Let $v \in \mathbb R^n$ be a random vector from $N(0,C)$, where $C$ is an a psd matrix of size $n$ such that $\lim_{n \to \infty} \mbox{tr}(C) = s \in [0, \infty)$. Consider the $d \times d$ psd matrix $R:=\|v\|^2 v v^\top$. Let $B$ be a deterministic $n \times n$ such that $\lim_{n \to \infty} \mbox{tr}(B) = \overline{t} \in (-\infty,\infty)$.
Question. In the limit $n \to \infty$, what is an asymptotic approximation for $\mbox{tr}(BR)$ valid in $L_2$ (or in probability) ?
Attempt
I'm not quite sure of my argument, so I'd greatly appreciate if someone could confirm / infirm.
So, we can write $R = \mbox{tr}(C)vv^\top + (\|v\|^2 - \mbox{tr}(C)vv^\top)$; the first term dorminates the second since $\|v\|^2$ has gaussian concentration around its expectation $\mbox{tr}(C)$. We deduce that if $\lim_{n \to \infty} \mbox{tr}(BC) = \overline{t} \in [0,\infty)$, then
$$ \mathbb E \mbox{tr}(BR) \to \mbox{tr}(C)\mbox{tr}(BC) \to s\overline{t}. $$
If, I could show that the variance of $\mbox{tr}(BR)$ is $o(1)$ (i.e converges to zero), then Chebychev's inequality would imply $\mbox{tr}(BR) \to s\overline{t}$ in $L_2$ (and therefore in probability too).