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.

  • $\begingroup$ Do you really mean "$\times$" and not "$+$" in the definition of $\zeta_t$? It looks like $\zeta_t = 0$ whenever $t$ is an integer. $\endgroup$ – Mateusz Kwaśnicki May 22 '20 at 10:41
  • $\begingroup$ 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 $t$? $\endgroup$ – I.Kiaan May 22 '20 at 10:46
  • $\begingroup$ 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? $\endgroup$ – Mateusz Kwaśnicki May 22 '20 at 10:50
  • 2
    $\begingroup$ 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 $\endgroup$ – mike May 22 '20 at 11:21
  • $\begingroup$ @MateuszKwaśnicki, sorry, I really misread your comment. Sure, you are right $\endgroup$ – I.Kiaan May 22 '20 at 11:52

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.