# Sum-to-zero constraint in one-way ANOVA

I'm trying to understand my lecture notes but am a bit stuck on the concept of identifiability. In one-way ANOVA, could someone please explain the reason for the constraint $\sum_{i=1}^{m} \beta_{j} = 0$ where we have m groups of observations, each group consisting of k observations with $Y_{ij}$ as the jth observation from the ith group, $E(Y_{ij}) = \mu + \beta_{i}, i = 1,...,m; j = 1,...,k, \text{Var}(Y_{ij}) = \sigma^{2}$and$H_{0} : \beta_{1} = \beta_{2} = ... = \beta_{m}$? I don't quite get the identifiability reason.

• Imagine someone asks you "Which two numbers add to 11?" – Glen_b Jan 24 '17 at 7:43
• @Glen_b I get the idea but do you mind elaborating in the context of ANOVA? – python_learner Jan 24 '17 at 17:12

Consider for simplicity that $m=2$ and compare the models

• $\mu=0,\beta_1=0,\beta_2=2$,

• $\mu=1,\beta_1=-1,\beta_2=1$,

• $\mu=2,\beta_1=-2,\beta_2=0$.

These models are all special cases of $(\mu,\beta_1,\beta_2)=(\mu,-\mu,2-\mu)$. You can see that whatever $\mu$ we choose, $\mu+\beta_1=0$ and $\mu+\beta_2=2$, so there's an infinite set of parameter-triples that match $E(Y_{1j})=0$ and $E(Y_{2j})=2$, and no way to distinguish between them.

Consequently, while data will allow you to estimate the two group-means, those two pieces of information (two df) - no matter how precisely estimated - are not going to be enough to estimate the three parameters (three df) in the model -- there's an extra degree of freedom that allows you to move all three parameters in particular ways relative to each other while keeping the group-means the same.

You need to restrict/constrain/regularize the situation in some way so that the model doesn't have more things to estimate than the design has the ability to identify.

• Ah, this answer gave me a light bulb moment - thank you! – python_learner Jan 25 '17 at 2:07