Let $A$ be a fixed $n$ by $n$ real symmetric positive definite matrix with eigenvalues $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_n > 0$, and let $f(A):=\sum_{i=1}^n\log(\lambda_i)$, and let $X$ be a random $n$ by $k$ matrix with real iid copies distributed according to $N(0,\sigma^2/k)$.

The regime
----
- $n$ is fixed (in particular, $n \not \to \infty$).
- $k \to \infty$ (in particular, $k \gg n$).


Question
-----
- How close is $f(A+XX^T)$ to $f(A)$ in espectation ?
- What is an upper-bound for  $\mathbb P(|f(A+XX^T)-f(A)|  > \epsilon)$?

Observations
======
I've observed that $f(A+XX^T)$ is approximately $\mathcal N(\mu,s^2)$, for some $\mu \in \mathbb R$, and $s > 0$.

Back-of-envelop calculation
----
By triangle inequality, one has
$$
\begin{split}
|f(A+XX^T) - f(A)| \le &|f(A+\sigma^2 I_n) - f(A)|\\
&\quad + |f(A+XX^T) - f(A+\sigma^2 I_n)|.
\end{split}
\tag{*}
$$

Note that as $k\rightarrow \infty$, $XX^T \rightarrow \sigma^2 I_n$ in probability. Thus, by the delta method, we know that $f(A+XX^T) - f(A+\sigma^2 I_n) \longrightarrow \mathcal N(0,s^2/k)$, where
$$
s^2 := \sigma^4\|(A+\sigma^2 I_n)^{-1}\|_F^2 \le \sigma^4(\sum_{j=1}^n\lambda_j((A + \sigma^2)^{-1})^2 \le (\sigma^2\eta(A))^2,
$$
where $\eta(A) := \text{trace}A^{-1} = \sum_{j=1}^n 1/\lambda_j \le n/\lambda_n$.
On the other hand,
$$
|f(A + \sigma^2 I_n) - f(A)| = \sum_{j=1}^n\log(1 + \sigma^2/\lambda_j) \le \sigma^2\sum_{j=1}^n\lambda_j^{-1} = \sigma^2 \text{trace}A^{-1} = \sigma^2\eta(A).
$$

Putting everything together then gives

$$
\begin{split}
E_X|f(A+XX^T) - f(A)| &\le \sigma^2\frac{n}{\lambda_n} + E_X|f(A+XX^T)-f(A+\sigma^2 I_n)|\\
&\le \sigma^2\eta(A) + \sigma^2 \eta(A)/\sqrt{k} \to \sigma^2\eta(A).
\end{split},
$$

Thus it appears that,
> To have $E|f(A+XX^T) - f(A)|$ small, it is sufficient to have $\sigma^2 \eta(A) \ll 1$ and $k \rightarrow \infty$.

This doesn't solve my problem, but it raises suspicion to what the important problem parameters could be; here, $\sigma$, $k$, and $\eta(A)$ (or $n/\lambda_n$, for an even cruder analysis).