# Strange sum of random variables

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 $a_1$.

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

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.