Let $A_n$ and $B_n$ be two sequences of random variables in $\mathbb{R}^K$ for $K\geq 1$. I would like to show that $plim V(A_n) = plim V(B_n)$. I have the following information
- $plim A_n = plim B_n$.
- $A_n - B_n$ is uniformly square integrable; that is, for each component of $A_n - B_n$, the sequence made of the square is uniformly integrable.
I succeeded to prove the statement for $K = 1$ but I am unable to generalize it. As $A_n-B_n$ is uniformly square integrable, the improved dominance convergence theorem (Lebesgue–Vitali theorem) implies that $plim V(A_n-B_n) = V(plim A_n-plim B_n) = 0$.
If $A_n$ and $B_n$ are scalars, then $V(A_n-B_n) = V(A_n) + V(B_n) - 2COV(A_n, B_n) \geq V(A_n) + V(B_n) - 2V(A_n)^{1/2}V(B_n)^{1/2}$, because $COV(A_n, B_n)\leq V(A_n)^{1/2}V(B_n)^{1/2}$. Thus, $V(A_n-B_n) \geq \left(V(A_n)^{1/2} - V(B_n)^{1/2}\right)^2 \geq 0$. Therefore $plim V(A_n) = plim V(B_n)$ since $plim V(A_n-B_n) = 0$.
I thought that it would be easy to generalize it to vector random variables. But I cannot do it. The condition $COV(A_n, B_n)\leq V(A_n)^{1/2}V(B_n)^{1/2}$ originates from Cauchy Schwarz inequality and the multivariate version seems completely different to me.