Prove that the product of 2 vectors Normally distributed converges for large dimensions to the full zero matrix

Question

Let $\mathbf{x}, \mathbf{y}$ $\in C^{M \times 1}$ are two i.i.d. vectors with distribution $\mathcal{CN(0,1)}$.
How we can prove by the strong law of large numbers that: $\lim_{M\rightarrow \infty} (\frac{\mathbf{x} \mathbf{y}^h}{M})$ converges almost surely to $\mathbf{0}_{M \times M}$, where $\mathbf{0}_{M \times M}$ is the full zero matrix of dimension $M \times M$.

Answer 1

Suppose that $X_M, Y_M \in \mathbb{C}^{M+1}$ are vectors whose components $X_{Mi}, Y_{Mi}$ are iid $\mathbb{C}\mathcal{N}(0,1)$ random variables. I want to show that $$\lim_{M \to \infty} \frac{X_{Mi}\overline{Y_{Mj}}}{M} = 0$$ for all $i, j$.

First suppose $M + 1 > \max(i, j)$ (so the vectors actually contain an $i$th and $j$th element). Then \begin{eqnarray*} |Var(X_{Mi}\overline{Y_{Mj}})| &=& |E[(X_{Mi}\overline{Y_{Mj}})^2] - E[X_{Mi}\overline{Y_{Mj}}]^2| \\ &=& |E[(X_{Mi}\overline{Y_{Mj}})^2]| \\ &\leq& E[(|X_{Mi}|^2]|E[\overline{|Y_{Mj}}|^2] \\ \end{eqnarray*} We used the means of the random variables are zero in the first line.

The modulus of the variance of $X_{Mi}\overline{Y_{Mj}}$ is therefore finite (since the final terms are finite) and independent of $M$. Therefore,

$$\lim_{M \to \infty} |Var(\frac{X_{Mi}\overline{Y_{Mj}}}{M})| = 0$$ Since their variances go to zero and their mean is always zero, the random variables $\frac{X_{Mi}\overline{Y_{Mj}}}{M}$ converge to the dirac measure at $0$.