Strictly Stationary

Definition

A stochastic process $\{Y_t\}$ is strictly stationary

If for any $n\in {\mathbb{Z}}^{+}$, any set of time points $\{t_1,t_2,\dots ,t_n\}$, lag $k$ the joint cumulative distribution function of $(Y_{t_1},Y_{t_2},\dots ,Y_{t_n})$ is identical to that of $(Y_{t_1+k},Y_{t_2+k},\dots ,Y_{t_n+k})$

Relationship with Weak Stationarity

Let $\{Y_t\}$ be strictly stationary process.

If $\text{Var}(Y_t)$ is finite for all $t$

The $\{Y_t\}$ is also weakly stationary

Converse is Not True

Weak stationarity does NOT imply strict stationarity. A process can have constant mean and covariance without having identical distributions at different times.

Procedure: Proving Strict Stationarity

The standard approach uses the i.i.d. white noise representation:

1. Express $Y_t$ as a measurable function of i.i.d. random variables

   • e.g., $Y_t=f(a_t,a_{t-1},a_{t-2},\dots )$

2. Write the shifted version $Y_{t+k}$ in terms of the same i.i.d. variables with shifted indices

   • e.g., $Y_{t+k}=f(a_{t+k},a_{t+k-1},a_{t+k-2},\dots )$

3. Argue that the joint distribution of $(Y_{t_1+k},\dots ,Y_{t_n+k})$ equals $(Y_{t_1},\dots ,Y_{t_n})$ because:

   • i.i.d. variables are exchangeable under index shifts
   • The transformation $f$ is the same for all time points

4. Conclude: All finite-dimensional distributions are time-invariant $⟹$ strictly stationary

Key Requirement

Strict stationarity requires:

• The process depends on i.i.d. (not just uncorrelated) variables
• Time enters only through indices of these i.i.d. variables
• No explicit time dependence (like $t\cdot a_t$ or $t+Y_{t-1}$)

Example: MA(1) Process

Process: $Y_t=a_t-\theta a_{t-1}$ where $\{a_t\}$ are i.i.d. white noise.

Claim: $\{Y_t\}$ is strictly stationary.

Proof:

1. $Y_t$ depends on $(a_t,a_{t-1})$
2. Shifted version: $Y_{t+k}=a_{t+k}-\theta a_{t+k-1}$
3. For any $(t_1,\dots ,t_n)$: $(Y_{t_1+k},\dots ,Y_{t_n+k})=(a_{t_1+k}-\theta a_{t_1+k-1},\dots ,a_{t_n+k}-\theta a_{t_n+k-1})$
4. Since $\{a_t\}$ are i.i.d., the joint distribution of $(a_{t_1+k},a_{t_1+k-1},\dots )$ equals that of $(a_{t_1},a_{t_1-1},\dots )$
5. Therefore $(Y_{t_1+k},\dots ,Y_{t_n+k})\stackrel{d}{=}(Y_{t_1},\dots ,Y_{t_n})$

$■$

Example: AR(1) Process with $|\varphi |<1$

Process: $Y_t=\varphi Y_{t-1}+a_t$ where $\{a_t\}$ are i.i.d. and $|\varphi |<1$.

Claim: $\{Y_t\}$ is strictly stationary.

Proof:

1. Solve recursively: $Y_t={\sum }_{j=0}^{\infty }{\varphi }^j{a}_{t-j}$ (converges since $|\varphi |<1$)
2. Shifted version: $Y_{t+k}={\sum }_{j=0}^{\infty }{\varphi }^j{a}_{t+k-j}$
3. For any $(t_1,\dots ,t_n)$ and lag $k$:
   • $(Y_{t_1+k},\dots ,Y_{t_n+k})$ depends on $(a_{t_1+k},a_{t_1+k-1},\dots )$
   • This has the same joint distribution as $(a_{t_1},a_{t_1-1},\dots )$ by i.i.d.
4. Therefore $(Y_{t_1+k},\dots ,Y_{t_n+k})\stackrel{d}{=}(Y_{t_1},\dots ,Y_{t_n})$

$■$

Example: Random Walk (NOT Strictly Stationary)

Process: $Y_t=Y_{t-1}+a_t={\sum }_{i=1}^t{a}_i$ where $\{a_t\}$ are i.i.d.

Claim: $\{Y_t\}$ is NOT strictly stationary.

Proof by Counterexample:

Consider the joint distribution of $(Y_1,Y_2)$: $Y_1=a_1{Y}_2=a_1+a_2$

Shifted by $k=1$: $(Y_2,Y_3)$: $Y_2=a_1+a_2{Y}_3=a_1+a_2+a_3$

The two distributions are different:

• $(Y_1,Y_2)$ depends on 2 random variables
• $(Y_2,Y_3)$ depends on 3 random variables

Joint distributions change with time $⟹$ not strictly stationary.

$■$

Related

• Weakly Stationary — requires only constant mean and covariance
• White Noise — i.i.d. white noise is strictly stationary
• Random Walk — example of non-stationary process