Question: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

The Mean Deviation formof any \(p \times N\)is given by:

\(B = \left( {\begin{array}{*{20}{c}}{{{{\bf{\hat X}}}_1}}&{{{{\bf{\hat X}}}_2}}&{........}&{{{{\bf{\hat X}}}_N}}\end{array}} \right)\)

Whose \(p \times p\) covariance matrixis:

\(S = \frac{1}{{N - 1}}B{B^T}\)

From the question, the\(w\)is a unit vector with all values equal to 1. Then,we have:

\(\begin{array}{c}\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = \left( {{{\bf{X}}_1},{{\bf{X}}_2}, \ldots ,{{\bf{X}}_n}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {{\bf{X}}_1} + ...... + {{\bf{X}}_N}\\ = 0\end{array}\)

The mean deviation formgiven is:

\(\begin{array}{c}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)w = \left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)w\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = {P^T}\left( {{{\bf{X}}_1} + ...... + {{\bf{X}}_N}} \right)\\ = 0\end{array}\)

Hence, this is the required proof.

From (a), thecovariance matrixcan be given as:

\(\begin{array}{c}{S_Y} = \frac{1}{{N - 1}}\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right){\left( {{{\bf{Y}}_1},................,{{\bf{Y}}_N}} \right)^T}\\ = \frac{1}{{N - 1}}\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right){\left( {{P^T}{{\bf{X}}_1},................,{P^T}{{\bf{X}}_N}} \right)^T}\\ = {P^T}\left\{ {\frac{1}{{N - 1}}\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right){{\left( {{{\bf{X}}_1},......,{{\bf{X}}_N}} \right)}^T}} \right\}P\\ = {P^T}SP\end{array}\)

Hence, this is the required proof.