카이 스퀘어 Identity만 잘 기억하자.
Let \(Y_{ij}=\theta_i+e_{ij}\), \((j=1,\ldots,n\), \(i=1,\ldots,p)\), where \(\theta_i\)’s are unknown and \(e_{ij}\stackrel{\text{iid}}\sim N(0,\sigma_0^2)\), where \(\sigma_0^2\) is unknown.
The minimal sufficient statistic for \(\theta=\left((\theta_1,\ldots,\theta_p)^T,\sigma_0^2\right)\) is \[ S=\left((\bar Y_1,\ldots, \bar Y_p)^T, \sum_{i=1}^p\sum_{j=1}^n(Y_{ij}-\bar Y_i)^2\right), \] where \(\bar Y_i=\frac{1}{n}\sum_j Y_{ij}\), \(i=1,\ldots,p\). Now writing \(X_i=\bar Y_i\), \(X\sim N(\theta, \sigma^2I_p)\) where \(\sigma^2=\frac{\sigma_0^2}{n}\). Also, the usual within mean squared error estimator of \(\sigma^2\) is \[ \hat \sigma^2=\frac{\sum_i \sum_j (Y_{ij}-\bar Y_i)}{n\cdot (n-1)p}, \] and \(\hat \sigma^2\) is distributed independent of \(X\). Let \(S=||X||^2\) and \(F=\frac{S}{\frac{m\hat \sigma^2}{m+2}}\), where \(m=p(n-1)\).
Suppose
\(0<\tau(F)<2(p-2)\), \(p\ge 3\);
\(\tau(F)\uparrow\) in \(F\);
\(E[\tau'(F)]<\infty\).
Then, by defining \(\phi_i(X)=-\frac{\tau(F)}{F}X_i\), \(1\le i \le p\), \(\left(\left(1-\frac{\tau(F)}{F}\right)X_1,\ldots,\left(1-\frac{\tau(F)}{F}\right)X_p \right)\) dominates \(X\) under the loss \(L(\theta,a)=\frac{||X-\theta||^2}{\sigma^2}\).
Let \(W\sim b \chi_m^2\), \(b\ge 0\). Then \[ E[Wh(W)]=bE[mh(W)+2Wh'(W)], \] provided \(E[|h'(W)|]<\infty\).
\[\begin{eqnarray*} E[Wh'(W)] &=& \int_0^\infty wh'(w)\exp\left(-\frac{w}{2b}\right)\frac{w^{\frac{m}{2}-1}}{(2b)^\frac{m}{2}\Gamma(\frac{m}{2})}dw\\ &=& \int_0^\infty h'(w)\exp\left(-\frac{w}{2b}\right)\frac{w^{\frac{m}{2}}}{(2b)^\frac{m}{2}\Gamma(\frac{m}{2})}dw\\ &=& \left[h(w)\exp\left(-\frac{w}{2b}\right)\frac{w^{\frac{m}{2}}}{(2b)^\frac{m}{2}\Gamma(\frac{m}{2})}dw\right]_{w=0}^{w=\infty}-\int_0^\infty h(w)\left\{-\frac{1}{2b}\exp\left(-\frac{w}{2b}\right)\frac{w^{\frac{m}{2}}}{(2b)^\frac{m}{2}\Gamma(\frac{m}{2})}+\frac{m}{2} \exp\left(-\frac{w}{2b}\right)\frac{w^{\frac{m}{2}-1}}{(2b)^\frac{m}{2}\Gamma(\frac{m}{2})} \right\}dw\\ &=&0 + \frac{1}{2b}E[Wh(W)]-\frac{m}{2}E[h(W)]. \end{eqnarray*}\] Hence, \(E[Wh(W)]=bE[mh(W)+2Wh'(W)]\).
Suppose \(X\sim N(\theta, \sigma^2I_p)\) distributed independently of \(\hat \sigma^2\sim\frac{\sigma^2}{m+2}\chi_m^2\). Then, under the loss \(L(\theta,a)=\frac{||\theta-a||^2}{\sigma^2}\), the estimator of \(\theta\) defined by \[ \hat \theta= \left(1-\frac{\tau(F)}{F}\right)X \] dominate \(X\) where \(F=\frac{||X||^2}{\hat\sigma^2}\) if
\(0<\tau(F)<2(p-2)\), \(p\ge 3\);
\(\tau(F)\) is differentiable and non-decreasing.
\(E[\tau'(F)]<\infty\).
\[\begin{eqnarray*} E\left[ \frac{||\hat\theta-\theta||^2}{\sigma^2} \right]&=& E\left[ \frac{||X-\theta-\frac{\tau(F)}{F}X||^2}{\sigma^2} \right]\\ &=& E\left[ \frac{1}{\sigma^2}||X-\theta||^2\right]-\frac{2}{\sigma^2}E\left[\frac{\tau(F)}{F}X^T (X-\theta) \right]+E\left[ \frac{\tau^2(F)}{F^2}\frac{||X||^2 }{\sigma^2}\right]\\ &=& p-\frac{2}{\sigma^2}E\left[\frac{\tau(F)}{F}X^T (X-\theta) \right]+E\left[ \frac{\tau^2(F)}{F}\frac{\hat \sigma^2 }{\sigma^2}\right]\\ E\left[\frac{\tau(F)}{F}X^T (X-\theta) \right]&=&E\left[E\left[\frac{\tau(F)}{F}X^T (X-\theta) \Big|\hat\sigma^2\right]\right]\\ &=&E\left[E\left[\sigma^2\sum_{i=1}^p\frac{d}{dX_i}\left\{\frac{\tau(F)}{F}X \right\} \Big|\hat\sigma^2\right]\right] \mbox{ }\mbox{ }\mbox{ by Stein's Identity}\\ &=&\sigma^2E\left[p\frac{\tau(F)}{F}+\left\{\frac{\tau'(F)}{F}-\frac{\tau(F)}{F^2}\right\}\sum_{i=1}^p\frac{dF}{dX_i}X_i\Big| \hat \sigma^2 \right]\\ &=& p\sigma^2E\left[\frac{\tau(F)}{F}\right]+\sigma^2 E\left[ \left\{\frac{\tau'(F)}{F}-\frac{\tau(F)}{F^2}\right\}\left(\sum_{i=1}^p\frac{2X_i}{\hat \sigma^2}X_i\right) \Big|\hat\sigma^2\right]\\ &=& p\sigma^2E\left[\frac{\tau(F)}{F}\right]+2\sigma^2 E\left[ \tau'(F)-\frac{\tau(F)}{F} \Big|\hat\sigma^2\right]\\ &=& \sigma^2E\left[2\tau'(F)+(p-2)\frac{\tau(F)}{F}\Big|\hat\sigma^2\right]. \end{eqnarray*}\] Finally, since \(\hat \sigma^2 \sim \frac{\sigma^2}{m+2}\chi_m^2\), by the chi-square identity, \[ E\left[\frac{\tau^2(F)}{F}\hat\sigma^2 \right]=\frac{\sigma^2}{m+2} E\left[m\frac{\tau^2(F)}{F}+2\hat\sigma^2\frac{d}{d\hat\sigma^2}\left\{\frac{\tau^2(F)}{F} \right\} \right]\tag{**} \] with \(W=\hat\sigma^2\) and \(h(W)=\frac{\tau^2(F)}{F}\) in the lemma.
But, \[ \frac{d}{d\hat\sigma^2}\left\{\frac{\tau^2(F)}{F}\right\}=\left(2 \frac{\tau(F)\tau'(F)}{F}-\frac{\tau^2(F)}{F^2} \right)\frac{dF}{d\hat\sigma^2}=\left(2 \frac{\tau(F)\tau'(F)}{F}-\frac{\tau^2(F)}{F^2} \right)\left(-\frac{||X||^2}{d\hat\sigma^4}\right). \] Hence from \((**)\), \[\begin{eqnarray*} E\left[\frac{\tau^2(F)}{F}\hat\sigma^2 \right]&=&\frac{\sigma^2}{m+2} E\left[m\frac{\tau^2(F)}{F} -2\left(2\tau(F)\tau'(F)-\frac{\tau^2(F)}{F} \right) \right]\\ &=&\frac{\sigma^2}{m+2} E\left[(m+2)\frac{\tau^2(F)}{F} -4\tau(F)\tau'(F)\right]. \end{eqnarray*}\] Combining all, \[\begin{eqnarray*} E\left[ \frac{||\hat\theta-\theta||^2}{\sigma^2} \right]&=&p-\{2(p-2)-\tau(F)\}\frac{\tau(F)}{F}-\tau'(F)\left(4+\frac{4}{m+2}\tau(F)\right)\\ &\le& p-\{2(p-2)-\tau(F)\}\frac{\tau(F)}{F}\mbox{ }\mbox{ }\mbox{ since }\tau(F)\uparrow \mbox{ in F}\\ &<&p \mbox{ }\mbox{ }\mbox{ for }0<\tau(F)<2(p-2). \end{eqnarray*}\]