Mathematics Asked by qp212223 on September 8, 2020
The question I’m trying to solve is shown below – from a practice qualifying exam in probability which has no posted solution (self study). I really don’t know where to start (even for part (a)) and am looking for some major help.
Am I supposed to apply one of Kolmogorov’s two/three series theorem? Thanks for any ideas you have!
a) is answered in the comments
b) For $phi in (0, 1/e)$ you can modify @Michael's hint and note that if
$$ sum_{n=1}^infty P(log^+(Z) > rn) < infty $$ for any $r > 0$, then $mathbb{E}[log^+(Z)] < infty$, which means you can just modify your calculation in the comment for $phi in (0, 1/e)$ by
$$sum_{n=1}^infty P(log^+(Z) > rn) = sum_{n=1}^infty P(Z > e^{rn} )$$ and now pick $r$ s.t. $e^r = 1/phi$.
c) Provided the indexing on $Z_t$s is from $-infty$ to $infty$, yes, this sequence is strictly stationary. To see that, note that shifting the indices of $Z_i$s does not change anything about the joint distribution of finitely many $X_{t_1}, X_{t_2}, ..., X_{t_n}$ and the joint distribution does not depend on the absolute values of $t_1, t_2, ..., t_n$, but only their differences.
I think$^*$ sequence, since $phi$ is a random variable, is not ergodic. To see that, formally, we have
$$lim_{n rightarrow infty} frac{1}{2n} sum_{t=-n}^{n} X_n = lim_{n rightarrow infty} frac{1}{2n} sum_{t=-n}^{n} sum_{j=0}^infty phi^j Z_{t-j} = lim_{nrightarrow infty} frac{1}{2n} sum_{k=-infty}^{infty} sum_{j= max(|k| - n, 0) }^{|k| + n} phi^j Z_k rightarrow frac{1}{1 - phi} mathbb{E} Z $$
where we can swap sums because everything is non-negative. Since this limit is not trivial provided $phi$ is not, by the ergodic theorem, this sequence is not ergodic.
$*$: I am not fully sure; in particular, I am very unsure about the limit. I am quite sure it isn't ergodic though, since the limit of this running average is going to depend on the value of $phi$.
Correct answer by E-A on September 8, 2020
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