Why is Chebyshev Bound stronger than Markov if it is an application of Markov? The Chebyshev bound is merely an application of the Markov bound. Why is it considered a strong / stricter / more powerful bound? I know that Chebyshev is two sided whereas Markov is only one sided, and that Chebyshev uses more information than Markov (needs to know both the 2nd moment and the 1st moment), but I don't see why Chebyshev is necessarily stronger.
I've not seen this explained anywhere. Specifically, this is always somehow stated as an obvious observation, with no justification.
 A: Suppose $X\ge 0$ with mean $\mu$ and standard deviation $\sigma$.  I presume you're talking about the Markov bound  $$ \mathbb P[X \ge a] \le \dfrac{\mu}{a} $$
and the Chebyshev bound 
$$\mathbb P[|X - \mu| \ge k \sigma] \le \frac{1}{k^2}$$ 
Thus if $a = \mu + k \sigma$ for $k > 0$,
$$ \mathbb P[X \ge a] \le \mathbb P[|X-\mu| \ge k \sigma] \le \frac{1}{k^2} = \frac{\sigma^2}{(a-\mu)^2}$$
Thus the Chebyshev bound here is stronger if
$$ \dfrac{\sigma^2}{(a-\mu)^2} < \dfrac{\mu}{a}$$ 
This is equivalent to 
$$ a^2 \mu - (2 \mu^2 + \sigma^2) a + \mu^3 > 0$$
That's certainly true if $a$ is sufficiently large, but not if $a$ is close to $\mu$: the Chebyshev bound is useless unless $k > 1$ (i.e. $a > \mu + \sigma$), while Markov gives you some useful information as soon as $a > \mu$.
Typically (at least in a theoretical context) we're mostly concerned with what happens when $a$ is large, so in such cases Chebyshev is indeed stronger.
A: A simpler way to look at this question is by considering the different bounds (due to Markov and Chebyshev) with the following inequalities for a positive random variable $X$ and a positive constant $t$.


*

*From Markov's Inequality:


\begin{equation}
\mathbb{P}[X \geq t] \leq \frac{\mathbb{E}[X]}{t}
\end{equation}
And we also know this for a fact that,
\begin{equation}
\mathbb{P}[X^2 \geq t^2] = \mathbb{P}[X \geq t]
\end{equation}
implying
\begin{equation}
\mathbb{P}[X \geq t] \leq \frac{\mathbb{E}[X^2]}{t^2}\,\, \text(Chebyshev\,Inequality)
\end{equation}
(From applying Markov's Inequality on $\mathbb{P}[X^2 \geq t^2]$)
Now Chebyshev gives a better (tighter) bound than Markov iff $\frac{\mathbb{E}[X^2]}{t^2} \leq \frac{\mathbb{E}[X]}{t}$ which in turn implies that $t \geq \frac{\mathbb{E}[X^2]}{\mathbb{E}[X]}$.
Thus, Markov bound is tighter (better) for the case $t \leq \frac{\mathbb{E}[X^2]}{\mathbb{E}[X]}$ (small values of $t$) otherwise Chebyshev bound fares better for larger values of $t$.
A: I will add the following counterexample to show that Chebyshev is not always better than Markov.
Consider the random uniform variable:
$X \sim U({0,1,2,3,...,10})$. $\ $ Then: $E[X]=5$, $V[X]=\frac{11^2-1}{12}=10$.

*

*Markov $P(X\ge 7)\le \frac{E[X]}{7} \approx 0.71 $

*Chebyshev $P(X \ge 7)=P(|X-5|\ge 2) \frac{10}{4}=2.5>1  \ !!  $
$ $

*

*So, in this case, the Chebyshev bound is totally unuseful.

*Nevertheless, Markov gives a relatively good bound (true probability is 4/11)

