Stochastic Process

A stochastic process is a family of random variables ${X_t, t\in\mathcal{T}}$, where $\mathcal{T}$ is an index set.

• Example: Amount of inventory in stock at the start of day $t$

1. Stationary Process

${X_t}$ is stationary IF $\forall s$, $(X_{s+t_1}, X_{s+t_2},\ldots, X_{s+t_k})$ is independent of $s$ when $t_1<t_2<\ldots<t_k$.

• Stationary implies equal mean and varaince $\mu = E(X_t)$ and $\sigma^2 = Var(X_t)$

1.1 Week Stationary

${X_t}$ is weakly stationary IF the $X_t$ have equal mean, variance $\sigma_X = Var(X_t) < \infty$, and $Cov(X_t, X_{t+s})$ depends only on “lag” $s$.

If the joint distributions of the $X_t$ are normal with nonsingular covariance matrices, weak stationarity implies stationarity because the joint distribution of the vector $(X_{s+t_1}, X_{s+t_2},\ldots, X_{s+t_k})$ uniquely defined by the respective covariance matrix whose elements are independent of s.

考虑如下示例，因为 week stationarity，我们已知 (1) $\mu_t=\mu_{t+s}=\mu$, $\sigma_t=\sigma_{t+s}=\sigma_X$ (2) $\rho$ 只与 $s$ 相关。因此 $(X_t, X_{t+s})$ 与 $(X_{t+k}, X_{t+s+k})$ 的联合分布完全相同，从而证明了 stationarity

[ \begin{bmatrix} X_t
X_{t+s} \end{bmatrix} \sim \mathcal{N} \left( \begin{bmatrix} \mu_t
\mu_{t+s} \end{bmatrix}, \begin{bmatrix} \sigma^2t & \rho\sigma_t\sigma{t+s}
\rho\sigma_t\sigma_{t+s} & \sigma^2_{t+s} \end{bmatrix} \right) ]

1.2 Autocovariance and Autocorrelation

Autocovariance: $C_k = Cov(X_t, X_{t+k})$ Autocorrelation: $\rho_k = Corr(X_t, X_{t+k}) = C_k/\sigma_X^2$

[ E\left(\sum_{i=1}^{n} a_i X_i + b\right) = \left(\sum_{i=1}^{n} a_i \mu\right) + b ]

[ \begin{aligned} \text{Var}\left(\sum_{i=1}^{n} a_i X_i + b\right) &= \sum_{i=1}^{n} a_i^2 \text{Var}(X_i) + 2 \sum_{i<j} a_i a_j \text{Cov}(X_i, X_j)
&= \left(\sum_{i=1}^{n} a_i^2\right) \sigma_X^2 + 2 \sum_{i<j} a_i a_j C_{j-i} \end{aligned} ]

[ \text{Cov}\left(\sum_{i=1}^{n} a_i X_i + b, \sum_{j=1}^{m} c_j Y_j + d\right) = \sum_{i=1}^{n} \sum_{j=1}^{m} a_i c_j \text{Cov}(X_i, Y_j) ]

1.3 Paramter Estimation

$\delta=\mu=E(X_t)$

(1) Sample mean $\delta_n = \bar{X}n = \frac{1}{n}\sum{i=1}^{n} X_i$

(2) Variance of sample mean \(\begin{aligned} Var(\bar{X}_n) &= \frac{1}{n^2} Var\left(\sum_{i=1}^{n} X_i\right) = \frac{1}{n^2}\left[\sum_{i=1}^{n} Var(X_i) + 2\sum_{i<j} C_{j-i}\right]\\ &= \frac{1}{n^2}\left[n\sigma_X^2 + 2\sum_{k=1}^{n-1} (n-k) C_k\right]\\ &= \frac{\sigma_X^2}{n}\left[1 + 2\sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right) \rho_k\right]\\ &= \frac{\sigma_X^2}{n}a_n \\ \end{aligned}\)

Since $Var(\bar{X}n) = \sigma^2_X/n$ if $X_t$ are independent, we can see $\frac{2}{n}\sum{k=1}^{n-1} \left(1-\frac{k}{n}\right) C_k$ as the autocorrelation penalty

(3) Variance [ \sigma^2 = \lim_ n \text{Var}(\bar{X}n) = \sum^{\infty} C_k = \sigma_X^2 + 2 \sum_^{\infty} C_k ]

If the following exists

[ \lim_ \sum_^{n-1} \left(1 - \frac{k}{n}\right) C_k < \infty \iff \sum_^{\infty} C_k < \infty ]

1.4 CLT for Stationary Processes

[ \frac{\bar{X}_n - \mu}{\sigma/\sqrt{n}} \xrightarrow{d} \mathcal{N}(0,1) ]

When $X_t$ are positively correlated, so that $a_n > 1$ [ E\left(\frac{S^2_n}{n}\right) = \frac{n/a_n - 1}{n - 1} \text{Var}(\bar{X}_n) < \text{Var}(\bar{X}_n) ]

then the confidence interval

[ \bar{X}n \pm t{n-1,1-\alpha/2} \frac{S_n}{\sqrt{n}} ]

2. Moving Average Process

Suppose ${\varepsilon_i}$ is an i.i.d. sequence with $E(\varepsilon_i)=0$ and $Var(\varepsilon_i)=\sigma^2_\varepsilon < \infty$. The MA Process of order $q$ is defined by

[ X_t = \beta_0\varepsilon_t + \beta_1\varepsilon_{t-1} + \ldots + \beta_q\varepsilon_{t-q} ]

where $\beta_0 \neq 0$ and $\beta_1, \ldots, \beta_q$ are constants.

We can easily know that $E(X_t)=0$, $Var(X_t)=\sigma^2\varepsilon\sum{i=0}^{q}\beta_i^2$

[C_k = Cov(X_t, X_{t+k}) = \begin{cases} \sigma^2\varepsilon\sum{i=0}^{q-k}\beta_i\beta_{i+k} & \text{if } 0 \leq k \leq q
0 & \text{if } k > q \end{cases}]

that’s because $Cov(\varepsilon_i, \varepsilon_j)=0$ for $i \neq j$.

3. Autoregressive Process

AR Process is defined as [X_t=\alpha_1X_{t-1} + \alpha_2X_{t-2} + \ldots + \alpha_pX_{t-p} + \varepsilon_t]

where $\alpha_1, \ldots, \alpha_p$ are constants and $\varepsilon_t$ is a white noise process with $E(\varepsilon_t)=0$ and $Var(\varepsilon_t)=\sigma^2_\varepsilon < \infty$.

3.1 AR(1) Process

[X_t=\alpha X_{t-1} + \varepsilon_t]

assuming zero mean $E(X_t)=0$. Therefore, [X_t=\alpha^tX_0 + \sum_{i=0}^{t-1}\alpha^i\varepsilon_{t-i}]

[X_t = \sum_{i=0}^{\infty}\alpha^i\varepsilon_{t-i}]

if non-zero mean $E(X_t)=\mu$, then $X_t-\mu$ is AR(1)

Because $\varepsilon_t$ are iid, then the marginal variance:

[\sigma_X^2 = Var(X_t) = \sigma^2\varepsilon(1+\alpha^2+\alpha^4+\ldots) = \frac{\sigma^2\varepsilon}{1-\alpha^2}]

If we multiply both sides of $X_t=\alpha X_{t-1} + \varepsilon_t$ by $X_{t-1}$ and take expectation, we have

[\begin{align} X_{t-k}X_t &= \alpha X_{t-k}X_{t-1} + X_{t-k}\varepsilon_t
E(X_{t-k}X_t) &= \alpha E(X_{t-k}X_{t-1}) + E(X_{t-k})E(\varepsilon_t) \quad \text{(independence)}
C_k &= \alpha C_{k-1} \end{align} ]

that’s because:

• $Cov(X,Y)=E[(X-E(X))(Y-E(Y))]$
• $E[X_t]=0$ and $E[\varepsilon_t]=0$

Plus that $C_0=Var(X_t)=\sigma^2_X$, we have

[C_k = \alpha^k\sigma^2_X\;\;\;\text{and}\;\;\;\rho_k=\alpha^k]

Some Facts:

• AR(1) is weakly stationary if $

  \alpha

  < 1$

3.1 Gaussian AR(1) Process

Assume $\varepsilon_t \sim \mathcal{N}(0, 1-\alpha^2)$ and $X_0 \sim \mathcal{N}(0, 1)$

Easily to show that

1. $X_t \sim \mathcal{N}(0, 1)$ (by take expectation of AR(1) expression)
2. $C_k = \alpha^k$ (since $\sigma_X^2=1$)

The asymptotic variance of this process