# Markov's inequality tight in general?

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.

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 $$P(|X-E(X)|\geq a)=\frac{V(X)}{a^2}$$.

• referencing as helpful: math.stackexchange.com/questions/708779/… Commented Sep 7, 2019 at 10:42
• Doesn't $E(X)=0$ imply a.s. $X=0$, If $X$ is a non-negative r.v. here? Commented Sep 7, 2019 at 10:51
• @charlieh_7 sure, but there's no reason for $E(X)=0$ to hold. Commented Sep 7, 2019 at 11:07
• @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. Commented Oct 17, 2019 at 17:08
• @rbs I'm assessing what happens in the equality case, i.e. when for some $a>0$, $P(X\geq a)=\frac{E(X)}a$. Commented Oct 17, 2019 at 19:33