# Multiplicative autocorrelation in stationary Markov processes

Proposed answer to the following question(s):

# Summary

A stationary process $$Z_t$$ has multiplicative autocorrelation when $\begin{eqnarray*} \operatorname{Cor}\!\left[{ Z_t}, { Z_r}\right] & = & \operatorname{Cor}\!\left[{ Z_t}, { Z_s}\right] \operatorname{Cor}\!\left[{ Z_s}, { Z_r}\right] \end{eqnarray*}$ for all $$t \le s \le r$$. Autocorrelation is defined as $\begin{eqnarray*} \operatorname{Cor}\!\left[{ Z_t}, { Z_s}\right] & := & \frac{ \operatorname{Cov}\!\left[{ Z_t}, { Z_s}\right] }{\sigma^2} \end{eqnarray*}$ with $$\sigma^2 = \operatorname{Var}({ Z_t})$$.

A stationary autoregressive process has multiplicative autocorrelation [1]. However, not all stationary Markov processes have multiplicative autocorrelation. See the section below about a real-valued 3-state Markov chain for a counterexample.

Among discrete-time stationary processes, only autoregressive processes have multiplicative autocorrelation. Some Markov processes are not obviously autoregressive processes even though technically they are. For example, all stationary real-valued two-state Markov chains are autoregressive (and thus also have multiplicative autocorrelation).

# Multiplicative autocorrelation implies autoregression

Consider any real-valued discrete-time stationary Markov process $$Z'_t$$ and translate it to $$Z_t := Z'_t - {\operatorname{E}\mathchoice{\!\left[{ Z'_t}\right]}{[{ Z'_t}]}{[{ Z'_t}]}{[{ Z'_t}]}}$$ without loss of generality.

Let $\begin{eqnarray*} \sigma^2 & := & \operatorname{Var}({ Z_t}) \\ \rho & := & \operatorname{Cov}\!\left[{ Z_t}, { Z_{t+1}}\right] / \sigma^2 \end{eqnarray*}$

Multiplicative autocorrelation implies $\begin{eqnarray*} \operatorname{Cor}\!\left[{ Z_t}, { Z_{t+n}}\right] & = & \rho^n \\ \operatorname{Cov}\!\left[{ Z_t}, { Z_{t+n}}\right] & = & \rho^n \sigma^2 \end{eqnarray*}$

Define what will be shown to be “white noise” of $$Z_t$$ as autoregressive process: $\begin{eqnarray*} \epsilon_t & := & Z_t - \rho Z_{t-1} \end{eqnarray*}$ By convenient translation, $\begin{eqnarray*} {\operatorname{E}\mathchoice{\!\left[{ Z_t}\right]}{[{ Z_t}]}{[{ Z_t}]}{[{ Z_t}]}} & = & 0 \\ \epsilon_t & = & 0 \\ \operatorname{Cov}\!\left[{ Z_t}, { Z_s}\right] & = & {\operatorname{E}\mathchoice{\!\left[{ Z_t Z_s}\right]}{[{ Z_t Z_s}]}{[{ Z_t Z_s}]}{[{ Z_t Z_s}]}} \\ {\operatorname{E}\mathchoice{\!\left[{ Z_t^2}\right]}{[{ Z_t^2}]}{[{ Z_t^2}]}{[{ Z_t^2}]}} & = & \sigma^2 \\ {\operatorname{E}\mathchoice{\!\left[{ Z_t Z_{t+1}}\right]}{[{ Z_t Z_{t+1}}]}{[{ Z_t Z_{t+1}}]}{[{ Z_t Z_{t+1}}]}} & = & \rho \end{eqnarray*}$

Consider any $$n > 0$$. $\begin{eqnarray*} {\operatorname{E}\mathchoice{\!\left[{ \epsilon_t \epsilon_{t+n}}\right]}{[{ \epsilon_t \epsilon_{t+n}}]}{[{ \epsilon_t \epsilon_{t+n}}]}{[{ \epsilon_t \epsilon_{t+n}}]}} & = & {\operatorname{E}\mathchoice{\!\left[{ (Z_t - \rho Z_{t-1})(Z_{t+n} - \rho Z_{t+n-1})}\right]}{[{ (Z_t - \rho Z_{t-1})(Z_{t+n} - \rho Z_{t+n-1})}]}{[{ (Z_t - \rho Z_{t-1})(Z_{t+n} - \rho Z_{t+n-1})}]}{[{ (Z_t - \rho Z_{t-1})(Z_{t+n} - \rho Z_{t+n-1})}]}} \\ & = & {\operatorname{E}\mathchoice{\!\left[{ Z_t Z_{t+n}}\right]}{[{ Z_t Z_{t+n}}]}{[{ Z_t Z_{t+n}}]}{[{ Z_t Z_{t+n}}]}} + \rho^2 {\operatorname{E}\mathchoice{\!\left[{ Z_{t-1} Z_{t+n-1}}\right]}{[{ Z_{t-1} Z_{t+n-1}}]}{[{ Z_{t-1} Z_{t+n-1}}]}{[{ Z_{t-1} Z_{t+n-1}}]}} - \rho ({\operatorname{E}\mathchoice{\!\left[{ Z_t Z_{t+n-1}}\right]}{[{ Z_t Z_{t+n-1}}]}{[{ Z_t Z_{t+n-1}}]}{[{ Z_t Z_{t+n-1}}]}} + {\operatorname{E}\mathchoice{\!\left[{ Z_{t-1} Z_{t+n}}\right]}{[{ Z_{t-1} Z_{t+n}}]}{[{ Z_{t-1} Z_{t+n}}]}{[{ Z_{t-1} Z_{t+n}}]}}) \\ & = & (1 + \rho^2) \rho^n \sigma^2 - \rho (\rho^{n-1} \sigma^2 + \rho^{n+1} \sigma^2) \\ & = & 0 \end{eqnarray*}$ thus $$\epsilon_t$$ satisfies the “white noise” condition for expressing $$Z_t$$ as the autoregressive process $\begin{eqnarray*} Z_{t+1} & = & \rho Z_t + \epsilon_t \end{eqnarray*}$ QED

# Real-valued 2-state Markov chain

For any stationary two-state Markov chain [1] $$Z_t$$, $\begin{eqnarray*} \operatorname{Cor}\!\left[{ Z_t}, { Z_0}\right] & = & \operatorname{Cor}\!\left[{ Z_t}, { Z_s}\right] \operatorname{Cor}\!\left[{ Z_s}, { Z_0}\right] \end{eqnarray*}$

Proof Let $\begin{eqnarray*} q_1 := \operatorname{P}({ Z_t = a_1 }) \\ q_0 := \operatorname{P}({ Z_t = a_0 }) \end{eqnarray*}$ Map $$Z_t$$ to a more convenient $\begin{eqnarray*} Y_t & := \frac{Z_t - a_0}{a_1 - a_0} \end{eqnarray*}$ Since $$Y_t$$ only equals $$0$$ or $$1$$: ${\operatorname{E}\mathchoice{\!\left[{ Y_t}\right]}{[{ Y_t}]}{[{ Y_t}]}{[{ Y_t}]}} = {\operatorname{E}\mathchoice{\!\left[{ Y_t^2}\right]}{[{ Y_t^2}]}{[{ Y_t^2}]}{[{ Y_t^2}]}} = q_1$ and thus $\operatorname{Var}({ Y_t}) = q_1 - q_1^2 = q_1 q_0$ For convenience let $\begin{eqnarray*} p_0 & := & \operatorname{P}({ Y_1 = 0 \mid Y_0 = 1 }) \\ p_1 & := & \operatorname{P}({ Y_1 = 1 \mid Y_0 = 0 }) \\ s & := & p_0 + p_1 \end{eqnarray*}$ Since $$Y_t$$ is stationary, it follows that $$q_i = p_i/s$$ for $$i \in \{0,1\}$$. In preparation for induction, assume $\begin{eqnarray*} \operatorname{P}({ Y_t = 1 \mid Y_0 = 1 }) & = & q_1 + q_0 (1-s)^t \\ \operatorname{P}({ Y_t = 1 \mid Y_0 = 0 }) & = & q_1 - q_1 (1-s)^t \end{eqnarray*}$ It must follow that $\begin{eqnarray*} \operatorname{P}({ Y_{t+1} = 1 \mid Y_0 = 1 }) & = & \operatorname{P}({ Y_{t+1} = 1 \mid Y_1 = 1 }) (1-p_0) + \operatorname{P}({ Y_{t+1} = 1 \mid Y_1 = 0 }) p_0 \\ & = & [q_1 + q_0 (1-s)^t] (1-p_0) + [q_1 - q_1 (1-s)^t] p_0 \\ & = & q_1 + [q_0 (1-p_0) - q_1 p_0](1-s)^t \\ & = & q_1 + [q_0 (1-p_0) - (1- q_0) p_0](1-s)^t \\ & = & q_1 + [q_0 - p_0](1-s)^t \\ & = & q_1 + [q_0 - q_0 s](1-s)^t \\ & = & q_1 + q_0 (1-s)^{t+1} \\ \end{eqnarray*}$ and $\begin{eqnarray*} \operatorname{P}({ Y_{t+1} = 1 \mid Y_0 = 0 }) & = & \operatorname{P}({ Y_{t+1} = 1 \mid Y_1 = 1 }) p_1 + \operatorname{P}({ Y_{t+1} = 1 \mid Y_1 = 0 }) (1-p_1) \\ & = & [q_1 + q_0 (1-s)^t] p_1 + [q_1 - q_1 (1-s)^t] (1-p_1) \\ & = & q_1 + [q_0 p_1 - q_1 (1-p_1)](1-s)^t \\ & = & q_1 + [(1 - q_1) p_1 - q_1 (1-p_1)](1-s)^t \\ & = & q_1 + [p_1 - q_1](1-s)^t \\ & = & q_1 + [q_1 s - q_1](1-s)^t \\ & = & q_1 - q_1 (1-s)^{t+1} \\ \end{eqnarray*}$ which completes induction, noting the base case of $$t=0$$ is true.

Due to the convenient mapping to $$Y_t$$, $\begin{eqnarray*} {\operatorname{E}\mathchoice{\!\left[{ Y_t Y_0}\right]}{[{ Y_t Y_0}]}{[{ Y_t Y_0}]}{[{ Y_t Y_0}]}} & = & \operatorname{P}({ Y_t = 1 \mid Y_0 = 1 }) \operatorname{P}({ Y_0 = 1}) \\ & = & (q_1 + q_0 (1-s)^t) q_1 \\ & = & q_1^2 + q_0 q_1 (1-s)^t \end{eqnarray*}$ thus $\begin{eqnarray*} \operatorname{Cov}\!\left[{ Y_t}, { Y_0}\right] & = & {\operatorname{E}\mathchoice{\!\left[{ Y_t Y_0}\right]}{[{ Y_t Y_0}]}{[{ Y_t Y_0}]}{[{ Y_t Y_0}]}} - {\operatorname{E}\mathchoice{\!\left[{ Y_t}\right]}{[{ Y_t}]}{[{ Y_t}]}{[{ Y_t}]}} {\operatorname{E}\mathchoice{\!\left[{ Y_0}\right]}{[{ Y_0}]}{[{ Y_0}]}{[{ Y_0}]}} \\ & = & q_1^2 + q_0 q_1 (1-s)^t - q_1^2 \\ & = & q_0 q_1 (1-s)^t \\ \operatorname{Cor}\!\left[{ Y_t}, { Y_0}\right] & = & (1-s)^t \end{eqnarray*}$ QED

# Counterexample of Real-Valued 3-State Markov Chain

Let $$Z_t$$ be a stationary Markov process such that $\operatorname{P}({ Z_t = -1}) = \operatorname{P}({ Z_t = 0}) = \operatorname{P}({ Z_t = 1}) = 1/3$ $\begin{eqnarray*} \operatorname{P}({ Z_{t+1} = 0 \mid Z_t = -1}) & = & 1/2 \\ \operatorname{P}({ Z_{t+1} = 1 \mid Z_t = 0}) & = & 1/2 \\ \operatorname{P}({ Z_{t+1} = -1 \mid Z_t = 1}) & = & 1/2 \end{eqnarray*}$ and for all $$i \in \{-1, 0, 1\}$$, $\begin{eqnarray*} \operatorname{P}({ Z_{t+1} = i \mid Z_t = i}) & = & 1/2 \\ \end{eqnarray*}$

Conveniently $${\operatorname{E}\mathchoice{\!\left[{ Z_t}\right]}{[{ Z_t}]}{[{ Z_t}]}{[{ Z_t}]}} = 0$$, thus $$\operatorname{Cov}\!\left[{ Z_t}, { Z_s}\right] = {\operatorname{E}\mathchoice{\!\left[{ Z_t Z_s}\right]}{[{ Z_t Z_s}]}{[{ Z_t Z_s}]}{[{ Z_t Z_s}]}}$$. For one time step we have $\begin{eqnarray*} \operatorname{P}({ Z_1 = 1 \wedge Z_0 = 1 }) & = & (1/3)(1/2) \\ \operatorname{P}({ Z_1 = -1 \wedge Z_0 = 1 }) & = & (1/3)(1/2) \\ \operatorname{P}({ Z_1 = 1 \wedge Z_0 = -1 }) & = & 0 \\ \operatorname{P}({ Z_1 = -1 \wedge Z_0 = -1 }) & = & (1/3)(1/2) \end{eqnarray*}$ thus autocorrelation of one time step must be positive: ${\operatorname{E}\mathchoice{\!\left[{ Z_1 Z_0}\right]}{[{ Z_1 Z_0}]}{[{ Z_1 Z_0}]}{[{ Z_1 Z_0}]}} = (1 \cdot 1) \frac{1}{6} + (-1 \cdot 1) \frac{1}{6} + (-1 \cdot -1) \frac{1}{6} = \frac{1}{6}$ For two time steps $\begin{eqnarray*} \operatorname{P}({ Z_2 = 1 \wedge Z_0 = 1 }) & = & (1/3) (1/2)^2 \\ \operatorname{P}({ Z_2 = -1 \wedge Z_0 = 1 }) & = & (1/3) [ (1/2)^2 + (1/2)^2 ] \\ \operatorname{P}({ Z_2 = 1 \wedge Z_0 = -1 }) & = & (1/3) (1/2)^2 \\ \operatorname{P}({ Z_2 = -1 \wedge Z_0 = -1 }) & = & (1/3) (1/2)^2 \end{eqnarray*}$ thus the autocorrelation for two time steps must be negative: ${\operatorname{E}\mathchoice{\!\left[{ Z_2 Z_0}\right]}{[{ Z_2 Z_0}]}{[{ Z_2 Z_0}]}{[{ Z_2 Z_0}]}} = (1 \cdot 1) \frac{1}{12} + (-1 \cdot 1) \frac{2}{12} + (1 \cdot -1) \frac{1}{12} + (-1 \cdot -1) \frac{1}{12} = - \frac{1}{12}$ thus $\begin{eqnarray*} \operatorname{Cor}\!\left[{ Z_2}, { Z_0}\right] & \not= & \operatorname{Cor}\!\left[{ Z_2}, { Z_1}\right] \operatorname{Cor}\!\left[{ Z_1}, { Z_0}\right] \end{eqnarray*}$

# References

1. Hamilton JD (1994) Time series analysis. Princeton University Press, Princeton, N.J