2
$\begingroup$

So guys, I'm having this hard proof to solve in probability. I don't really know how to tackle it! Hope that someone can help.

Let $\{Z_i\}_{i\in\mathbb{Z}}$ be i.i.d. random variables with zero mean and unit standard deviation. For $(a_0, a_1, ..., a_r)$ a sequence of $r$ real numbers and $j\in\mathbb{Z}$, let

$$\begin{align} Y_j & = \sum_{i=0}^ r a_i Z_{j-i} \end{align}$$

For $r=1$, suppose $a_0=1$ and $|a_1|<1$. Show that $Y_j$ can also be written

$$\begin{align} Y_j & = Z_j - \sum_{i<j} \rho^{(j-i)} Y_i \end{align}$$

For a constant $\rho$ that you should determine in terms of $a_1$.

$\endgroup$
  • $\begingroup$ what is $\rho$? $\endgroup$ – Slug Pue May 30 '15 at 23:33
  • $\begingroup$ Sorry, I've forget to write it... I'll edit the post and add it now $\endgroup$ – james42 May 30 '15 at 23:34
5
$\begingroup$

For $r=1$ and $a_0=1$ we have $$ Y_j = Z_j + a_1 Z_{j-1} $$ Now note that $$ Y_{j-1} = Z_{j-1} + a_1 Z_{j-2} $$ so $Z_{j-1} = Y_{j-1} - a_1 Z_{j-2}$ and substituting this into the first equation you get $$ Y_j = Z_j + a_1 (Y_{j-1} - a_1 Z_{j-2}) $$ Doing the same operation for $Z_{j-2}=Y_{j-2}-a_1 Z_{j-3}$ and substituting again you get $$ Y_j = Z_j + a_1 Y_{j-1}-a^2_1Y_{j-2}+ a_1^3Z_{j-3} $$ and so on. This can generally be written $$Y_j = Z_j - \sum^n_{i=1}(-a_1)^i Y_{j-i} + a_1^nZ_{j-n}$$ and taking $\lim n \rightarrow \infty$ the last term vanishes since $|a_1|<1$. With $\rho = -a_1$ you get $$Y_j = Z_j - \sum^{\infty}_{i=1}\rho^i Y_{j-i}$$ which is the same as $$ Y_j = Z_j - \sum^{-\infty}_{i=j-1}\rho^{j-i} Y_{i} $$


As a technical note, one can show that $a_1^n Z_n \rightarrow 0$ by the fact that $Z_n$ has a finite variance: For any $\epsilon>0$, by Chebyshev's inequality $$ \mathbb P (|a^n_1 Z_n| \geq \epsilon) \leq \frac{a^{2n}\text{Var}(Z_n)} {\epsilon^2} =a_1^{2n}\epsilon^{-2} $$ So $$ \sum^{\infty}_{n=1} P (|a^n_1 Z_n| \geq \epsilon) \leq \epsilon^{-2}\sum^{\infty}_{n=1} (a_1^2)^n = \frac{1}{\epsilon^2 (1-a^2)} < \infty $$ hence by Borel-Cantelli Lemma, $\mathbb P (|a^n_1 Z_n| \leq \epsilon \text{ eventually}) = 1$ and hence $a^n Z_n \rightarrow 0$ almost surely.

$\endgroup$

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.