I have a question regarding standardized regression coefficients and the correlation matrix.
I have a two part problem that I am working on, and I need to show that the correlation matrix R is equal to:
$$R = \frac{1}{N-1}X*'X$$
I'm not quite sure how to set this up. I will post the problem I am working on and how far along I have gotten.
Q: To obtain numerically stable computation of regression coefficients it is often a good idea to first standardize all variables as follows:
\begin{align} y_i* &= \frac{y_i-\bar{y}}{s_y} \\ &\text{ and } \\ x_{ij}* &= \frac{x_{ij}-\bar{x}_j}{s_{x_j}}, \\ &(1 \leq i \leq N, 1 \leq j \leq p) \end{align}
where $\bar{x}$ and $\bar{y}$ are the sample means and $s_y$ and $s_{x_j}$ are the sample standard deviations of the corresponding variables. Denote the estimated regression coefficients (called the standardized regression coefficients) for these data by $\hat{\beta_j}*$ $(0\leq j \leq p)$. Note that the $\hat{\beta_j}*$ are unitless and can be compared with each other to assess the relative importance of the $x_j$'s in their strength of linear relationship of $y$ in the presence of other predictor variables.
a) Show that $\hat{\beta}^*_0=0$. Hence we may define the model matrix $X*$ for the standardized data to be an $N \times p$ matrix by excluding the first column of $1$'s; similarly, let $\hat{\beta}^* = (\hat{\beta}_1^*,,...,\hat{\beta}_p^*)'$ to be a $p \times 1$ vector.
b) Show that the correlation matrix $R = (r_{x_jx_k})$ among the $x_j$'s and the correlation vector $ r = (r_{yx_1},...,r_{yx_p})'$ among $y$ and $x_j$ for $j = 1,2,...,p$ are given by
\begin{align} R &= \frac{1}{N-1}X*'X*$ \\ &\text{ and } \\ r &= \frac{1}{N-1}X*'y*' \end{align}
For part a,
\begin{align} Y_i &= \hat{\beta}_0^* + \hat{\beta}_1^*X_{i1}+...+\hat{\beta}_p^*X_{1p}+\hat{e}_i \\ & \\ \frac{y_i-\bar{y}}{s_y} &= (\hat{\beta}_1\frac{s_1}{s_y})\frac{x_i1-\bar{x}_1}{s_1}+...+(\hat{\beta}_p\frac{s_p}{s_y})\frac{x_ip-\bar{x}_p}{s_1} \\ & \\ \frac{y_i-\bar{y}}{s_y} &= (\hat{\beta}_1\frac{s_1}{s_y})\frac{x_i1-\bar{x}_1}{s_1}+...+(\hat{\beta}_p\frac{s_p}{s_y})\frac{x_ip-\bar{x}_p}{s_p} \\ & \\ \hat{\beta}_0 &= \frac{y_i-\bar{y}}{s_y} - (\hat{\beta}_1\frac{s_1}{s_y})\frac{x_i1-\bar{x}_1}{s_1}+...+(\hat{\beta}_p\frac{s_p}{s_y})\frac{x_ip-\bar{x}_p}{s_p} \\ & \\ \hat{\beta}_0 &= (\hat{\beta}_1\frac{s_1}{s_y})\frac{x_i1-\bar{x}_1}{s_1}+...+(\hat{\beta}_p\frac{s_p}{s_y})\frac{x_ip-\bar{x}_p}{s_p} \\ &\quad -\bigg[(\hat{\beta}_1\frac{s_1}{s_y})\frac{x_i1-\bar{x}_1}{s_1}+...+(\hat{\beta}_p\frac{s_p}{s_y})\frac{x_ip-\bar{x}_p}{s_p}\bigg] \end{align}
Therefore, $\hat{\beta}_0 = 0$.
For part b, I'm able to multiply out the matrices $ X*' X*$, but I'm not entirely sure how to interpret the meaning.
