# Convergence rate of the smallest eigenvalue of an integral of a multivariate squared Brownian Motion

I am interested in deriving the convergence rate of the smallest eigenvalue of a sequence of random matrices with diverging dimension. More precisely, let $$W_n(r)$$ represent an $$n$$-dimensional standard brownian motion at time $$r$$, and define $$\lambda_1(A)$$ as the minimum eigenvalue of A. Then, I would like to know how fast does $$\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)$$ converge to zero in probability, as $$n \to \infty$$.

First of all, the reason that I believe that the minimum eigenvalue converges to zero in probability is given by the following argument, which was kindly explained to me by fellow MathOverflow user Nate Eldredge. Let $$E = \lbrace e_1,\ldots,e_n\rbrace$$ represent the collection of basis vectors spanning $$\mathbb{R}^n$$. Then, for any $$\epsilon > 0$$,

$$\mathbb{P}\left(\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right) > \epsilon \right) = \mathbb{P}\left(\underset{x \in \mathbb{R}^n\setminus \lbrace 0 \rbrace}{\inf}\frac{x^\prime\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)x}{x^\prime x} > \epsilon \right) \leq \mathbb{P}\left(\underset{x \in E}{\inf} x^\prime\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)x > \epsilon \right) = \mathbb{P}\left(\underset{1 \leq i \leq n}{\min} \int_0^1 W_{n,i}^2(r)dr > \epsilon \right) = \mathbb{P}\left(\int_0^1 W_{n,1}^2(r)dr > \epsilon \right)^n \to 0,$$

as $$n \to \infty$$. The last inequality follows from the $$i.i.d.$$-ness of the elements in $$W_n(r) = (W_{n,1}(r),\ldots,W_{n,n}(r))^\prime$$. Then, what I would like to know essentially is if there exists an increasing function in $$n$$, say $$f(n)$$, such that $$\mathbb{P}\left(f(n)\times\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right) > \epsilon \right) \to 1$$, and, if so, how does $$f(n)$$ look like?

Here is my thought process so far:

From Theorem 10 in Tolmatz (2002), it can be shown that $$\mathbb{P}\left(n^a \times \int_0^1 W_{n,1}^2(r)dr > \epsilon \right)^n \to 1$$, such that $$f(n) = n^a$$ would be sufficient to show that the upper bound that I used before converges to zero in probability slower than $$n^a$$. Unfortunately, what I am really after is showing that the minimum eigenvalue converges to zero in probability sufficiently slow. Hence, if I could find an analytically manageable lower bound for the minimum eigenvalue and show the rate at which it goes to zero (from above), my problem would be solved. The minimum eigenvalue bounds that I could find, such as the one based on Gershgorin's circle theorem, are not strict enough, as they lead to negative lower bounds.

Hope somebody can help me with this issue. Any suggestions, either in the form of a nice minimum eigenvalue bound or perhaps in terms of a different strategy to derive my answer, would be sincerely appreciated.

Best,

Etienne

Cited article: Tomatz (2002) On the distribution of the square integral of the Brownian Bridge. The Annals of Probability,30, 253-269