Chebyshev's Inequality: Computing the maximum probability of a return falling 3.5 σs from the mean

Code: https://github.com/dashn9/learning_quant/blob/main/src/lessons/downside_tail_risk.rs

Chebyshev’s inequality is an extension of Markov’s Inequality.

A bit about Markov’s Inequality: it applies to non-negative random variables, i.e., X >= 0. It states that P(X >= t) <= (E[X] / t). Using the law of total expectation of E [X], it states that the weighted sum of all values below and above t.E[X]= E(X|X < t) . P(X < t) + E(X|X >= t) . P(X >= t).

The differences between Markov’s and Chebyshev’s inequality.

1. Chebyshev does not care about the direction of the values, just the magnitude of σ from the mean.

2. Chebyshev requires σ^2, an extension of Markov that just requires μ.

3. Markov inequality states that a value of x cannot exceed an upper bound, whereas Chebyshev’s inequality tells you that, for a given value x, it cannot stray away from the mean too much to exceed the upper bound.Basically, both say that the biggest value cannot be too big and many beyond a point.

Chebyshev’s Inequality: P((x - μ)^2 > t^2) <=σ^2 /t^2The reason why both sides of this inequality are squared is to eliminate negatives and make sure the distance can be calculated neutrally, i.e it gauges in both number line directions — (μ - t and μ + t). This depicts the threshold value on either end. I’m still trying to understand how σ^2 /t^2 determines the upper bound, but I guess I will soon discover.σ^2 = E[(x - μ)^2 ||x - μ| < t] * P(|x - μ| < t) + E[(x - μ)^2 ||x - μ| >= t] * P(|x - μ| >= t) = (E(x - μ) ^ 2) / nLol, I have not fully proved the equation above for myself.

Resourceshttps://www.youtube.com/watch?v=mlelI1LA9o4

https://www.youtube.com/watch?v=e-nAr3MkAII

Note: These are notes from my learning journey; please do not implement the strategy with real money unless you know what you are doing.