is this process a Markov one?

Question

Here is the problem I can't solve.

Let $\xi_n$ $(n=1,2,3,\dots)$ be a sequence of i.i.d. random variables on $\mathbb{R}$ with density $p(x)>0$, let $\eta_n=\sum_{i=1}^{n}\xi_i^2$. Define $$\zeta_t = \eta_{[t]}(t-[t]) \times \eta_{[t]+1}(-t+[t]+1),$$ where $[t]$ denotes the integer part of $t$. Is $\zeta_t$ a Markov process?

I tried to prove that given process is Markov by definition: $$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t}) =\mathbb{P}(\zeta_{t+s}\in A| \mathcal{F_{=t}}),$$ where $A \in \mathcal{F}_{\ge t}$, $\mathcal{F}_{\le t} = \sigma(\zeta_s: s\le t)$, $\mathcal{F}_{\ge t} = \sigma(\zeta_s: s\ge t)$, $\mathcal{F}_{= t} = \sigma(\zeta_t)$.

Here is my first attempt so far

$$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t})=\mathbb{P}(\eta_{[t+s]}(t+s-[t+s]) \times \eta_{[t+s]+1}(-t-s+[t+s]+1)\in A|\mathcal{F}_{\le t})=$$ $$=\mathbb{P}\left(\left(\sum_{k=1}^{[t+s]}\xi_k^2(t+s-[t+s])\right)\times\left(\sum_{i=1}^{[t+s]+1}\xi_i^2([t+s]+1-t-s)\right)\in A|\mathcal{F}_{\le t}\right)=$$
$$=\sum_{k=1}^{[t+s]}\sum_{i=1}^{[t+s]+1}\mathbb{E}(\mathbb{1}_A\xi_k^2(t+s-[t+s])\xi_i^2([t+s]+1-t-s)|\mathcal{F}_{\le t}),$$ using that $\mathbb{P}(A|\mathcal{F})=\mathbb{E}(\mathbb{1}_A|\mathcal{F})$.

My second attempt was to make an example of suitable random variables and probability space, such that $\zeta_t$ is not Markov. I began to think it is not Markov as $\zeta_t$ depends on $\eta_{[t]+1}$, so it depends on the future in some sense because $[t]+1>t$.

I understand that it is kind of "not Mathoverflow question", but I did not receive answer on MathStackexchange and I faced this problem during my term paper, so may be this is a reason to post it here.

Answer 1

The process is not Markov in general. Indeed, let $X_i:=\xi_i$, $S_n:=\eta_n=\sum_1^n X_i^2$, and $$Z_t:=\zeta_t=(t-[t])([t]+1-t)S_{[t]}S_{[t]+1},$$ where $P(X_i=0)=P(X_i=1)=1/2$. Then $$Z_{3/2}=\tfrac14\,X_1(X_1+X_2),\quad Z_2=0,\quad Z_{5/2}=\tfrac14\,(X_1+X_2)(X_1+X_2+X_3).$$ So, the conditional distribution of $Z_{5/2}$ given $Z_2$ is the same as the unconditional distribution of $Z_{5/2}$. On the other hand, the conditional distribution of $Z_{5/2}$ given $Z_{3/2},Z_2$ is not the same as the unconditional distribution of $Z_{5/2}$, because $Z_{5/2}$ depends on $Z_{3/2}$: $P(Z_{3/2}=0)=P(X_1=0)=1/2$ and $P(Z_{5/2}=0)=P(X_1+X_2=0)=1/4$, whereas $$P(Z_{3/2}=0,Z_{5/2}=0)=P(X_1+X_2=0)=1/4\ne P(Z_{3/2}=0)P(Z_{5/2}=0).$$

So, $(Z_t)$ is not Markov.

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If you insist that the distribution of $X_1$ be absolutely continuous with a strictly positive density, then you can appropriately approximate the discrete distribution by such an absolutely continuous one.

Alternatively, you may e.g. assume that $X_1\sim N(0,1)$. Then $EZ_{3/2}=1$, $EZ_{3/2}=5/2$, but $EZ_{3/2}Z_{3/2}=27/2\ne EZ_{3/2}EZ_{5/2}$, so that here too $Z_{5/2}$ depends on $Z_{3/2}$.

Answer 2

This is a good exercise. As far as I understand the OP, the situation is as follows. Given a deterministic non-negative function $\phi$ on the interval $[0,1]$ with $\phi(t)=0\iff t \in\{0,1\}$ and a sequence of random variables $Z_n$, one defines $$ \zeta(t)=\phi(t-n) \cdot Z_n \;, \qquad n\le t\le n+1 \;. $$ The question is when $\zeta(t)$ is Markov, and the answer to this question is pretty obvious: if and only if the variables $Z_n$ are independent (look at the Markov condition at integer times).

In the original question $$ \phi(t)=t(1-t) $$ and $$ Z_n=(\xi_1^2+\dots+\xi_n^2)(\xi_1^2+\dots+\xi_{n+1}^2) \;, $$ which are pretty obviously not independent unless $\xi_i^2$ are a.s. constant.