I'm trying to find a lower bound on $\text{Var}(X / Y)$ for dependent random variables $X, Y \in [0, 1]$ with $X \leq Y$. More specifically, $X$ and $Y$ are defined as follows:
Let $h, n \in \mathbb{N}$ be positive integers and $\{Z_k, k \in \mathbb{N}_0\}$ be a strictly stationary time series whose margins are absolutely continuous random variables with support $(0, \infty)$. Then, it holds that $$ X = \frac{1}{n} \sum_{k = 1}^n \mathbb{1}\{X_k > u, X_{k + h} > u\}, \quad Y = \frac{1}{n} \sum_{k = 1}^n \mathbb{1}\{X_k > u\}, $$ where $u > 0$ is some threshold. Furthermore, assume that the correlation between $X$ and $Y$ lies in (-1, 1). Also, we know that the ratio $\mathbb{P}(Z_0 > u, Z_h > u) / \mathbb{P}(X_0 > u)$ converges to some constant in $(0, 1)$.
Online, I found the following approximation for general random variables $X$ and $Y$ with support $(0, \infty)$:
$$ \text{Var}(X / Y) \approx \frac{(\mathbb{E}X)^2}{(\mathbb{E}Y)^2} \bigg( \frac{\text{Var}(X)}{(\mathbb{E}X)^2} - 2 \frac{\text{Cov}(X, Y)}{\mathbb{E}X\mathbb{E}Y} + \frac{\text{Var}(Y)}{(\mathbb{E}Y)^2} \bigg) \tag{1} $$
The underlying proof of this approximation seems to be a Taylor expansion. However, it appears that all remainder terms of the expansion were just removed and equality signs $=$ turned to approximation $\approx$. Is there any justification why the remainder terms are negligible in the variance? This is non-trivial to me.
In terms of my desired lower bound, the RHS of (1) is something I can work with, so I'd like to show something along the lines of $$ \text{Var}(X / Y) \geq C \frac{(\mathbb{E}X)^2}{(\mathbb{E}Y)^2} \bigg( \frac{\text{Var}(X)}{(\mathbb{E}X)^2} - 2 \frac{\text{Cov}(X, Y)}{\mathbb{E}X\mathbb{E}Y} + \frac{\text{Var}(Y)}{(\mathbb{E}Y)^2} \bigg) $$ for some constant $C > 0$. I tried proving this using a Taylor expansion which yields
$$ \begin{align} \text{Var}(X / Y) &= \frac{(\mathbb{E}X)^2}{(\mathbb{E}Y)^2} \bigg( \frac{\text{Var}(X)}{(\mathbb{E}X)^2} - 2 \frac{\text{Cov}(X, Y)}{\mathbb{E}X\mathbb{E}Y} + \frac{\text{Var}(Y)}{(\mathbb{E}Y)^2} \bigg) \\ &+ \text{Var}(R_1) + \frac{2}{\mathbb{E}Y} \mathbb{E} \big[(X - \mathbb{E}X) R_1 \big] - \frac{2\mathbb{E}X}{(\mathbb{E}Y)^2} \mathbb{E} \big[(Y - \mathbb{E}Y) R_1 \big], \end{align} $$ where $R_1$ is the remainder term $$ R_1 = \frac{Y - \mathbb{E}Y}{((1 - \theta) \mathbb{E}Y + \theta Y)^2} \bigg( \frac{(1 - \theta) \mathbb{E}X + \theta X}{(1 - \theta) \mathbb{E}Y + \theta Y} - (X - \mathbb{E}X) \bigg), $$ for some $\theta \in [0, 1]$.
Consequently, it would be desirable if one could show that $$ \text{Var}(R_1) + \frac{2}{\mathbb{E}Y} \mathbb{E} \big[(X - \mathbb{E}X) R_1 \big] - \frac{2\mathbb{E}X}{(\mathbb{E}Y)^2} \mathbb{E} \big[(Y - \mathbb{E}Y) R_1 \big] \geq 0. $$ (greater than some other non-negative constant would work too.) I tried showing this using the boundedness of $X$ and $Y$ and their monotonic relationship but so far I failed to make this work.
Any ideas on
- how to make this proof work? or
- how to find a lower bound for the variance some other way? or
- why the remainder terms in (1) were negligible?
