# Help understanding the step $\sum_{j=0}^n\sum_{k=0}^ng_jg_k\text{Cov}(\epsilon_{n-1},\epsilon_{n+h-k})=\sum_{j=0}^ng_j^2+h\sigma^2$

Given is that $$\epsilon_n$$ is a white noise process with $$\text{Var}(\epsilon_n)=\sigma^2$$ and that $$g_j\in\mathbb{R}$$. There is a step in my lecture notes that I don't get. It says the following

$$\sum_{j=0}^n\sum_{k=0}^ng_jg_k\text{Cov}(\epsilon_{n-j},\epsilon_{n+h-k})=\sum_{j=0}^ng_j^2+h\sigma^2 \quad \text{for} \quad h\ge0,$$

with the motivation "need $$k=h+j$$ otherwise the covariance is zero, we use this to remove the sum over $$k$$".

I understand that the sum over $$k$$ gets removed and that we want to avoid zero covariance, but how does $$+h\sigma^2$$ pop up? Doing the substitution $$k=h+j$$ then the covariance is just the variance which is onl $$\sigma^2$$ (no $$h$$), and then it's multiplied to the sum and not added.

• You seem to have a typo in the double summation $\epsilon_{n-1}$ (subscript should be $n-j$, I think). Also, do you think the formula is correct if $h=0$ (it looks like $\sigma$ completely disappears from the right hand side)?
– ir7
Commented Apr 18, 2021 at 17:04
• Thank you, I corrected it. Yes it does disappear completely if $h=0$. Maybe I'm missing something but this is the link to the slides, see slide 4: ionides.github.io/531w20/04/notes04-annotated.pdf Commented Apr 18, 2021 at 17:11
• What I'm actually trying to do is calculating the autocovariance function of a $MA(q)$ process. Commented Apr 18, 2021 at 17:20
• I see. You have a typo i the right hand side (see my answer).
– ir7
Commented Apr 18, 2021 at 17:22
• Ahh.. I misstok the $h$ of beeing outside the index setting. So it's actually $\sigma^2$ multiplied with the sum. Commented Apr 18, 2021 at 17:31

$${\rm Cov} (\epsilon_{n-j}, \epsilon_{n+h-k}) =\gamma (h-k+j) = \sigma^2 1_{k=h+j}$$

where $$1_A$$ is the indicator function, set to $$1$$ if statement $$A$$ is true, and set to $$0$$ otherwise.

So the double summation is:

$$\sum_{j=0}^\infty\sum_{k=0}^{\infty}g_jg_k\text{Cov}(\epsilon_{n-j},\epsilon_{n+h-k})$$ $$= \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} g_jg_k \sigma^2 1_{k=h+j}$$ $$= \sum_{j=0}^{\infty} g_jg_{h+j} \sigma^2$$

as

$$\sum_{k=0}^{\infty} g_jg_k \sigma^2 1_{k=h+j} = g_jg_{h+j} \sigma^2$$

As $$k$$ runs from $$0$$ to $$\infty$$, the terms in the last summation are $$0$$ except when $$k$$ hits $$h+j$$, which will happen for any given, but fixed, $$h$$.

If the summation is finite, $$k$$ runs from $$0$$ to $$n$$, then

$$\sum_{k=0}^{n} g_jg_k \sigma^2 1_{k=h+j} = g_jg_{h+j} \sigma^2 1_{h+j \leq n}$$

• How is it infinity the index goes to $n$? And what happens to the sum in the last step why does it disappear? Commented Apr 18, 2021 at 17:28
• I used the slides summations to $\infty$. I added an explanation for the finite case when the sum could be 0 is $h$ is too high compared to $n$.
– ir7
Commented Apr 18, 2021 at 17:59