On the relation of the expected squared norm of a random vector with its dimensionality

Question

Let $\mathbf{x}=(x_1, \ldots,x_n)^\top\in\Bbb{R}^n$ be a random vector with mean $\mathbf{\bar{x}}=(\bar{x}_1, \ldots,\bar{x}_n)^\top\in\Bbb{R}^n$ and covariance matrix $\Sigma\in\Bbb{S}_{++}^n$, where $\Bbb{S}_{++}^n$ denotes the space of symmetric positive definite $n\times n$ matrices with entries in $\Bbb{R}$.

I'm interested to prove (or disprove) that the expected value of $\Vert\mathbf{x}\Vert^2$ is linear to $\mathbf{x}$'s dimensionality $n$, i.e., $\Vert\mathbf{x}\Vert^2 = \lambda\cdot n$, where $\lambda>0$.

What I have so far is as follows: $$ \Bbb{E}\left[\Vert\mathbf{x}\Vert^2\right] = \Bbb{E}\left[\mathbf{x}^\top\mathbf{x}\right] = \sum_{i=1}^{n}\Bbb{V}\left[x_i\right] + \sum_{i=1}^{n}\left(\Bbb{E}\left[ x_i \right]\right)^2 = tr(\Sigma) + \sum_{i=1}^{n}\left(\Bbb{E}\left[ x_i \right]\right)^2. $$

Now, if $\Sigma=\sigma^2I_n$ and $\bar{\mathbf{x}}=\mathbf{0}$, we have that $$ \Bbb{E}\left[\Vert\mathbf{x}\Vert^2\right] = tr(\Sigma) + \sum_{i=1}^{n}\left(\Bbb{E}\left[ x_i \right]\right)^2 = n\cdot\sigma^2 + 0, $$ which is what I want. But what if $\Sigma$ is not a multiple of the identity matrix? In other words, how does the trace of an $n\times n$ matrix relate to $n$?

Another approach would be to "standarize" the random vector $\mathbf{x}$, i.e., to create a new random vector $\mathbf{y}=C(\mathbf{x}-\bar{\mathbf{x}})$, where $C = B^{-1}$ and $B$ such that $\Sigma_x = BB^\top$. Then could we work with $\mathbf{y}$ instead of $\mathbf{x}$? In such a case, it would be $\Bbb{E}\left[\Vert\mathbf{y}\Vert^2\right]=n$. What do you think?

Answer 1

My answer focuses on the part of $tr(\Sigma)$. Since covariance matrix $\Sigma$ is symmetric, $\Sigma$ can be diagonalized as

$$ \Sigma = U^{-1}DU $$ with an orthogonal matrix $U$ and a diagonal matrix $D$. From the property of trace, $$ tr(\Sigma) = tr(U^{-1}DU) = tr(DUU^{-1}) = tr(D). $$

Importantly, the (diagonal) elements of $D$ correspond to the eigenvalues of $D$, such that

$$ D = \begin{pmatrix} \lambda_1 & 0 & 0 & 0& 0 \\ 0 & \lambda_2 & 0 & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & 0 & \lambda_{n-1} & 0 \\ 0 & 0 & 0 & 0 & \lambda_n \end{pmatrix}, $$ and thus $$ tr(\Sigma) = \sum_{i=1}^{n} \lambda_i . $$ So my understanding is:

• there is no direct connection between $\mathop{\mathbb{E}} [\Vert x \Vert^2] $ and the dimensionality of $x$.
• but if a set of $x$ is highly biased in some directions, the variance is higher in those directions and $\lambda$s bocomes larger, possibly resulting in a larger $\mathop{\mathbb{E}} [\Vert x \Vert^2] $.