### 统计代写|应用随机过程代写Stochastic process代考| Forecasting stationary behavior

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|应用随机过程代写Stochastic process代考|Forecasting stationary behavior

Often interest lies in the stationary distribution of the chain. For a low-dimensional chain where the exact formula for the equilibrium probability distribution can be derived, this is straightforward.

Example 3.5: Suppose that $K=2$ and $\boldsymbol{P}=\left(\begin{array}{cc}p_{11} & 1-p_{11} \ 1-p_{22} & p_{22}\end{array}\right)$. Then the equilibrium probability of being in state 1 can easily be shown to be
$$\pi_{1}=\frac{1-p_{22}}{2-p_{11}-p_{22}}$$
and the predictive equilibrium distribution is
$$E\left[\pi_{1} \mid \mathbf{x}\right]=\int_{0}^{1} \int_{0}^{1} \frac{1-p_{22}}{2-p_{11}-p_{22}} f\left(p_{11}, p_{22} \mid \mathbf{x}\right) \mathrm{d} x$$
which can be evaluated by simple numerical integration techniques.
Example 3.6: In the Sydney rainfall example, we have
$$E\left[\pi_{1} \mid \mathbf{x}\right]=E\left[\frac{1-p_{22}}{2-p_{11}-p_{22}} \mid \mathbf{x}\right]=0.655$$
so that we predict that it does not rain on approximately $65 \%$ of the days at this weather center.

For higher dimensional chains, it is simpler to use a Monte Carlo approach as earlier so that given a Monte Carlo sample $\boldsymbol{P}^{(1)}, \ldots, \boldsymbol{P}^{(S)}$ from the posterior distribution of $\boldsymbol{P}$, then the equilibrium distribution can be estimated as
$$E[\pi \mid \mathbf{x}] \approx \frac{1}{S} \sum_{s=1}^{s} \pi^{(s)}$$
where $\pi^{(s)}$ is the stationary distribution associated with the transition matrix $\boldsymbol{P}^{(s)}$.

## 统计代写|应用随机过程代写Stochastic process代考|Model comparison

One may often wish to test whether the observed data are independent or generated from a first (or higher) order Markov chain. The standard method of doing this is via Bayes factors (see Section 2.2.2).

Example 3.7: Given the experiment proposed at the start of section 3.2, suppose that we wish to compare the Markov chain model $\left(\mathcal{M}_{1}\right)$ with the assumption that the data

are independent and identically distributed with some distribution $\mathbf{q}=\left(q_{1}, \ldots, q_{K}\right)$, $\left(\mathcal{M}{2}\right)$ where we shall assume a Dirichlet prior distribution, $$\mathbf{q} \sim \operatorname{Dir}\left(a{1}, \ldots, a_{K}\right)$$
Then,
\begin{aligned} f\left(\mathbf{x} \mid \mathcal{M}{1}\right) &=\int{f=1} f(\mathbf{x} \mid \boldsymbol{P}) f\left(\boldsymbol{P} \mid \mathcal{M}{1}\right) \mathrm{d} \boldsymbol{P} \ &=\prod{i=1}^{k} \frac{\Gamma\left(\alpha_{i}\right)}{\Gamma\left(n_{i-}+\alpha_{i}\right)} \prod_{j=1}^{k} \frac{\Gamma\left(\alpha_{i j}+n_{i j}\right),}{\Gamma\left(\alpha_{i j}\right)} \end{aligned}
where $n_{i=}=\sum_{j=1}^{k} n_{i j}$ and $\alpha_{i-}=\sum_{j=1}^{K} \alpha_{i j}$. Also, under the independent model, we have
$$f\left(\mathbf{x} \mid \mathcal{M}{2}\right)=\frac{\Gamma(a)}{\Gamma(a+n)} \prod{i=1}^{K} \frac{\Gamma\left(a_{i}+n_{i}\right)}{\Gamma\left(a_{i}\right)}$$
where $a=\sum_{i=1}^{K} a_{i}$ and $n_{i}$ is the number of times that event $i$ occurs (discounting the initial state $X_{0}$ ). The Bayes factor can now be calculated as the ratio of the two marginal likelihood functions, as illustrated.

## 统计代写|应用随机过程代写Stochastic process代考|Unknown initial state

When the initial state, $X_{0}$, is not fixed in advance, to implement Bayesian inference, we need to define a suitable prior distribution for $X_{0}$. The standard approach is simply to assume a multinomial prior distribution, $P\left(X_{0}=x_{0} \mid \theta\right)=\theta_{x_{0}}$ where $0<\theta_{k}<1$ and

$\sum_{k=1}^{K} \theta_{k}=1$. Then, we can define a Dirichlet prior for the multinomial parameters, $\operatorname{say} \theta \sim \operatorname{Dir}(\gamma)$ so that, a posteriori, $\theta \mid \mathbf{x} \sim \operatorname{Dir}\left(\gamma^{\prime}\right)$, with $\gamma_{x_{0}}^{\prime}=\gamma_{x_{0}}+1$ and, otherwise, $\gamma_{i}^{\prime}=\gamma_{i}$ for $i \neq x_{0}$. Inference for $\boldsymbol{P}$ then proceeds as before.

An alternative approach, which may be reasonable if it is assumed that the chain has been running for some time before the start of the experiment, is to assume that the initial state is generated from the equilibrium distribution, $\pi$, of the Markov chain. Then, making the dependence of $\pi$ on $\boldsymbol{P}$ obvious, the likelihood function becomes
$$l(\boldsymbol{P} \mid \mathbf{x})=\pi\left(x_{0} \mid \boldsymbol{P}\right) \prod_{i=1}^{K} \prod_{j=1}^{K} p_{i j}^{n_{i j}} .$$
In this case, simple conjugate inference is impossible but, given the same prior distribution for $\boldsymbol{P}$ as above, it is straightforward to generate a Monte Carlo sample of size $S$ from the posterior distribution of $\boldsymbol{P}$ using, for example, a rejection sampling algorithm as follows:
For $s=1, \ldots, S$ :
For $i=1, \ldots, K$, generate $\tilde{\mathbf{p}}{i} \sim \operatorname{Dir}\left(\alpha^{\prime}\right)$ with $\alpha^{\prime}$ as in (3.4). Set $\tilde{\boldsymbol{P}}$ to be the transition probability matrix with rows $\tilde{p}{1}, \ldots, \tilde{p}{K}$. Calculate the stationary probability function $\tilde{\pi}$ satisfying $\tilde{\pi}=\tilde{\pi} \tilde{\boldsymbol{P}}$. Generate $u \sim \mathrm{U}(0,1)$. If $u<\tilde{\pi}\left(x{0}\right)$, set $\mathbf{P}^{(s)}=\tilde{\boldsymbol{P}}$. Otherwise repeat from step $1 .$

## 统计代写|应用随机过程代写Stochastic process代考|Model comparison

F(X∣米1)=∫F=1F(X∣磷)F(磷∣米1)d磷 =∏一世=1ķΓ(一种一世)Γ(n一世−+一种一世)∏j=1ķΓ(一种一世j+n一世j),Γ(一种一世j)

F(X∣米2)=Γ(一种)Γ(一种+n)∏一世=1ķΓ(一种一世+n一世)Γ(一种一世)

## 统计代写|应用随机过程代写Stochastic process代考|Unknown initial state

∑ķ=1ķθķ=1. 然后，我们可以为多项式参数定义 Dirichlet 先验，说⁡θ∼目录⁡(C)因此，后验，θ∣X∼目录⁡(C′)， 和CX0′=CX0+1并且，否则，C一世′=C一世为了一世≠X0. 推断磷然后像以前一样进行。

l(磷∣X)=圆周率(X0∣磷)∏一世=1ķ∏j=1ķp一世jn一世j.

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