# Why does absolutely-summable weights ensures a linear series itself summable (convergent)? Some questions on def'n of Linear Series

A "linear series" $$y_t$$ is the linear combination $$y_t - \mu = \sum_{i=-\infty}^{\infty}\psi_iL^i\nu_t = \sum_{i=-\infty}^{\infty}\psi_i\nu_{t-i}=S(L)\nu_t$$

of weighted (by $$\psi_i$$ weights) lags and forwards of a white noise series $$ν_t \thicksim wn(0,σ^2 )$$

[$$S(L)\equiv \sum_{i=-\infty}^{\infty}\psi_iL^i$$; $$\mu \equiv E(y_t)$$; $$\psi_i$$ weights are absolutely-summable: $$\sum_{i=-\infty}^{\infty}|\psi_i|<\infty$$].

(Shumway, 2017:25).

The absolutely-summability condition of $$\psi_i$$ weights in the definition of linear series ensures that the infinite sum in the definition to be convergent (the series itself becomes summable).

Proof:
WLOG assume $$μ≡0$$. By the def'n of a linear series, $$E(ν_t )=0$$. The convergence is obvious since $$E(\nu_t) \leq\sigma$$, and

$$E(y_t)=E(\sum_{i=-\infty}^{\infty}\psi_i\nu_{t-i})$$ $$\leq \sum_{i=-\infty}^{\infty}(|\psi_i| E(\nu_{t-i}))$$ $$\leq \left (\sum_{i=-\infty}^{\infty}|\psi_i| \right)\sigma<\infty$$ QED.

Source of the context:
Brockwell 2016; Introduction to Time Series and Forecasting. Page 44.

My questions are:
1. By $$\sigma$$, what is meant? (my guess: $$\sigma_{\nu_t}$$)

2. Why the author employs expected value in the proof?
(though what we have to show is just summability).
Why does it materialize in the proof?

3. In the proof, passing from 1st line to 2nd line, and 2nd to 3rd line is not clear to me.

1. It is the deviation of $$v_t$$. You've also written it in the beginning: $$v_t\sim wn(0,\sigma^2)$$.

2. $$y_t,v_t$$ are modelled as stochastic variables; there are no bounds over the input values $$v_t$$, but there are bounds on the statistics of it, i.e. its mean, variance. We can't put a bound on the exact sum expression, but expectation of it. A similar analysis without any stochasticity argument looks for BIBO stability (i.e. bounded input - bounded output $$|v_t| ) in signal processing literature, but we have explicit bounds on the input variable there.

3. First of all, you've some typos: it should be $$E[|v_t|]\leq \sigma$$ (because $$\operatorname{var}(|v_t|)=E[v_t^2]-E[|v_t|]^2\geq 0$$, $$E[v_t^2]$$ is already $$\sigma^2$$, and variance should be non-negative, which means $$E[|v_t|]\leq\sigma$$), and we're interested in finding the expected value of the absolute value of $$y_t$$, i.e. $$E[|y_t|]$$, otherwise both expectations are already $$0$$, i.e. $$E[v_t]=E[y_t]=0$$. The flow of proof follows as below: \begin{align}E[|y_t|]&=E\left[\left\lvert\sum \psi_i v_{t-i}\right\rvert\right]\underbrace{\leq}_{\text{triangle ineq.}} E\left[\sum \left\lvert\psi_i \right\rvert \left\lvert v_{t-i}\right\rvert\right]\\&=\sum E[|\psi_i||v_{t-i}|]=\sum|\psi_i|\underbrace{E[|v_{t-i}|]}_{\leq \sigma}\\&\leq\sum|\psi_i|\sigma=\sigma\sum|\psi_i|<\infty\end{align}

Note: the summation index is $$i$$, which is omitted due to notational simplicity.

Edit: Here, we assumed that expectation and summation are exchangeable, i.e. $$E[\sum |\psi_i| |\nu_{t-i}|]=\sum E[|\psi_i| |\nu_{t-i}|]$$ which is correct if $$\sum |E[|\psi_i| |\nu_{t-i}|]|=\sum E[|\psi_i| |\nu_{t-i}|]<\infty$$ And, this is correct since $$\sum E[|\psi_i \nu_{t-i}|]=\sum |\psi_i| E[|\nu_{t-i}|]\leq \sum |\psi_i| \sigma=\sigma \sum |\psi_i|<\infty$$

Being this correct we can change the summation and expectation.

• You say: "we're interested in finding the expected value of the absolute value of $y_t$, i.e. $E[|y_t|]$, otherwise both expectations are already $0$, i.e. $E[v_t]=E[y_t]=0$". Could you clarify this part further? $E[y_t]=0$ (in this your sentence) is not clear to me: $E[y_t]=E(\sum_{i=-\infty}^{\infty}\psi_i\nu_{t-i})$. To reach $E[y_t]=0$, we need the interchange of sum and expectation here; this requires condition for Fubini-Tonelli theorem must be met. Here that condition is $\sum |\psi_i\nu_{t-i}|<\infty$. However, we don't know the validity of this condition! – Erdogan CEVHER Aug 17 '19 at 19:11
• This is a very good point. Actually, the condition should be $\sum E[|\psi_i \nu_{t-i}|]<\infty$ and it is indeed satisfied. Here, a similar discussion: math.stackexchange.com/questions/575974/… – gunes Aug 17 '19 at 19:19
• Just you wrote when I correcting it as you said :) Since 5 minutes passed, SOF system did not allow me to edit my comment. – Erdogan CEVHER Aug 17 '19 at 19:20
• Theorem 3.1.2 in prepas.org/2013/Info/Liesse/Telecom/Scilab/docs/… states that $E(|y_t|)<\infty$ must be given in order to apply Tonelli theorem (see in Proof of Theorem 3.1.2 in page 39). So, that source implicitely assumes $E(|y_t|)<\infty$ must be given. Whereas you pass from $E[\Sigma|\psi_i||v_{t−i}|]$ to $\Sigma E[|\psi_i||v_{t−i}|]$ without $E(|y_t|)<\infty$ already in your hand. – Erdogan CEVHER Aug 19 '19 at 6:24
• You say: $\sum E[|\psi_i \nu_{t-i}|]<\infty$ is indeed satisfied. I can't see this. Also, the link of the discussion you provided does not reply this. That link refers to MITZENMACHER 2005 Probability and Computing_randomized Algorithms and Probabilistic Analysis, page 23. In that page, there only appears a simple statement about Tonelli. The proof of $\sum E[|\psi_i \nu_{t-i}|]<\infty$ is still missing for me unfortunately. By the way, I know the need of absolute convergence for the well-defined of expectation of countably many random variables, Riemann rearranagement theorem etc. – Erdogan CEVHER Aug 19 '19 at 6:54