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


Question
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- 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
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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
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By triangle inequality, one has
$$
|f(A+XX^T) - f(A)| \le |f(A+\sigma^2 I_n) - f(A)| + |f(A+XX^T) - f(A+\sigma^2 I_n)|
$$

Note that as $k\rightarrow \infty$, $XX^T \rightarrow \sigma^2 I_n$ in probability. Thus, by the delta method, we know that $f(A+XX^T) - f(A+I_n) \longrightarrow \mathcal N(0,s^2/k)$, for some $s > 0$.

On the other hand,
$$
|f(A + \sigma^2 I_n) - f(A)| = \sum_{j=1}^n\log(1 + \sigma^2/\lambda_j) \le \sigma^2\sum_{j=1}^n\lambda_j^{-1} = \sigma^2 \text{trace}(A^{-1}) \le \sigma^2\frac{n}{\lambda_n}.
$$

Putting everything together then gives

$$
\begin{split}
E|f(A+XX^T) - f(A)| &\le |f(A+\sigma^2 I_n)-f(A)| + \mathcal O(1/\sqrt{k}) \le \sigma^2\frac{n}{\lambda_n} + \mathcal O(1/\sqrt{k}).
\end{split}
$$

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