统计代写|随机过程代写stochastic process代考|STAT4061

统计代写|随机过程代写stochastic process代考|Partially observed data

Assume now that the Markov chain is only observed at a number of finite time points. Suppose, for example, that $x_0$ is a known initial state and that we observe $\mathbf{x}o=\left(x{n_1}, \ldots, x_{n_m}\right)$, where $n_1<\ldots<n_m \in N$. In this case, the likelihood function is
$$l\left(\boldsymbol{P} \mid \mathbf{x}o\right)=\prod{i=1}^m p_{n_{i-1} n_i}^{\left(t_i-t_{i-1}\right)}$$
where $p_{i j}^{(t)}$ represents the $(i, j)$ th element of the $t$ step transition matrix, defined in Section 1.3.1. In many cases, the computation of this likelihood will be complex. Therefore, it is often preferable to consider inference based on the reconstruction of missing observations. Let $\mathbf{x}m$ represent the unobserved states at times $1, \ldots$, $t_1-1, t_1+1, \ldots, t{n-1}-1, t_{n-1}+1, \ldots, t_n$ and let $\mathbf{x}$ represent the full data sequence. Then, given a matrix beta prior, we have that $\boldsymbol{P} \mid \mathbf{x}$ is also matrix beta. Furthermore, it is immediate that
$$P\left(\mathbf{x}_m \mid \mathbf{x}_o, \boldsymbol{P}\right)=\frac{P(\mathbf{x} \mid \boldsymbol{P})}{P\left(\mathbf{x}_o \mid \boldsymbol{P}\right)} \propto P(\mathbf{x} \mid \boldsymbol{P}),$$
which is easy to compute for given $\boldsymbol{P}, \mathbf{x}_m$. One possibility would be to set up a Metropolis within Gibbs sampling algorithm to sample from the posterior distribution of $\boldsymbol{P}$.

Such an approach is reasonable if the amount of missing data is relatively small. However, if there is much missing data, it will be very difficult to define an appropriate algorithm to generate data from $P\left(\mathbf{x}m \mid \mathbf{x}_o, \boldsymbol{P}\right)$ in (3.5). In such cases, one possibility is to generate the elements of $\mathbf{x}_m$ one by one, using individual Gibbs steps. Thus, if $t$ is a time point amongst the times associated with the missing observations, then we can generate a state $x_t$ using $$P\left(x_t \mid \mathbf{x}{-t}, \boldsymbol{P}\right) \propto p_{x_{t-1} x_t} p_{x_t x_{t+1}}$$
where $\mathbf{x}_{-t}$ represents the complete sequence of states except for the state at time $t$.

统计代写|随机过程代写stochastic process代考|Higher order chains and mixtures of Markov chains

Bayesian inference for the full $r$ th order Markov chain model can, in principle, be carried out in exactly the same way as inference for the first-order model, by expanding the number of states appropriately, as outlined in Section 3.2.2.

Example 3.11: In the Australian rainfall example, Markov chains of orders $r=2$ and 3 were considered. In each case, $\operatorname{Be}(1 / 2,1 / 2)$ priors were used for the first nonzero element of each row of the transition matrix and it was assumed that the initial $r$ states were generated from the equilibrium distribution. Then, the predictive equilibrium probabilities of the different states under each model are as follows The log marginal likelihoods are $-30.7876$ for the second-order model and $-32.1915$ for the third-order model, respectively, which suggest that the simple Markov chain model should be preferred.

Bayesian inference for the MTD model of (3.1) is also straightforward. Assume first that the order $r$ of the Markov chain mixture is known. Then, defining an indicator variable $Z_n$ such that $P\left(Z_n=z \mid \mathbf{w}\right)=w_z$, observe that the mixture transition model can be represented as
$$P\left(X_n=x_n \mid X_{n-1}=x_{n-1}, \ldots, X_{n-r}=x_{n-r}, Z_n=z, P\right)=p_{x_{n-z} x_n} .$$
Then, a posteriori,
$$P\left(Z_n=z \mid X_n=x_n, \ldots, X_{n-r}=x_{n-r}, Z_n=z, P\right)=\frac{w_z p_{x_{n-z} x_n}}{\sum_{j=1}^r w_j p_{x_{n-j} x_n}}$$

随机过程代考

统计代写|随机过程代写随机过程代考|部分观测数据

$$l\left(\boldsymbol{P} \mid \mathbf{x}o\right)=\prod{i=1}^m p_{n_{i-1} n_i}^{\left(t_i-t_{i-1}\right)}$$
，其中$p_{i j}^{(t)}$表示在1.3.1节中定义的$t$阶跃跃迁矩阵的第$(i, j)$个元素。在许多情况下，这种可能性的计算将是复杂的。因此，考虑基于缺失观察的重建的推断通常是可取的。让$\mathbf{x}m$表示每次$1, \ldots$、$t_1-1, t_1+1, \ldots, t{n-1}-1, t_{n-1}+1, \ldots, t_n$未观察到的状态，让$\mathbf{x}$表示完整的数据序列。然后，给定一个矩阵先验，我们有$\boldsymbol{P} \mid \mathbf{x}$也是矩阵。此外，
$$P\left(\mathbf{x}_m \mid \mathbf{x}_o, \boldsymbol{P}\right)=\frac{P(\mathbf{x} \mid \boldsymbol{P})}{P\left(\mathbf{x}_o \mid \boldsymbol{P}\right)} \propto P(\mathbf{x} \mid \boldsymbol{P}),$$

统计代写|随机过程代写随机过程代考|高阶链和马尔可夫链的混合物

$$P\left(X_n=x_n \mid X_{n-1}=x_{n-1}, \ldots, X_{n-r}=x_{n-r}, Z_n=z, P\right)=p_{x_{n-z} x_n} .$$

$$P\left(Z_n=z \mid X_n=x_n, \ldots, X_{n-r}=x_{n-r}, Z_n=z, P\right)=\frac{w_z p_{x_{n-z} x_n}}{\sum_{j=1}^r w_j p_{x_{n-j} x_n}}$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: