Interpretation of intercept in Helmert coding I have a general, presumably simple, question, but I couldn't find a conclusive answer so far. 
Assume I have a simple case of a General Linear Model with one categorical predictor variable, that has 3 levels. This corresponds to a one-way ANOVA with 3 groups. 
The Linear Model would then be: $Y_i = \beta_0 + \beta_1X_1 + \beta_2X_2$
I use the following standard Helmert contrast codes (I, II) for the three groups A, B, C: 
     A    B    C
I   -2    1    1
II   0   -1    1

Based on these contrast codes, will the intercept of the linear model represent the mean of all group means? More generally, does this always apply to Helmert contrast coded variables, regardless of the number of predictor variables or variable levels? 
Obviously, the intercept represents the predicted value, when all predictor variables (in this case $X_1$ and $X_2$) equal zero, but I can't figure out if that represents the mean of the group means. 
 A: Yes. You can easily verify this by carrying out the following steps:
First, express the means of $A$, $B$, and $C$, in terms of the model with the specified contrast:
\begin{eqnarray*}
1\hat{\beta}_{0}-2\hat{\beta}_{1}+0\hat{\beta}_{2} & = & \hat{\mu}_{A}=E(Y_{A})\\
1\hat{\beta}_{0}+1\hat{\beta}_{1}-1\hat{\beta}_{2} & = & \hat{\mu}_{B}=E(Y_{B})\\
1\hat{\beta}_{0}+1\hat{\beta}_{1}+1\hat{\beta}_{2} & = & \hat{\mu}_{C}=E(Y_{C})
\end{eqnarray*}
Here, each $\hat{\mu}_i$ represents the group mean of group $i$, $i=A, B, C$.
Next, place each beta coefficient into a matrix augmented with the
means on the right, and place the matrix in row-reduced echelon form
using Guass-Jordan elimination:
\begin{eqnarray*}
\begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
1 & 1 & -1 & | & \hat{\mu}_{B}\\
1 & 1 & 1 & | & \hat{\mu}_{C}
\end{bmatrix} & \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 3 & 1 & | & \hat{\mu}_{C}-\hat{\mu}_{A}
\end{bmatrix}\\
 & \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 0 & 2 & | & \left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)
\end{bmatrix}\\
 & \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 3 & -1 & | & \hat{\mu}_{B}-\hat{\mu}_{A}\\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}\\
 & \sim & \begin{bmatrix}1 & -2 & 0 & | & \hat{\mu}_{A}\\
0 & 1 & 0 & | & \frac{1}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}\\
 & \sim & \begin{bmatrix}1 & 0 & 0 & | & \hat{\mu}_{A}+\frac{2}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 1 & 0 & | & \frac{1}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
0 & 0 & 1 & | & \frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]
\end{bmatrix}
\end{eqnarray*}
So, now, we know that the first pivot position corresponds to:
\begin{eqnarray*}
\hat{\beta}{}_{0} & = & \hat{\mu}_{A}+\frac{2}{3}\left\{ \left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)+\frac{1}{2}\left[\left(\hat{\mu}_{C}-\hat{\mu}_{A}\right)-\left(\hat{\mu}_{B}-\hat{\mu}_{A}\right)\right]\right\} \\
 & = & \hat{\mu}_{A}-\frac{2}{3}\hat{\mu}_{A}-\frac{1}{3}\hat{\mu}_{A}+\frac{1}{3}\hat{\mu}_{A}+\frac{2}{3}\hat{\mu}_{B}-\frac{1}{3}\hat{\mu}_{B}+\frac{1}{3}\hat{\mu}_{C}\\
 & = & \frac{1}{3}\hat{\mu}_{A}+\frac{1}{3}\hat{\mu}_{B}+\frac{1}{3}\hat{\mu}_{C}\\
 & = & \frac{\hat{\mu}_{A}+\hat{\mu}_{B}+\hat{\mu}_{C}}{3}
\end{eqnarray*}
The final expression indicates that $\hat{\beta}{}_{0}$, the intercept,
represents the simple mean of the group means.
