Let \(X\sim N(\theta, \Sigma)\), where \(\lambda_{\text{max}}(\Sigma)\) denote the largest eigenvalue of \(\Sigma\). Then, writing \(S=||X||^2\), the estimator \(\delta(X)=\left(1-\frac{\tau(S)}{S}\right)X\) dominates \(X\) under \(L(\theta,a)=||\theta-a||^2\) if
\(\tau(\cdot)\) is differentiable, non-decreasing;
\(0<\tau(S)<2\{tr(\Sigma)-2\lambda_{\text{max}}(\Sigma)\}\) and \(\lambda_{\text{max}}(\Sigma)\}<\frac{tr(\Sigma)}{2}\).
\[\begin{eqnarray*} E\left[ \left|\left|\left(X-\frac{\tau(S)}{S}X \right)-\theta\right|\right|^2\right]&=&E[||X-\theta||^2]+E\left[\frac{\tau^2(S)}{S}\right]-2E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]\\ &=&tr(\Sigma)+E\left[\frac{\tau^2(S)}{S}\right]-2E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]. \end{eqnarray*}\] 여기서 \(Y=\Sigma^{-\frac{1}{2}}X\), \(\phi=\Sigma^{-\frac{1}{2}}\theta\)라고 하자. 그렇다면 \(Y\sim N(\phi,I)\)이다. 또한 \[ E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]= E\left[ \frac{\tau(S)}{S} Y^T\Sigma(Y-\phi)\right],\mbox{ }\mbox{ }\mbox{ where }S=Y^T\Sigma Y. \] 또한, \[ Y^T\Sigma (Y-\phi)=\left(\sum_{j=1}^p\sigma_{1j}Y_j\cdots\sum_{j=1}^p\sigma_{pj}Y_j\right) \begin{pmatrix}Y_1-\phi_1\\Y_2-\phi_2\\ \vdots \\Y_p-\phi_p \end{pmatrix}=\sum_{i=1}^p\sum_{j=1}^p \sigma_{ij}Y_j(Y_i-\phi_i). \] 그러므로, Stein’s Identity 에 의해서 \[\begin{eqnarray*} E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]&=& E\left[ \frac{\tau(S)}{S} Y^T\Sigma(Y-\phi)\right]\\ &=& E\left[ \frac{\tau(S)}{S} \sum_{i=1}^p\sum_{j=1}^p \sigma_{ij}Y_j(Y_i-\phi_i)\right]\\ &=& \sum_{i=1}^p E\left[ (Y_i-\phi_i)\frac{\tau(S)}{S} \sum_{j=1}^p \sigma_{ij}Y_j\right]=\sum_{i=1}^p E\left[ (Y_i-\phi_i)h(Y)\right]\\ &=&\sum_{i=1}^p E\left[ \frac{dh(Y)}{dY_i}\right]=\sum_{i=1}^p E\left[ \frac{d}{dY_i}\left(\frac{\tau(S)}{S} \sum_{j=1}^p \sigma_{ij}Y_j\right) \right]. \end{eqnarray*}\] 또한 \(S=Y^T\Sigma Y\)에 대해 \(\frac{dS}{dY_i}=2Y^T\Sigma_{(i)}=2\sum_{k=1}^p \sigma_{ik}Y_k\)이므로 \[\begin{eqnarray*} E\left[ \frac{d}{dY_i}\left(\frac{\tau(S)}{S} \sum_{j=1}^p \sigma_{ij}Y_j\right) \right]&=& E\left[ \frac{d}{dY_i}\left(\frac{\tau(S)}{S} \sum_{j=1}^p \sigma_{ij}Y_j\right) \right]\\ &=&E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)\left(2\sum_{k=1}^p \sigma_{ik}Y_k\right)\left(\sum_{j=1}^p \sigma_{ij}Y_j\right)+\frac{\tau(S)}{S}\sigma_{ii}\right]. \end{eqnarray*}\] 그러므로 \[\begin{eqnarray*} E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]&=& \sum_{i=1}^p E\left[ \frac{d}{dY_i}\left(\frac{\tau(S)}{S} \sum_{j=1}^p \sigma_{ij}Y_j\right) \right]\\ &=&\sum_{i=1}^pE\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)\left(2\sum_{k=1}^p \sigma_{ik}Y_k\right)\left(\sum_{j=1}^p \sigma_{ij}Y_j\right)+\frac{\tau(S)}{S}\sigma_{ii}\right]\\ &=&2E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)\sum_{k=1}^p \sum_{i=1}^pY_kY_j \left( \sum_{j=1}^p\sigma_{ik}\sigma_{ij}\right)\right]+tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]\\ &=&2E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)\sum_{k=1}^p \sum_{i=1}^pY_kY_j \left( \sum_{j=1}^p\sigma_{ik}\sigma_{ij}\right)\right]+tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]\\ &=&2E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)Y^T\Sigma^2Y\right]+tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]\\ &=&2E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)X^T\Sigma X\right]+tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]\\ \end{eqnarray*}\] 종합하면 \[\begin{eqnarray*} E\left[ \left|\left|\left(X-\frac{\tau(S)}{S}X \right)-\theta\right|\right|^2\right]&=&E[||X-\theta||^2]+E\left[\frac{\tau^2(S)}{S}\right]-2E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right]\\ &=&tr(\Sigma)+E\left[\frac{\tau^2(S)}{S}\right]-2E\left[\frac{\tau(S)}{S}X^T(X-\theta)\right].\\ &=&tr(\Sigma)+E\left[\frac{\tau^2(S)}{S}\right]-4E\left[ \left(\frac{\tau'(S)}{S}-\frac{\tau(S)}{S^2} \right)X^T\Sigma X\right]-2tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]\\ &\le&tr(\Sigma)+E\left[\frac{\tau^2(S)}{S}\right]+4E\left[ \left(\frac{\tau(S)}{S} \right)\frac{X^T\Sigma X}{S}\right]-2tr(\Sigma)E\left[\frac{\tau(S)}{S}\right]. \end{eqnarray*}\] 또한, \(X^T \Sigma X\le\lambda_{\text{max}}(\Sigma)X^TX=\lambda_{\text{max}}(\Sigma)S\)이므로, \[\begin{eqnarray*} E\left[ \left|\left|\left(X-\frac{\tau(S)}{S}X \right)-\theta\right|\right|^2\right]\ &\le&tr(\Sigma)-4E\left[\frac{\tau(S)}{S} (2tr(\Sigma)-4\lambda_{\text{max}}(\Sigma)-\tau(S)) \right]\\ &<&tr(\Sigma)=E[||X-\theta||^2]\mbox{ }\mbox{ }\mbox{ }\mbox{ by the second assumption} \end{eqnarray*}\]