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From here:

Even though Markov’s and Chebyshev’s Inequality only use information about the expectation and thevariance of the random variable under consideration, they are essentially tight for a general random variable.

So for a non-negative random variable $X$, if $P(X\geq a)$ is bound tightly by Markov's equation, we have: $$\mathbb{E}(X) = \int_0^a xf(x) dx + \int_a^\infty xf(x)dx = \int_a^\infty af(x)dx $$ $$0\leq\int_a^\infty (x-a) f(x)dx=-\int_0^a xf(x) dx \leq 0$$ $$\int_a^\infty (x-a) f(x)dx = \int_0^a xf(x) dx = 0$$ This doesn't seem to hold in general at all.

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Let $a>0$ be fixed. Note that $X-a1_{X\geq a}\geq 0$. In the equality case of Markov's inequality, this non-negative r.v has expectation $0$, thus $X-a1_{X\geq a} = 0$ a.s, that is $X=a1_{X\geq a}$ a.s. Hence almost surely $X\in \{0,a\}$.

Consider $X$ a discrete r.v. such that $P(X=a)=\lambda$ and $P(X=0)=1-\lambda$. It is non-constant and you can check that $P(X\geq a)=\frac{E(X)}a$.

Equality in Chebyshev’s Inequality implies equality in Markov's inequality for the random variable $|X-E(X)|$. So almost surely $|X-E(X)|\in \{0,a\}$.

Consider $X$ a discrete r.v. such that $P(X=a)=P(X=-a)=\frac 12$. It is non-constant and you can check that $P(|X-E(X)|\geq a)=\frac{V(X)}{a^2}$.

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  • $\begingroup$ referencing as helpful: math.stackexchange.com/questions/708779/… $\endgroup$
    – charlieh_7
    Commented Sep 7, 2019 at 10:42
  • $\begingroup$ Doesn't $E(X)=0$ imply a.s. $X=0$, If $X$ is a non-negative r.v. here? $\endgroup$
    – charlieh_7
    Commented Sep 7, 2019 at 10:51
  • $\begingroup$ @charlieh_7 sure, but there's no reason for $E(X)=0$ to hold. $\endgroup$ Commented Sep 7, 2019 at 11:07
  • $\begingroup$ @GabrielRomon It is a bit confusing. Should the Markov inequality not hold for any a. Why assume it is fixed. It should also hold for 2a, or 3a. i.e. P(X>2a) = E(X)/2a etc. $\endgroup$
    – user715918
    Commented Oct 17, 2019 at 17:08
  • $\begingroup$ @rbs I'm assessing what happens in the equality case, i.e. when for some $a>0$, $P(X\geq a)=\frac{E(X)}a$. $\endgroup$ Commented Oct 17, 2019 at 19:33

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