A stationary process Zt has multiplicative autocorrelation when Cor[Zt,Zr]=Cor[Zt,Zs]Cor[Zs,Zr] for all t≤s≤r. Autocorrelation is defined as Cor[Zt,Zs]:=Cov[Zt,Zs]σ2 with σ2=Var(Zt).

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).

Consider any real-valued discrete-time stationary Markov process Z′t and translate it to Zt:=Z′t−E[Z′t] without loss of generality.

Let σ2:=Var(Zt)ρ:=Cov[Zt,Zt+1]/σ2

Multiplicative autocorrelation implies Cor[Zt,Zt+n]=ρnCov[Zt,Zt+n]=ρnσ2

Define what will be shown to be "white noise" of Zt as autoregressive process: ϵt:=Zt−ρZt−1 By convenient translation, E[Zt]=0ϵt=0Cov[Zt,Zs]=E[ZtZs]E[Zt2]=σ2E[ZtZt+1]=ρ

Consider any 0]]>n>0. E[ϵtϵt+n]=E[(Zt−ρZt−1)(Zt+n−ρZt+n−1)]=E[ZtZt+n]+ρ2E[Zt−1Zt+n−1]−ρ(E[ZtZt+n−1]+E[Zt−1Zt+n])=(1+ρ2)ρnσ2−ρ(ρn−1σ2+ρn+1σ2)=0 thus ϵt satisfies the "white noise" condition for expressing Zt as the autoregressive process Zt+1=ρZt+ϵt QED

For any stationary two-state Markov chain 1 Zt, Cor[Zt,Z0]=Cor[Zt,Zs]Cor[Zs,Z0]

Proof Let q1:=P(Zt=a1)q0:=P(Zt=a0) Map Zt to a more convenient Yt:=Zt−a0a1−a0 Since Yt only equals 0 or 1: E[Yt]=E[Yt2]=q1 and thus Var(Yt)=q1−q12=q1q0 For convenience let p0:=P(Y1=0∣Y0=1)p1:=P(Y1=1∣Y0=0)s:=p0+p1 Since Yt is stationary, it follows that qi=pi/s for i∈{0,1}. In preparation for induction, assume P(Yt=1∣Y0=1)=q1+q0(1−s)tP(Yt=1∣Y0=0)=q1−q1(1−s)t It must follow that P(Yt+1=1∣Y0=1)=P(Yt+1=1∣Y1=1)(1−p0)+P(Yt+1=1∣Y1=0)p0=[q1+q0(1−s)t](1−p0)+[q1−q1(1−s)t]p0=q1+[q0(1−p0)−q1p0](1−s)t=q1+[q0(1−p0)−(1−q0)p0](1−s)t=q1+[q0−p0](1−s)t=q1+[q0−q0s](1−s)t=q1+q0(1−s)t+1 and P(Yt+1=1∣Y0=0)=P(Yt+1=1∣Y1=1)p1+P(Yt+1=1∣Y1=0)(1−p1)=[q1+q0(1−s)t]p1+[q1−q1(1−s)t](1−p1)=q1+[q0p1−q1(1−p1)](1−s)t=q1+[(1−q1)p1−q1(1−p1)](1−s)t=q1+[p1−q1](1−s)t=q1+[q1s−q1](1−s)t=q1−q1(1−s)t+1 which completes induction, noting the base case of t=0 is true.

Due to the convenient mapping to Yt, E[YtY0]=P(Yt=1∣Y0=1)P(Y0=1)=(q1+q0(1−s)t)q1=q12+q0q1(1−s)t thus Cov[Yt,Y0]=E[YtY0]−E[Yt]E[Y0]=q12+q0q1(1−s)t−q12=q0q1(1−s)tCor[Yt,Y0]=(1−s)t QED

Let Zt be a stationary Markov process such that P(Zt=−1)=P(Zt=0)=P(Zt=1)=1/3 P(Zt+1=0∣Zt=−1)=1/2P(Zt+1=1∣Zt=0)=1/2P(Zt+1=−1∣Zt=1)=1/2 and for all i∈{−1,0,1}, P(Zt+1=i∣Zt=i)=1/2

Conveniently E[Zt]=0, thus Cov[Zt,Zs]=E[ZtZs]. For one time step we have P(Z1=1∧Z0=1)=(1/3)(1/2)P(Z1=−1∧Z0=1)=(1/3)(1/2)P(Z1=1∧Z0=−1)=0P(Z1=−1∧Z0=−1)=(1/3)(1/2) thus autocorrelation of one time step must be positive: E[Z1Z0]=(1⋅1)16+(−1⋅1)16+(−1⋅−1)16=16 For two time steps P(Z2=1∧Z0=1)=(1/3)(1/2)2P(Z2=−1∧Z0=1)=(1/3)[(1/2)2+(1/2)2]P(Z2=1∧Z0=−1)=(1/3)(1/2)2P(Z2=−1∧Z0=−1)=(1/3)(1/2)2 thus the autocorrelation for two time steps must be negative: E[Z2Z0]=(1⋅1)112+(−1⋅1)212+(1⋅−1)112+(−1⋅−1)112=−112 thus Cor[Z2,Z0]≠Cor[Z2,Z1]Cor[Z1,Z0]