6

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.

kjetil b halvorsen
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Herman Autore
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1 Answers1

4

There are plenty of algebraic ways to show this. But I find that often one can verify assertions of this nature with a quick differentiation. In this case it leads to the result with no actual calculation.

Because all the means are linear combinations of the $y_{ij},$ it's obvious that their derivatives are constants. In particular, the derivative of $\bar y_i$ does not depend on $i$ and the derivative of $\bar y_j$ does not depend on $j.$ (This follows from the fact that this is a balanced ANOVA with no missing cells.) The Chain Rule implies

$$\frac{\partial}{\partial y_{kl}} S_{BE} = \sum_{i,j} (y_{ij} - \bar y_j - \bar y_i)\frac{\partial}{\partial y_{kl}}(\bar y_i - \bar y) + (\bar y_i - \bar y)\frac{\partial}{\partial y_{kl}}(y_{ij} - \bar y_j - \bar y_i).$$

Stare at this for a few seconds until you realize this decomposes as two sums,

$$\frac{\partial}{\partial y_{kl}} S_{BE} = \text{Constant}_1 \sum_{i,j} (y_{ij}-\bar y_j - \bar y_i) + \text{Constant}_2\sum_{i,j} (\bar y_i - \bar y).$$

(The constants are easy to compute but you don't even need to do that.)

Those sums you already know equal zero. Conclude that $S_{BE}$ is constant and evaluate that constant at the data $y_{ij}=0$ for all $i,j$ to conclude $S_{BE}=0.$

Interchanging the roles of $i$ and $j$ implies $S_{TE}=0.$

whuber
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