Let $\mathcal{H}$ be a Hilbert space over $\mathbb{R}$ with inner product $\langle \cdot , \cdot \rangle$ and let $(X_n)_{n \in \mathbb{N}}$ and $(Y_n)_{n \in \mathbb{N}}$ be i.i.d sequences of random variables with values in $\mathcal{H}$. Assume that $E\langle X_n, Y_n \rangle = 0$, $0 < E( \lVert X_n \rVert^2 \lVert Y_n \rVert^2)< \infty$ and that the sequence of inner products is independent. Does

$$ \frac{1}{\sqrt{n}} \sum_{i=1}^n \langle X_i, Y_i \rangle \overset{\mathcal{D}}{\to} N(0, E((\langle X_1, Y_1 \rangle)^2)) $$

then follow from the classical CLT?


Let $Z_n = \langle X_n, Y_n\rangle$. We know that the sequence $(Z_n)$ is independent, and that $E Z_n = 0$ and $\text{Var}(Z_n) \le E(|X_n|^2 |Y_n|^2) < \infty$. If the $Z_n$ also are identically distributed (which you seem to assume), then your result is simply an application of the classical CLT on the sequence $(Z_n)$.

  • $\begingroup$ I don't see where OP states that $Z_n$'s are identically distributed. Classical CLT does not apply. $\endgroup$ – Kavi Rama Murthy Feb 11 at 23:19
  • $\begingroup$ I added the iid assumption. I see how this holds but I'm just uncomfortable with the simplicity of it. I can hardly believe that it is this simple. I presume I could use LLN for the inner products aswell? $\endgroup$ – Lundborg Feb 12 at 8:14

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