Remark: Where $e^{(j)}\in \mathbb{R}^k$ is the standard basis.
This is from the book, "The Coordinate Free Approach to Linear Models", by Michael Wichura. I don't quite understand what he means by "taking the value $e^j$ with probability $p_j$. Does he mean that $E(X_j)=p_j$ or what?
EDIT: I still have some questions. How do we show that the dispersion operator is projection onto the orthogonal complement of r? The only information I know is that $E(X)=0$ and $E(X_i)=p_i$.I tried computing it but I didn't get anywhere.
What is meant that the probability that $X=e^{(j)}$ is $p_j$. The only possible values of $X$ are the basis vectors. The fact that the probabilities add to one supports this interpretation; there is no probability left for other values.
The expected value of $Y$ should be computed component by component. The value of $X_j$ is either one (probability $p_j$) or zero (probability $1-p_j$). The possible values of $Y_j$ (with corresponding probabilities) can be computed from the formula for $Y_j$. A quick calculation, which I leave as an exercise, shows that the expected value of $Y_j$ is indeed zero.