# is this process a Markov one?

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.

• Do you really mean "$\times$" and not "$+$" in the definition of $\zeta_t$? It looks like $\zeta_t = 0$ whenever $t$ is an integer. – Mateusz Kwaśnicki May 22 '20 at 10:41
• Yes, there really should be $\times$ in the definition of $\zeta_t$. Can you tell, please, how do you know that $\zeta_t=0$ for integer $t$? – I.Kiaan May 22 '20 at 10:46
• I thought $\eta_{[t]} (t - [t])$ is the product of $\eta_{[t]}$ and $t - [t]$, and the latter is zero when $t$ is an integer? – Mateusz Kwaśnicki May 22 '20 at 10:50
• No, it's not markov. If you see the process at times 1.25, 1.5 then you know $\eta_1, \eta_2$ and so you know the process perfectly in the near future, which is not the case if you only see the process at time 1.5 – mike May 22 '20 at 11:21
• @MateuszKwaśnicki, sorry, I really misread your comment. Sure, you are right – I.Kiaan May 22 '20 at 11:52

## 2 Answers

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.

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 $$Z_{3/2}$$.

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.