I'm already tr(r)

I shall begin by saying that I thought about this today in my Detection and Estimation Class while looking at the mse calculations for the MMSE estimator.

Let’s suppose we have a vector of discrete $RV$s $player$ of elements $player_{i} \$ with the following rule:

$$i \ne j \implies Pr\{player_{i}=player_{j}\} = 0$$

which means

$$i \ne j \implies player_{i} \ne player_{j}$$

Suppose we are given a $N\times N$ square matrix $R$.
A square matrix has the operator Trace ($tr$) defined as the sum of the elements on its main diagonal:

$$tr({R}) = \sum_{i=1}^{N}{r_{ii}}$$

Suppose that the value $tr(R)$ is part of the set of values that $player_{i}$ can assume.

Now suppose that the variable $player_{A}$ assumes the value of $tr(R)$. Indeed:

$$player_{A} = tr(R)$$

So $player_{A}$ is Trace(R).

Let’s now suppose that also another random variable $player_{x}$, with $x \ne A$ wants to assume the role of Trace(R).

However,

$$Pr\{player_{x}=tr(R)|player_{A}=tr(R)\} = 0$$

Therefore, the variable $player_{x}$ cannot assume the value of Trace(R) and should proceed to choose another value, since $player_{A}$ is already Trace(R).

Sorry about this stupid meme.

Meru.