# Which is the correct application of Chebyshev's inequality in the below case?

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 $$(1)$$?

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.

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