# Understanding summation identity

I'm currently working on proposition 6.3.1 from {Introduction to Statistical Time Series and I have the following equality (consider $$\theta_0 = 1$$)

$$\begin{equation*} \sum_{j = 1}^n \left(\theta_0 u_j + \theta_1 u_{j-1} + \cdots + \theta_q u_{j - q} \right) = \sum_{k = 0}^q \theta_k \sum_{j = 1}^{n} u_j + \sum_{s = 1}^q \sum_{j = s}^q \theta_j u_{j - s} - \sum_{s = 0}^{q-1}\sum_{j = s + 1}^q \theta_j u_{n - s}. \end{equation*}$$

I didn't manage to find if this right. Any tip on how to proof that this works will be appreciated!

There is a typo in the RHS. The $$u_{j-s}$$ in the middle sum should instead be $$u_{s-j}$$: \begin{align} &\quad \sum_{k=0}^q \theta_k \sum_{j=1}^n u_j + \sum_{s=1}^q \sum_{j=s}^q \theta_j \color{red}{u_{s-j}} - \sum_{s=0}^{q-1}\sum_{j=s+1}^q \theta_j u_{n-s} \\ &= \sum_{k=0}^q \theta_k \sum_{j=1}^n u_j + \sum_{s=1}^q \sum_{k=s}^q \theta_k u_{s-k} - \sum_{s=0}^{q-1}\sum_{k=s+1}^q \theta_k u_{n-s} \\ &= \left(\theta_0 \sum_{j=1}^n u_j + \sum_{k=1}^q \theta_k \sum_{j=1}^n u_j\right) + \sum_{k=1}^q \theta_k \sum_{s=1}^k u_{s-k} - \sum_{k=1}^q \theta_k \sum_{s=0}^{k-1} u_{n-s} \\ &= \theta_0 \sum_{j=1}^n u_j + \color{red}{\sum_{k=1}^q \theta_k \sum_{j=1}^n u_j} + \sum_{k=1}^q \theta_k \sum_{j=1-k}^0 u_j \color{red}{- \sum_{k=1}^q \theta_k \sum_{j=n-k+1}^n u_j} \\ &= \theta_0 \sum_{j=1}^n u_j + \color{red}{\sum_{k=1}^q \theta_k \sum_{j=1}^{n-k} u_j} + \sum_{k=1}^q \theta_k \sum_{j=1-k}^0 u_j \\ &= \theta_0 \sum_{j=1}^n u_j + \sum_{k=1}^q \theta_k \sum_{j=1-k}^{n-k} u_j \\ &= \theta_0 \sum_{j=1}^n u_j + \sum_{k=1}^q \theta_k \sum_{j=1}^n u_{j-k} \\ &= \sum_{k=0}^q \theta_k \sum_{j=1}^n u_{j-k} \\ &= \sum_{j=1}^n \sum_{k=0}^q \theta_k u_{j-k} \end{align}