, n}(r, t)=p_{k, n-1}(r, t) \lambda_{n}(t)-p_{k, n}(r, t) \lambda_{n+1}(t)$$
with initial condition $p_{k, n}(r, r)=0$ has at most one solution.

# 测量理论和概率 Measure Theory and Probability MATH365

0

Now for each $n \geq 1$, let
Then, $\mu\left(A_{n}\right)<\frac{1}{2^{n}}$ and hence $$\sum_{n=1}^{\infty} \mu\left(A_{n}\right)<\infty$$ By the MCT, this implies that $\int_{[a, b]}\left(\sum_{n=1}^{\infty} I_{A_{n}}\right) d \mu<\infty$ and hence $$\sum_{n=1}^{\infty} I_{A_{n}}<\infty \text { a.e. } \mu \text {. }$$ Thus $h_{n} \rightarrow f_{K}$ a.e. $\mu$ on $[a, b]$. By Egorov’s theorem for any $\epsilon>0$, there is a set $A_{e} \in \mathcal{B}([a, b])$ such that
$$\mu\left(A_{\epsilon}^{c}\right)<\epsilon / 2 \text { and } h_{n} \rightarrow f_{K} \text { uniformly on } A_{\tau} \text {. }$$

## MATH365 COURSE NOTES ：

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f: \Omega \rightarrow[0, \infty]$ be such that it admits two representations
$$f=\sum_{i=1}^{k} c_{i} I_{A_{i}} \text { and } f=\sum_{j=1}^{\ell} d_{j} I_{B_{j}},$$
where $c_{i}, d_{j} \in[0, \infty]$, and $A_{i}$ and $B_{j} \in \mathcal{F}$ for all $i, j$. Show that
$$\sum_{i=1}^{k} c_{i} \mu\left(A_{i}\right)=\sum_{j=1}^{\ell} d_{j} \mu\left(B_{j}\right) .$$

# 医学统计 Medical Statistics MATH364

0

$$\varphi \sim \operatorname{Beta}\left(a_{0}, b_{0}\right),$$
with density
$$p(\varphi)=\frac{1}{\mathrm{~B}\left(a_{0}, b_{0}\right)} \varphi^{a_{0}-1}(1-\varphi)^{b_{0}-1}$$
The advantage is that the posterior is also a Beta distribution (hence the name conjugate), with density
$$p(\varphi \mid \text { data }) \propto \ell(\varphi \mid \text { data }) \times p(\varphi) \propto \varphi^{a_{0}+a-1}(1-\varphi)^{b_{0}+b-1}$$
The prior weights $a_{0}$ and $b_{0}$ are added to the observed counts $a_{1}$ and $b_{1}$, so that at the interim analysis
$$\varphi \mid \text { data } \sim \varphi \mid a_{1} \sim \operatorname{Beta}\left(a_{1}+a_{0}, b_{1}+b_{0}\right) .$$
The predictive distribution, which is a mixture of Binomial distributions, is naturally called a Beta-Binomial distribution
$$a_{2} \mid a_{1} \sim \operatorname{Beta}-\operatorname{Bin}\left(a_{1}+a_{0}, b_{1}+b_{0} ; n_{2}\right) .$$

## ECON364 COURSE NOTES ：

$$\delta(\mathbf{x})=\frac{\partial E[y \mid \mathbf{x}]}{\partial \mathbf{x}}=\exp \left(\boldsymbol{\beta}^{\prime} \mathbf{x}\right) \times \boldsymbol{\beta}$$
As in any regression model, this measure is a function of the data point at which it is evaluated. For analysis of the Poisson model, researchers typically use one of the two approaches: The marginal effects, computed at the mean, or the center of the data are
$$\delta(\overline{\mathbf{x}})=\frac{\partial E[y \mid \overline{\mathbf{x}}]}{\partial \overline{\mathbf{x}}}=\exp \left(\boldsymbol{\beta}^{\prime} \overline{\mathbf{x}}\right) \times \boldsymbol{\beta},$$
where $\overline{\mathbf{x}}=(1 / N) \sum_{i=1}^{N} \mathbf{x}{i}$ is the sample mean of the data. An alternative, commonly used measure is the set of average partial effects, $$\bar{\delta}(\mathbf{X})=\frac{1}{N} \sum{i=1}^{N} \frac{\partial E\left[y \mid \mathbf{x}{i}\right]}{\partial \mathbf{x}{i}}=\frac{1}{N} \sum_{i=1}^{N} \exp \left(\boldsymbol{\beta}^{\prime} \mathbf{x}_{i}\right) \times \boldsymbol{\beta} .$$

# 拓扑学 Topology MATH346

0

When $W$ is finite, $L_{T}$ is a simplex for each $T \neq S$ and $L_{S}=\emptyset$. Hence,
$$1-\chi\left(L_{T}\right)= \begin{cases}0 & \text { if } T \neq S, \ 1 & \text { if } T=S .\end{cases}$$
So when $W$ is finite the theorem is the tautology $1 / W(\mathbf{t})=1 / W(\mathbf{t})$.
Suppose $W$ is infinite. We can rewrite Corollary (ii) as
$$\frac{1}{W(t)}=-\varepsilon(S) \sum_{T \subsetneq S} \frac{\varepsilon(T)}{W_{T}(t)}$$
The proof is by induction on $\operatorname{Card}(T)$. For any $T \subset S$, let $\mathcal{S}(T)$ be the set of spherical subsets of $T$ and for any $U \in \mathcal{S}(T)$, let $L_{U}(T)$ be the simplicial complex corresponding to $\mathcal{S}(T), U$. Using (17.11) and the inductive hypothesis, we get
$$\frac{1}{W(\mathbf{t})}=-\varepsilon(S) \sum_{T \subsetneq S} \varepsilon(T) \sum_{U \in \mathcal{S}(T)} \frac{1-\chi^{\left(L_{U}(T)\right)}}{W_{U}(\mathbf{t})}$$
The coefficient of $1 / W_{U}(t)$ on the right hand side is
$$-\varepsilon(S) \sum_{U \subset T \subseteq S} \varepsilon(T)\left[1-\chi\left(L_{U}(T)\right)\right] .$$

(Right-angled polygon groups.) Suppose $W$ is right angled with nerve a $k$-gon, $k \geqslant 4$, and that $\mathbf{t}$ is a single indeterminate $t$. Using Theorem as before, we get
$$\frac{1}{W(t)}=1-\frac{k t}{1+t}+\frac{k t^{2}}{(1+t)^{2}}=\frac{t^{2}+(2-k) t+1}{(1+t)^{2}} .$$
The roots of the numerator are $\rho$ and $\rho^{-1}$; so
$$\rho^{\pm 1}=\frac{(k-2) \mp \sqrt{k^{2}-4 k}}{2},$$
e.g., $\rho=\frac{3-\sqrt{5}}{2}$ when $k=5$.