Convergence of a quadratic form in probability

Question

Suppose $x$ is a random vector in $R^n$ with positive definite covariance matrix, $R$. I'm interested in the convergence in probability of the quadratic form, $\dfrac{1}{n}x^TR^{-1}x$. If $x$ were Gaussian, this would be simple enough since were can write, $x^TR^{-1}x = \sum\limits_{i=1}^ny_i^2$ where $y_i \sim N(0,1)$ i.i.d. Then, $\dfrac{1}{n}x^TR^{-1}x \xrightarrow{P} 1$ by the law of large numbers. How about for a non-Gaussian $x$? The only thing that I can calculate is,

$$ E\left[\dfrac{1}{n}x^TR^{-1}x\right] = 1, \;\;\; \forall n$$

But I don't think this is enough to ensure that $\frac{1}{n}x^TR^{-1}x \xrightarrow{P} 1$ (I don't think I can apply Chebyshev?). Are there other conditions on $x$ that I can impose here?

EDIT: I guess a more general way of asking the question would be, let $\{X_n\}$ be a non-negative sequence of random variables such that, $$ \lim\limits_{n\rightarrow\infty}E[X_n] = 1 \; (\text{or} \;E[X_n] = 1 \;\; \forall n)$$ What conditions ensure that, $X_n \xrightarrow{P} 1$?

Answer 1

Suppose the covariance matrix of $X$ is $R\in \mathbb{R}^{n\times n}$ and the expectation of $X$ is $\mu \in \mathbb{R}^{n}$

Because $R$ is a symetric definite positive matrix, then its square root matrix $R^\frac{1}{2}$ defined as $R^\frac{1}{2} R^\frac{1}{2} =R $ is also a symetric definite positive matrix. The matrix $R^\frac{1}{2}$ is also invertible.

Denote $Y = R^{-\frac{1}{2}}X$, then $X = R^\frac{1}{2}Y$. We prove that the covariance matrix of $Y$ is the identity matrix $I$. Indeed, we have \begin{align} Cov(Y) &= E((R^{-\frac{1}{2}}X)(R^{-\frac{1}{2}}X)^T) - E(R^{-\frac{1}{2}}X)(E(R^{-\frac{1}{2}}X))^T \\ &= E(R^{-\frac{1}{2}}XX^TR^{-\frac{1}{2}}) - R^{-\frac{1}{2}}E(X)E(X)^TR^{-\frac{1}{2}} \\ &= R^{-\frac{1}{2}}E(XX^T)R^{-\frac{1}{2}} - R^{-\frac{1}{2}}E(X)E(X)^TR^{-\frac{1}{2}} \\ &= R^{-\frac{1}{2}} Cov(X) R^{-\frac{1}{2}} \\ &= R^{-\frac{1}{2}}R R^{-\frac{1}{2}} \\ &= I \\ \end{align} In particular, we have $V(Y_i) = E(Y_i^2)-E(Y_i)^2 =1\tag{1}$ with $Y_i$ is the i-th element of the random vector $Y$.

Because $Y = R^{-\frac{1}{2}}X$, hence $E(Y)= R^{-\frac{1}{2}}\mu \tag{2}$

From (1) and (2), we deduce that $$\sum_{i=1}^n E(Y_i^2) = n- (\mathbf{1}^T R^{-\frac{1}{2}}\mu)$$ with $\mathbf{1} \in \mathbb{R}^{n}$ is the all-ones vector. For information, the term $\mathbf{1}^T R^{-\frac{1}{2}}\mu$ is the sum of all elements of the $R^{-\frac{1}{2}}\mu$ (the expectation vector of $Y$).

We have now \begin{align} \frac{1}{n}X^TR^{-1}X &= \frac{1}{n}(R^\frac{1}{2}Y)^TR^{-1}(R^\frac{1}{2}Y) \\ &= \frac{1}{n} Y^T R^\frac{1}{2}R^{-1}R^\frac{1}{2}Y \\ &= \frac{1}{n} Y^T Y \\ &= \frac{1}{n} \sum_{i=1}^n Y_i^2 \\ \end{align}

By denoting $S_n = \frac{1}{n} \sum_{i=1}^n Y_i^2 $ the problem becomes: what is the condition which ensures that $S_n$ converges in probability given that $E(S_n)= 1 - \frac{1}{n}(\mathbf{1}^T R^{-\frac{1}{2}}\mu)$.

For sufficient conditions, for example, if we have 2 following conditions:

1. the term $\frac{1}{n}(\mathbf{1}^T R^{-\frac{1}{2}}\mu)$ converges (to $a \in \mathbb{R}$ for example)
2. $E(|S_n - (1-a)|) \xrightarrow{n\rightarrow +\infty} 0$ (convergence in $L_1$)

Then $S_n$ converges in probability to $1-a$, as the convergence in $L_1$ implies the convergence in probability.