Suppose that $V_i$, for $i \in \mathbb{N}$, are i.i.d. standard normal random variables and $Y_i = \sum_{k=1}^i V_k$ for $i \in \mathbb{N}$ with $Y_0 = 0$. Let $X_n = (\sum_{i=1}^n V_i Y_{i-1})^2 Y_n^2$ and $Z_n = (\sum_{i=1}^n Y_{i-1}^2)^2$.
How can I show that $\lim_{n \rightarrow \infty} p(Z_n - X_n \geq 0) = 1$? From simulations, I have seen that this probability converges to 1 very fast.
Intuitively, this makes sense because the we have $E[Z_n] = \Theta(n^4)$ and $\sigma(Z_n) = \Theta(n^4)$ but $E[X_n] = \Theta(n^3)$ and $\sigma(Z_n) = \Theta(n^3)$. Chebyshev inequality implies that $X_n$ samples are mostly around $n^3$ but $Z_n$ samples are mostly around $n^4$. Thus, there is a higher chance that $Z_n$ is greater than $X_n$.