I would like to find maximum of the following function:

$$I = \max_{a\in \mathbb{R}^p} \frac{(a'\hat{\beta})^2}{S^2a'(X'X)^{-1}a},$$

where $X$ is a design matrix and of course $Y$ is normally distributed $N(X\beta, \sigma^2I).$

Trying to do it the standard way, i.e. differentiating the quotient doesn't work well, I can't solve the resulting equation for $a$. Is it manageable to do it through differentiation?

I tried to do it using Cauchy-Schwarz inequality. I obtained:

$$I = \frac{1}{S^2} \max_{a\in \mathbb{R}^p} \frac{a'(X'X)^{-1}X'YY'X(X'X)^{-1}a}{a'(X'X)^{-1}X'X(X'X)^{-1}a}.$$

And now by Cauchy-Schwarz inequality what I get is

$$\frac{1}{S^2} \frac{(c'Y)^2}{c'c} \leq \frac{1}{S^2} \frac{c'cY'Y}{c'c} = \frac{Y'Y}{S^2},$$

where $c = X(X'X)^{-1}a$

But the equality in Cauchy-Schwarz inequality holds, when there exists $d\in \mathbb{R}$ such that $c = dY$ but it requires invertibility of $(XX').$

I would appreciate any input.

  • $\begingroup$ $(XX’)$ can’t be invertible because it is not full rank. $\endgroup$
    – treskov
    Nov 17, 2019 at 23:01

2 Answers 2


Using that $\hat\beta= (X^TX)^{-1} X^T Y$ so $(a^T \hat\beta )^2=a^T (X^TX)^{-1} X^T Y Y^T X (X^TX)^{-1} a$, you can write $I$ as $$ I= max_{a\in\mathbb{R}^p}\frac{a^T C a}{a^T B a} $$ where $C=(X^TX)^{-1} X^T Y Y^T X (X^TX)^{-1}$ and $B=(X^TX)^{-1}$, where we left out $S^2$ to simplify, since it is just a constant so does not influence the solution. Then you can apply the answer given to your other question.


Consider the following result, discussed at https://math.stackexchange.com/q/1226524/321264 :

If $B$ is a symmetric positive definite matrix and $x,d$ are real vectors, then $$\max_{x\ne 0}\frac{(x^Td)^2}{x^TBx}=d^T B^{-1}d \tag{1}$$

The proof is based on the following generalization of Cauchy-Schwarz inequality:

$$(x^T d)^2\le (x^TBx)(d^TB^{-1}d) \tag{2}$$

And to prove $(2)$, one can apply the usual Cauchy-Schwarz inequality to $B^{1/2}x$ and $B^{-1/2}d$.

I assume $X$ is of full column rank, whence $X^TX$ is a positive definite matrix.

Using $(1)$, your $I$ reduces to



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