# Background

For the model $$y_{ti} = \beta_i + \tau_j + \epsilon_{ij} + \mu,$$

where $\beta_i$ is the $i^{th}$ block effect, $\tau_j$ is the $j^{th}$ treatment effect, $\epsilon_{ij}$ is the $(ij)^{th}$ error, and $\mu$ is the overall mean,

the sum of squares $S$ is: $$S_O = S_B + S_T + S_E + S_{BT} + S_{BE} + S_{TE}$$

O is for observation, B for block, T for treatment, and E for error. Also,

$$S_{BE} = \sum_{i}^{m}\sum_{j}^{n}{\big{(}\bar{y}_i-\bar{y}\big{)}\big{(}y_{ij}-\bar{y}_j-\bar{y}_i \big{)}}\\ S_{TE} = \sum_{i}^{m}\sum_{j}^{n}{\big{(}\bar{y}_j-\bar{y}\big{)}\big{(}y_{ij}-\bar{y}_j-\bar{y}_i \big{)}}$$

# Question

How do I show that $S_{BE}$ and $S_{TE}$ equal 0?

I have already shown that $S_{BT}$ is zero because of the property mentioned in this comment: Partitioning sum of squares. However, in that case the sum is in the form

$$\sum_{i \in P}\sum_{j \in Q}{x_i x_j}=\sum_{i \in P}{x_i}\sum_{j \in Q}x_j$$

whereas for $S_{BE}$ and $S_{TE}$, the sum is of the form

$$\sum_{i \in P}\sum_{j \in Q}{x_i x_{ij}}$$

and I am unsure if I can split the product into two sums.