Here is the objective function to be maximized: $$ E_{v}(\log(1+v^{\mathsf T} \Lambda v) ) $$ where $v$ is a Gaussian distributed random variable vector $v ∼ \mathrm{CN}(M,I)$ with its mean vector $M$ and covariance matrix $I$ (identity matrix). $\Lambda$ is a given diagonal matrix whose elements are non-negative and in decreasing order.

I want to find $M$ that maximize the objective function where $M$ is subject to power constraint: $$ \Vert M \Vert^2 \le P\ . $$

I have done some simulations on this problem. It turns out that $M=[\sqrt{P},0,0,\ldots,0]$ is optimal. However the proof is somewhat difficult because it isn't a concave function respect to $v$.

I wonder that whether a strict proof can be given or anyone can give me some hints on it?

Thank you!


You can relax the $log$ since it is monotonically increasing, and then you have

$$f = \mathbb{E}\left[1 + \sum_{i=1}^n\lambda_iv_i^2\right] = 1 + \sum_{i=1}^n\lambda_i\mathbb {E}[v_i^2] = 1 + \sum_{i=1}^n\lambda_i (1-M_i^2)$$

Removing all constant terms, you get:

$$f = -\sum_{i=1}^n \lambda_iM_i^2$$

At this point, you can transform your problem into a linear programming one and use some methods you like (i.e. simplex method). Just impose that:

$$x_i = M_i^2$$

and you have the following:

$$\left\{\begin{array}{l} \max_{x} \left(-\sum_{i=1}^n\lambda_i x_i\right)\\ \text{s.t.}\\ \sum_{i=1}^n x_i \leq P \\ x_{i} \geq 0 ~~\forall i \in \{1, \ldots, n\} \end{array} \right. $$

Remember that the solutions you get in $x$ will produce solutions in $M$ since $M_i = \pm \sqrt{x_i}$.

  • $\begingroup$ I doubt that whether you can relax log. Because E(log(f)) is not equal to log(E(f)), and I am not sure that the same M can achieve maximum in both expression. Could you give some further explanation about this? Thank you :) $\endgroup$ – NalRa Dec 7 '14 at 11:59
  • $\begingroup$ Never said that $$E(\log(f)) = log(E(f))$$. I'm just saying that $$\arg \max f(x) = \arg \max g(f(x))$$ for any strictly monotonically increasing function $g$ like $\log$. $\endgroup$ – the_candyman Dec 7 '14 at 17:07
  • $\begingroup$ I admit that $$\arg\max f(x)=\arg \max g(f(x)) $$ for any monotonically increasing function for $g$. But what I concerned about is that whether $$\arg \max E(f(x))=\arg\max E(g(f(x)))$$ can still hold for monotonically increasing function for $g$. After all, $x$ is random variable vector, what we can determine is parameter of it(i.e its mean vector). $\endgroup$ – NalRa Dec 8 '14 at 9:28

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