relation variance and covariance

Question

Let’s assume a (weakly) stationary process $\{u_t\}$, such that the mean of $u_t$ and the covariance $(u_t, u_{t+h})$ do not depend on $t$. For simplicity, we assume that $E(u_t) = 0$. (Don't know how whether this is relevant.). Then my book makes the following step:

$$var(\sum^T_{t=1}u_t)=\sum^T_{t=1}\sum^T_{s=1}cov(u_s,u_t).$$ I don't understand this step. Isn't the variance the covariance with itself?

Answer 1

If the mean is zero or not (you can forget about all the right hand side term if it is) :\begin{align*} \mathrm{Var}\left[\sum_{t}^T u_t \right] &= \mathbb E\left[\left(\sum_{t}^T u_t\right)^2 \right]-\mathbb E\left[\sum_{t}^T u_t \right]^2\\ &=\mathbb E\left[\sum_{t}^T\sum_{\tau}^T u_t u_\tau \right]-\left(\sum_{t}^T \mathbb E\left[u_t \right]\right)^2\\ &= \sum_{t}^T\sum_{\tau}^T \mathbb E\left[u_t u_\tau \right]-\sum_{t}^T\sum_{\tau}^T\mathbb E\left[ u_t \right]\mathbb E[u_\tau]\\ &= \sum_{t}^T\sum_{\tau}^T\left[ \mathbb E\left[u_t u_\tau \right]-\mathbb E\left[ u_t \right]\mathbb E[u_\tau]\right]\\ &= \sum_{t}^T\sum_{\tau}^T \mathrm{Cov}[u_t, u_\tau]\\ \end{align*}

Note that this is independent of the definition of the $u_t$, in particular it works even if you don't have stationarity.

Answer 2

$$var (\sum\limits_{t=1}^{T}u_t)=E((\sum\limits_{t=1}^{T}u_t))^{2}$$ $$=E((\sum\limits_{t=1}^{T}u_t)(\sum\limits_{s=1}^{T}u_s)).$$ This can be written as $$E(\sum\limits_{t=1}^{T}\sum\limits_{s=1}^{T}u_tu_s)$$ which gives $$\sum\limits_{t=1}^{T}\sum\limits_{s=1}^{T}E(u_tu_s)$$ or $$\sum\limits_{t=1}^{T}\sum\limits_{s=1}^{T}cov(u_t,u_s)$$.

Answer 3

In general if the variance of a random variable $X$ is well defined then: $$\mathsf{Var}X=\mathsf{Cov}\left(X,X\right)$$Secondly: $$\text{covariance is bilinear}$$

Applying that we find:

$$\mathsf{Var}\left(\sum_{t=1}^{T}u_{t}\right)=\mathsf{Cov}\left(\sum_{t=1}^{T}u_{t},\sum_{t=1}^{T}u_{t}\right)=\mathsf{Cov}\left(\sum_{t=1}^{T}u_{t},\sum_{s=1}^{T}u_{s}\right)=\sum_{t=1}^{T}\sum_{s=1}^{T}\mathsf{Cov}\left(u_{t},u_{s}\right)$$

For this to be valid it is not necessary that the $u_t$ have mean $0$.