Let $y_{ij}$, $i=1,2,\cdots,n_j$ be a random sample from $N_p(\mu_j,\Sigma_j)$, $j=1,2$. Let $$\overline{y}_j=\frac{1}{n_j}\sum_{i=1}^{n_j}y_{ij} \hspace{2mm} \mathrm{and}$$ $$S_j=\frac{1}{n_j-1}\sum_{i=1}^{n_j}(y_{ij}-\overline{y}_j)(y_{ij}-\overline{y}_j)'$$ be the sample mean and variance, respectively. Let $$n=n_1+n_2,$$ $$\rho_j=\sqrt{\frac{n_j-1}{n-2}}\hspace{2mm} (j=1,2),$$ $$\overline{\Sigma}=\frac{n_2}{n}\Sigma_1+\frac{n_1}{n}\Sigma_2$$ $$\Omega_1=\sqrt{\frac{n_2}{n}}\overline{\Sigma}^{-1/2}\Sigma_1^{1/2},$$ $$\Omega_2=\sqrt{\frac{n_1}{n}}\overline{\Sigma}^{-1/2}\Sigma_2^{1/2},$$ $$V_j=\sqrt{n_j-1}(\Sigma_j^{-1/2}S_j\Sigma_j^{-1/2}-I_p)\hspace{2mm} (j=1,2).$$ Here, $I_p$ denotes the $p \times p$ identity matrix. Let $$\overline{V}=\rho_1^{-1}\Omega_1 V_1 \Omega_1' + \rho_2^{-1}\Omega_2 V_2 \Omega_2'.$$ If $$z=\sqrt{\frac{n_1n_2}{n}}\overline{\Sigma}^{-\frac{1}{2}}(\overline{y}_1-\overline{y}_2),$$ show that $E[z'\overline{V}^2z]$ equals $\psi_1+\psi_2$, where $$\psi_1=\frac{n_2^2(n-2)}{n^2(n_1-1)}(tr(\Sigma_1\overline{\Sigma}^{-1}))^2+\frac{n_1^2(n-2)}{n^2(n_2-1)}(tr(\Sigma_2\overline{\Sigma}^{-1}))^2,$$ $$\psi_2=\frac{n_2^2(n-2)}{n^2(n_1-1)}tr(\Sigma_1\overline{\Sigma}^{-1}\Sigma_1\overline{\Sigma}^{-1})+\frac{n_1^2(n-2)}{n^2(n_2-1)}tr(\Sigma_2\overline{\Sigma}^{-1}\Sigma_2\overline{\Sigma}^{-1}).$$

I have tried to use the identity $E[AZB + C] = AE[Z]B + C$ where $A,B,C$ are constant matrices, $E(S_j)=\Sigma_j$ and $E[z'\overline{V}^2z]=E_z[E_{\overline{V}}[z'\overline{V}^2z\mid z]]$. Considering that $\psi_1$ and $\psi_2$ are 'complicated' expressions, they should give a clear hint as to how to proceed with the proof. However, I am still stuck and can't get anywhere.

Any help is very appreciated. Thank you very much!

  • 1
    $\begingroup$ Sometimes, complicated identities become trivial when they can be interpreted statistically. Could you then perhaps explain what motivated this question and what these expressions might mean? $\endgroup$ – whuber Aug 21 '18 at 21:34
  • $\begingroup$ Thanks for your reply @whuber. Actually this question is from the following paper: tandfonline.com/doi/abs/10.1080/03610910500308396. I am trying to understand all the steps involved in this paper, yet I am stuck at showing this. The paper only states the following result without mentioning any proof. $\endgroup$ – wibxyz Aug 21 '18 at 23:00

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