Inequalities in Chebyshev’s Inequality and Chernoff Bounds proofs Following this book
Probability and Computing: Randomization and Probabilistic Techniques in Algorithms and
Data Analysis Second Edition; Michael Mitzenmacher and Eli Upfal
they show Chebyshev’s Inequality (page 51)
$$
\Pr(|X - \mathbb{E}[X]| \geq a) \leq \frac{\mathbb{Var}[X]}{a^2}
$$
and start the proof by this "trick"

We first observe that

$$
\Pr(|X - \mathbb{E}[X]| \geq a) = \Pr((X - \mathbb{E}[X])^2 \geq a^2)
$$
If we ignore the mean and assume non-negative values of X, it basically says
$$
\Pr(X \geq a) = \Pr(X^2 \geq a^2)
$$
Later on, they introduce Chernoff Bounds (p. 68) by this equality
$$
\Pr(X \geq a) = \Pr(e^{tX} \geq e^{ta})
$$
for some "well-chosen" $t$.
It seems like the general rule would be
$$
\Pr(X \geq a) = \Pr(g(X) \geq g(a))
$$
for some function $g: \mathbb{R} \to \mathbb{R}$
But I haven't seen such a general rule in any textbook. Does it always hold true? How can we prove it?
Update:
So it turns out that if $g$ is increasing, the events $\{X \geq a\}$ and $\{g(X) \geq g(a)\}$ are the same. How can one prove it?
 A: This is true as long as $g$ is increasing. Indeed, if $g$ is increasing, the events $\{X \geq a\}$ and $\{g(X) \geq g(a)\}$ are the same, so of course they have the same probability:
$$\begin{align}\omega \in \{X \geq a\} & \Leftrightarrow X(\omega) \geq a  \quad \text{by defintion of the set} \\ & \Leftrightarrow g(X(\omega)) \geq g(a) \quad \text{since } g \text{ is increasing} \\ & \Leftrightarrow \omega \in \{g(X) \geq g(a)\}. \quad\text{by defintion of the set} \end{align}$$
For instance, in the first case, $|X-\mathbb{E}(X)|$ is nonnegative and the function $g : y \mapsto y^2$ is increasing in $[0, +\infty)$, so as long as $a \geq 0$, we can apply the observation above to get
$$\mathbb{P} (|X-\mathbb{E}(X)| \geq a) = \mathbb{P} (|X-\mathbb{E}(X)|^2 \geq a^2).$$
Note that this equality is false in general if $a < 0$ (exercise : find a counter-example), and our argument fails because $g : y \mapsto y^2$ is not increasing on $\mathbb{R}$.
The case of the exponential is even simpler, but requires $t > 0$ so that $g : y \mapsto e^{ty}$ is increasing.
While I don't recall seeing this principle written explicitely, it is useful to remember it in full generality. The main idea is that, the stronger constraints you have on the distribution $X$, the stronger bounds you get on its tail. Chernov bounds requires exponential moments, but gets you an exponential bound on the tail. In the other direction, I have sometimes used a version such as (for $X$, $a \geq 0)$
$$\mathbb{P} (X \geq a) = \mathbb{P} (\ln (1+X) \geq \ln(1+a)) \leq \frac{\mathbb{E} (\ln (1+X))}{\ln(1+a)},$$
which requires a very weak condition on $X$ (its logarithm needs to be integrable), but gives you a weak bound on its tail. So this family of inequalities is quite flexible.
