I've been struggling on the following exercice lately :
Consider $(X_n)_{n\ge 1}$ a sequence of mutually independent real-valued random variables, standardized, that follow any given law. Let $Y$ be a real random variable such that $\mathbb{E}[Y]$ is finite. What is the limit of $(\mathbb{E}[Y X_n])_{n\ge 1}$ ?
Let's write the variance-covariance matrix of the vector $(Y,X_1,\dots,X_n)$. We will denote this $(n+1) \times (n+1)$ matrix by $\Sigma_n$: $$\Sigma_n = \begin{pmatrix} \operatorname{Var}[Y] & \operatorname{Cov}(Y, X_{1}) & \cdots & \operatorname{Cov}(Y,X_{n}) \\ \operatorname{Cov}(Y, X_{1}) & \ddots & \cdots & \vdots\\ \vdots & \vdots & \ddots & \vdots\\ \operatorname{Cov}(Y, X_{n}) & \cdots & \cdots& \operatorname{Var}[X_n] \end{pmatrix} = \begin{pmatrix} \operatorname{Var}[Y] & \mathbb{E}[YX_1] & \cdots & \cdots & \mathbb{E}[YX_n] \\ \mathbb{E}[YX_1] & 1 & 0 & \cdots & 0\\ \vdots & 0 & \ddots & \ddots & \vdots\\ \mathbb{E}[YX_{n-1}] & \vdots & \ddots & 1 & 0\\ \mathbb{E}[YX_n] & 0 & \cdots & 0 & 1 \end{pmatrix}$$
We used the fact that the $X_i$ are normalized and mutually independent.
One thing that we know with this type of matrix is that, to be a covariance matrix, it has to be positive semi-definite. In particular, its eigenvalues are positive. Thus we can have a look at the roots of the $\lambda$-polynomial $\text{det}(\Sigma_n - \lambda I_{n+1}).$ I do end up with a recursive expression for this determinant by developing the last line, but it doesn't seem to be enough to conclude. We also have $\displaystyle \sum \operatorname{eigenvalues} = \operatorname{Tr}(\Sigma_n) = \operatorname{Var}[Y] + n.$
Do you have any other ideas to solve this ? Thanks in advance.