EA答辩

今天历时一个学期的EA答辩结束了，随手写几点吧。

1.对一个方向特别感兴趣，和真正钻进去研究和学习还是很不一样的。有时候我们满足于自己在某些方面有点见解，可是真正投入到一个方向上，那绝对是一个全新的战场。自己是菜鸟，而其他人都是遍地老手。这个时候才是检验是否是真爱吧？

2.我就是这样一个例子。想着要做随机几何想了好久，终于有机会上手试试了，上述就是我的一些真实感受。然而，当中有那么一段时间全身心投入，拼了命想折腾点东西的劲头还是感动了自己。以及最后写报告时发现好些证明似是而非，只能一一自己补全，也算是一种科研锻炼吧。

3.终了，还是非常喜欢这个方向。有机会让我再去画些奇奇怪怪的东西。以后还要加油干呢。导师也说现在既然已经略窥门径了，要再接再厉啊。比如说做个什么问题或者证明吧，不能说是等着人家告诉你能不能证明和怎么证明，要是这样岂不是变成了DM了么？

4.答辩的时候，PPT要少弄一点。老师说一分钟看一张，快了大家就不开心了……哦不开心了额。

5.以后读论文，粗读一遍看大意，略读一遍掌握思路，然后必须要精读一遍（如果是钻研论文的话）验算过程啊！！！

6.毕竟后来没有拍照合影。我觉得我还是需要一个自己的成果才能填满欲望。定个小目标，3A结束前写出一篇论文吧。

就说这么多了，加油！

Erlang and Jackson Network

This is a note for reviewing the MAP554 and some points about network.

M/M/1, M/M/$\infty$, birth and death

The basic model of queue theory. M/M/1 has just one server and has an invariant measure like geometric law,
$$\pi(n) = p^n(1-p), p = \frac{\lambda}{\mu}$$
M/M/$\infty$ has infinite server and Poisson law
$$\pi(n) = e^{-p} \frac{p^n}{n!}, p = \frac{\lambda}{\mu}$$
where $\lambda$ is the rate of arrival and $\mu$ the rate of waiting. A more general case can be done like change of power.

Erlang network:

This is just an application for truncated technique. That is if we have already a network with reversible invariant measure, we can generate a new by changing the power of that part. That is  

$$\begin{eqnarray*} \tilde{q}(x,y) &=& C q(x,y), \forall x \in \mathcal{A}, y \in \mathcal{S} - \mathcal{A}\\ \tilde{q}(x,y) &=& q(x,y) , \text{ otherwise } \end{eqnarray*}$$
Then the new invariant measure becomes
$$\begin{eqnarray*} \tilde{\pi}(x) &=& K\pi(x) , \forall x \in \mathcal{A} \\ \tilde{\pi}(y) &=& KC\pi(y) , \forall y \in \mathcal{S} -\mathcal{A}\\ K &=& \frac{1}{\pi(\mathcal{A}) + \pi(\mathcal{S} - \mathcal{A})} \end{eqnarray*}$$
The application is that we make $C = 0$ then the network is defined in just the part $\mathcal{A}$. For example, in the network of route with restriction $\mathcal{R}$, we can just do the case without restriction to get $\pi$, which is just the case of several M/M/1 independent, then we do restriction and normalization.
$$\tilde{\pi}(x) = K\pi(x) , \forall x \in \mathcal{A}, K = \frac{1}{\sum_{x \in \mathcal{R}} \pi(x)}$$

Jackson network

A more general model of network is like that. Each station has rate $\lambda_i$ of arrival and $\phi_i(n_i)\mu_i$ rate to tackling the service. Here $\phi_i(n_i)$ can be considered as the power of server, in the case M/M/1 it is always 1 and M/M/$\infty$ it is always $\phi_i(n_i) = n_i$. However, the difference is that after each service of station $i$, it has possibility $r_{ij}$ to go to the station j

The key is to find a equivalent $\tilde{\lambda}_i$ which satisfies that
$$\tilde{\lambda}_i = \lambda_i + \sum_j \tilde{\lambda}_j r_{ji}$$
then the station looks like independent and has the invariant measure
$$\pi(n) = \Pi_i \frac{\tilde{p}_i^{n_i}}{\Pi_{m=1}^{n_i}\phi_i(m)}, \tilde{p}_i = \frac{\tilde{\lambda}_i}{\mu_i}$$

Levy characterization, representation of martingale and change of probability

I am preparing for the final, so I write some notes for the course maths finance.

Levy characterization for Brownian motion: 
If $\phi_s^T \phi_s = Id$, then

$B_t = \int^t_0 \phi_s dW_s$

is a standard Brownian motion

This theorem is very useful and it describes the nature that after a con-formal transform, the BM keeps its properties.

Representation of martingale:

This theorem has different version. The most general version is that for a $\mathcal{F}_t$ adapted local martingale $M_t$, it can be written as 

$M_t = \mathbb{E}[M_t] + \int_0^t H_s dWs$

where $H_t \in \mathbb{H}^2_{loc}$.

This is a mathematical version of perfect duplication theorem. The proof starts from the case $L^2 \text{martingale} \rightarrow L^1 \text{martingale} \rightarrow \text{local martingale}$. It has many application in the stochastic calculus.

Change of probability: 

First we define $Z_T = \exp{(\int_0^T \phi_s dW_s - \frac{1}{2}\phi_s^2 ds)}$. Generally, it's only a local martingale and if it satisfies $\mathbb{E}[Z_T] = 1$, we can define a change of probability

$\frac{d\mathbb{Q}}{d\mathbb{P}} = Z_T$ 

then under the new probability $\mathbb{Q}$, we can define a new BM in the form

$\tilde{B}_t = B_t - \int_0^t \phi_s ds$

We remark that in the case $\phi$ is deterministic, then the there is no problem since in this case, $Z_T$ is well defined of expectation 1. Otherwise, the expectation is not so clear but there is a theorem Novikov, says that if $\exp{(\int_0^T \frac{1}{2}\phi_s^2 ds)} < \infty$, then all the condition is satisfied.

The change of probability  can simplify the formula and has important applications on Monte-Carlo algorithms.