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Q7.5-11E

Expert-verified
Found in: Page 395

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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 $${Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N$$. 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 $${\mathbb{R}^p}$$).) Show that if the covariance matrix of $${X_1},................,{X_N}$$ is $$S$$, then the covariance matrix of $${Y_1},................,{Y_N}$$ is $${P^T}SP$$.

It is verified that:

1. $$\left( {{{\bf{X}}_1},................,{{\bf{X}}_N}} \right)w = 0$$
2. The covariance matrix is: $${S_Y} = {P^T}SP$$
See the step by step solution

## Step 1: Mean Deviation form and Covariance Matrix

The Mean Deviation form of 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 matrix is:

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

## Step 2: The Mean Deviation Form (a)

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 form given 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.

## Step 3: The Covariance Matrix (b)

From (a), the covariance matrix can 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.