0
$\begingroup$

Problem
We are given two sequences of i.i.d random variables $A_1, A_2,\dots $ and $B_1, B_2, \dots$ with known expected values $\mu_A$, $\mu_B$ and variances $\sigma^2_A$, $\sigma^2_B$. Define new variable $C_t$ that at time $t$ takes a value $$C_t = \sum_{i=1}^t A_i\cdot B_i$$ Is $C_t$ a Markov process and when it is a martinagle?

My ideas
I guess that $C_t$ is indeed a Markov process because the probability of $C_t$ taking some value depends only the value of the previous sum, $\sum_{i=1}^{t-1} A_i\cdot B_i $ which is $C_{t-1}$. However, the proof is more intuitive than formal.

As for the second part, here I'm less certain. If I'm not mistaken, $E(C_t | C_{t-1}) = C_{t-1} + \mu_A\cdot \mu_B $. So, $C_t$ is martingale if $E(C_t | C_{t-1}) = C_{t-1} $ or $ \mu_A\cdot \mu_B = 0$?

$\endgroup$
3
  • 4
    $\begingroup$ Your guesses are right. It is always Markov and it is a martingale iff $EA_iEB_i=0$. $\endgroup$ Oct 11, 2022 at 11:23
  • $\begingroup$ Is there a more formal way to show that it's Markov? $\endgroup$
    – student
    Oct 11, 2022 at 12:25
  • $\begingroup$ All you need is the fact that $A_{t+1}B_{t+1}$ is independent of $C_t$ $\endgroup$ Oct 11, 2022 at 12:30

0

You must log in to answer this question.