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 $$ s^2 := n\sigma^4\|(A+\sigma^2 I_n)^{-1}\|_F^2 \le \sigma^4(\sum_{j=1}^n\lambda_j((A + \sigma^2)^{-1})^2 \le (\sqrt{n}\sigma^2\eta(A))^2, $$ where $\eta(A) := \text{trace}(A+\sigma^2 I_n)^{-1} = \sum_{i=1}^n (\lambda_i + \sigma^2)^{-1} \le n/\lambda_n$. On the other hand, $$ |f(A + \sigma^2 I_n) - f(A)| = \sum_{i=1}^n\log(1 + \sigma^2/\lambda_i) \le \sigma^2\sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1} = \sigma^2\eta(A). $$

Putting everything together then gives

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

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).

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 $$ s^2 := n\sigma^4\|(A+\sigma^2 I_n)^{-1}\|_F^2 \le \sigma^4(\sum_{j=1}^n\lambda_j((A + \sigma^2)^{-1})^2 \le (\sqrt{n}\sigma^2\eta(A))^2, $$ where $$ \begin{split} \eta(A) &= \eta(A; \sigma^2) := \text{trace}(A+\sigma^2 I_n)^{-1} = \sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1} \le n\min(\sigma^{-2},\lambda_n(A)^{-1}). \end{split} $$

On the other hand, $$ |f(A + \sigma^2 I_n) - f(A)| = \sum_{i=1}^n\log(1 + \sigma^2/\lambda_i) \le \sigma^2\sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1} = \sigma^2\eta(A). $$

Putting everything together then gives

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

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).

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).

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 $$ s^2 := n\sigma^4\|(A+\sigma^2 I_n)^{-1}\|_F^2 \le \sigma^4(\sum_{j=1}^n\lambda_j((A + \sigma^2)^{-1})^2 \le (\sqrt{n}\sigma^2\eta(A))^2, $$ where $\eta(A) := \text{trace}(A+\sigma^2 I_n)^{-1} = \sum_{i=1}^n (\lambda_i + \sigma^2)^{-1} \le n/\lambda_n$. On the other hand, $$ |f(A + \sigma^2 I_n) - f(A)| = \sum_{i=1}^n\log(1 + \sigma^2/\lambda_i) \le \sigma^2\sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1} = \sigma^2\eta(A). $$

Putting everything together then gives

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

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).

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 $$ |f(A+XX^T) - f(A)| \le |f(A+\sigma^2 I_n) - f(A)| + |f(A+XX^T) - f(A+\sigma^2 I_n)| $$

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

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).