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Starting from a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$, a standard normal random variable $Z\sim\mathsf{N}\left(0,1\right)$ and a positive constant quantity $\varepsilon$, if I have to say something about the following probability: $$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\tag{1}$$ By applying Chebyshev's inequality, doesn't one have that (since $Z$ is a standard normal random variable, its second moment equals $1$): $$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\leq 1-\dfrac{1}{\dfrac{\varepsilon}{\Delta t^2}\left(1+\Delta t-\dfrac{\Delta t}{\varepsilon}\right)^2}\tag{2}$$ ?
If so, how can $(2)$ be furtherly expanded/simplified? My goal is to be able to say how $\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)$ behaves as $\varepsilon$ increases/decreases and as $\Delta t$ increases/decreases.


If my reasoning is wrong, which would be the correct application of Chebyshev's inequality to $(1)$?

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Let $M := M(\epsilon,\Delta t) = \frac{\sqrt{\epsilon}}{\Delta t}\left(1 + \left(1 - \frac{1}{\epsilon}\right)\Delta t\right)$. Recall that for any random variable $X$ with a finite second moment and non-negative number $x$, Chebyshev's inequality states:

$$\mathbb{P}(|X| > x) \leq \frac{\mathbb{E}[X^2]}{x^2}.$$

I'm guessing your application of the inequality roughly followed these steps:

$$\mathbb{P}(Z < -M) = 1 - \mathbb{P}(Z > -M) \leq 1 - \frac{1}{M^2}.$$

If I got that right, then the inequality is wrong for two reasons. First, Chebyshev's inequality should be applied to $\mathbb{P}(|Z| > M)$ instead of $\mathbb{P}(Z > -M)$. Second, the direction of the inequality is wrong as $a \leq b$ implies that $1 - a \geq 1 - b$. Instead, note that by symmetry of the standard Gaussian, $\mathbb{P}(Z < -M) = \frac{1}{2}\mathbb{P}(|Z| > M)$. By Chebyshev,

$$\mathbb{P}(Z < -M) = \frac{1}{2}\mathbb{P}(|Z| > M) \leq \frac{1}{2M^2} = \frac{(\Delta t)^2}{2\epsilon\left(1 + \left(1 + \frac{1}{\epsilon}\right)\Delta t\right)^2}.$$

From here you can use standard calculus to analyze the expression as $\epsilon$ and $\Delta t$ increase, decrease and converge to $0$ or whatever else you want to study.

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Chebyshev bounds are "loose" bounds and useful when you don't know the distribution. Here you know the distribution of $Z$. Why not directly work with the distribution function?

For example, let $y = -\frac{\sqrt{\epsilon}}{\Delta t}(1 + \frac{\epsilon - 1}{\epsilon}\Delta t)$. Then, $$ N(y) = P(Z < y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y}e^{-\frac{x^2}{2}}\;dx $$ $$\frac{\partial N(y)}{\partial \epsilon} = N'(y) \frac{\partial y}{\partial \epsilon}$$ $$\frac{\partial N(y)}{\partial \Delta t} = N'(y) \frac{\partial y}{\partial \Delta t}$$

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