From Two Population We have two $p$ dimensional sampels $\mathbf{X}_1$ and $\mathbf{X}_2$ with sample size $n_1$ and $n_2$ respectively, Let We have $\mathbf{\Lambda} = \mathrm{diag}\{(\sigma_{11}^2 + \gamma\sigma_{21}^2)^{-1/2},\dots,(\sigma_{1p}^2+ \gamma\sigma^2_{2p})^{-1/2}\}$ which $\gamma$ is $\frac{n_1}{n_2}$, $\boldsymbol{\mu}_1$ and $\boldsymbol{\mu}_2$ are the $p \times 1$ mean vectors and $\Sigma_1$ and $\Sigma_2$ are the $p \times p$ Variance-Covariance Matrices of two populations. if we set $\frac{n_1}{n_1 + n_2} \rightarrow \kappa$ and $\tilde{\boldsymbol{\Sigma}} = (1-\kappa)\boldsymbol{\Sigma}_1 + \kappa\boldsymbol{\Sigma}_2$

Q: How can I show this result using the Cauchy-Schwarz inequality: $$ \frac{\mathrm{tr}(\boldsymbol{\Lambda}^2)}{p} \sqrt{\frac{\mathrm{tr}(\tilde{\boldsymbol{\Sigma}}^2)}{\mathrm{tr}((\tilde{\boldsymbol{\Lambda}\boldsymbol{\Sigma}\boldsymbol{\Lambda}})^2)}}>1$$ My Source is this article. Any Help will be appreciated.


1 Answer 1


I am able to use Cauchy-Schwartz inequality, but I not quite getting the same result. I may have made a mistake, so here are all the steps.

Note that for positive semi-definite matrices, trace defines an inner product. That is $tr(AB) = \langle B^T, A \rangle$.

Then by Cauchy-Schwarz for $A$ and $B$ positive semi-definite symmetric matrices, $$tr(AB) = \langle B, A \rangle \leq \sqrt{\langle B,B\rangle \langle A,A \rangle } = \sqrt{tr(B^2) tr(A^2)} \,.$$

In addition, for positive semi-definite matrices, all eigenvalues are non-negative, so $tr(B^2) \leq tr(B)^2$. Also, the product of positive semi-definite matrices is also a positive semidefinite if the product is symmetric (so $\Lambda \tilde{\Sigma} \Lambda$ is PSD). Finally, for matrices $A,B,C$, $tr(ABC) = tr(CAB)$. Putting all of this together.

\begin{align*} \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr((\Lambda \tilde{\Sigma} \Lambda)^2)}}& \geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr(\Lambda \tilde{\Sigma} \Lambda)^2}}\\ &= \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{tr(\Lambda^2 \tilde{\Sigma})^2}}\\ & = \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{\langle \tilde{\Sigma}, \Lambda^2\rangle^2}}\\ &\geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{tr(\tilde{\Sigma}^2)}{ tr(\tilde{\Sigma}^2)tr(\Lambda^4)}}\\ & \geq \dfrac{tr(\Lambda^2)}{p} \sqrt{ \dfrac{1}{tr(\Lambda^2)^2}}\\ & = \dfrac{1}{p} \end{align*}

I don't get that this is greater than 1.

  • $\begingroup$ thanks dude! But shame on me. I just copy and paste the articles bad notation and writing. the denominator below to sqrt was written bad (in real was $tr((\Lambda\tilde{\Sigma}\Lambda)^2))$. I edited that. Is that make any difference? check that out again please. $\endgroup$
    – SirSaleh
    Feb 7, 2018 at 20:05
  • $\begingroup$ @SirSaleh That is horrible usage of notation in the paper. I show the steps now for the changes made. I still get the same result, but now I can't get the simulation to work. So I suspect their result holds, and at some point I am using a weak bound. $\endgroup$ Feb 7, 2018 at 20:32
  • 1
    $\begingroup$ Your result is the correct one: $1/p$ is the minimum, not $1$. You can establish this by computing things explicitly for identity matrices $\Sigma_i$. (You will have to work in terms of $\Lambda^{-1}$ because $\Lambda$ isn't even defined!) I suspect there are typos either in the question or the paper. $\endgroup$
    – whuber
    Feb 7, 2018 at 21:13
  • $\begingroup$ I don't get why with simulation in all different settings it's become more than 1! while p become bigger it's become more close to 1! $\endgroup$
    – SirSaleh
    Feb 7, 2018 at 21:55
  • $\begingroup$ sorry for late bounty. I've been waiting if anybody find sharper one. But looks like this is sharpest :)) $\endgroup$
    – SirSaleh
    Feb 11, 2018 at 12:43

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