I've been idly watching Probability and Stochastics for finance: NPTEL , and I came across this nice way to think about the markov and chebyshev inequality. I wonder whether Chernoff bounds also fall to this viewpoint.

§ Markov's inequality

In markov's inequality, we want to bound P(X≥A)P(X \geq A). Since we're in measure land, we have no way to directly access P(⋅)P(\cdot). The best we can do is to integreate the constant function 11, since the probability is "hidden inside" the measure. This makes us compute:

P(X≥A)≡=∫{X≥A}1dμ P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu

Hm, how to proceed? We can only attempt to replace the 11 with the XX to get some non-trivial bound on XX. But we know that X≥AX \geq A. so we should perhaps first introduce the AA:

P(X≥A)≡=∫{X≥A}1dμ=1/A∫{X≥A}Adμ P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = 1/A \int_{\{X \geq A\}} A d \mu

Now we are naturally led to see that this is always less than XX:

P(X≥A)≡=∫{X≥A}1dμ=1/A∫{X≥A}Adμ<1/A∫{X≥A}Xdμ=1/AE[X] \begin{aligned} &P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = \\ & 1/A \int{\{X \geq A\}} A d \mu < 1/A \int_{\{X \geq A\}} X d \mu = 1/A \mathbb{E}[X] \end{aligned}

This completes marov's inequality:

P(X≥A)≤E[X]/A P(X \geq A) \leq \mathbb{E}[X]/A

So we are "smearing" the indicator 11 over the domain {X≥A}\{X \geq A\} and attempting to get a bound.