In Wikipedia and many textbooks, they explain Markov's inequality for $X \geq a$ for a non-negative random variable $X$ and any $a > 0$ as follows: $$ \text{Pr}(X \geq a) \leq \frac{\mathbb{E}(X)}{a}$$ where $\mathbb{E}(X)$ is the expectation of $X$. Then, how to get the bound of $\text{Pr}(X > a)$ (i.e., without the equality in $\text{Pr}(X \geq a)$)? If the bound exists, is it meaningful to obtain the bound?

My guesses are:

  • under a p.d.f., $\text{Pr}(X>a)$ has the same bound as $\text{Pr}(X \geq a)$ according to Proof 1.
  • under a p.m.f., for a non-negative integer random variable X, the bound of $\text{Pr}(X>a)$ is represented as follows:
    • B-PMF-1) $\text{Pr}(X \geq a+1)=\text{Pr}(X>a) \leq \frac{\mathbb{E}(X)}{a+1}$, or
    • B-PMF-2) $\text{Pr}(X \geq a)= \text{Pr}(X > a) + \text{Pr}(X=a) \leq \frac{\mathbb{E}(X)}{a} \Leftrightarrow \text{Pr}(X > a) \leq \frac{\mathbb{E}(X)}{a} - \text{Pr}(X=a)$
    • B-PMF-2 will has the same bound as B-PMF-1 or tighter one than B-PMF-1.

Are those claims true?

Proof 1) Suppose that we have very small $\delta \geq 0$. Then, $\mathbb{E}(X)$ is $$ \mathbb{E}(X) = \int_{0}^{\infty}xp(x)dx = \int_{0}^{a-\delta}xp(x)dx + \int_{a-\delta}^{\infty}xp(x)dx \geq \int_{a-\delta}^{\infty}xp(x)dx \geq (a - \delta)\int_{a-\delta}^{\infty}p(x)dx = (a-\delta)\text{Pr}(X \geq a - \delta)$$ Note that $\text{Pr}(X \geq a - \delta) \geq \text{Pr}(X > a)$. Hence, $$ \mathbb{E}(X) \geq (a-\delta)\text{Pr}(X > a) \Leftrightarrow \text{Pr}(X > a) \leq \frac{\mathbb{E}(X)}{a - \delta}$$ As $\delta \rightarrow 0$, the bound converges to $\frac{\mathbb{E}(X)}{a }$.

Another way is that $\text{Pr}(X > a)=\text{Pr}(X \geq a) - \text{Pr}(X = a) \leq \frac{\mathbb{E}(X)}{a} - \text{Pr}(X = a)$, and $\text{Pr}(X = a) = 0$ when $p(\cdot)$ is a p.d.f. Thus, $\text{Pr}(X > a) \leq \frac{\mathbb{E}(X)}{a}$.

  • 2
    $\begingroup$ Huh? But wait, $$\{X>a\}\subseteq\{X\geqslant a\}$$ hence $$P(X>a)\leqslant P(X\geqslant a)$$ right? And since Markov inequality says that $$P(X\geqslant a)\leqslant E(X)/a$$ you are done, no? $\endgroup$ – Did Nov 12 '17 at 16:16
  • $\begingroup$ Thanks for your comment. You're right, and your answer solved my curiosity clearly. $\endgroup$ – 정진홍 Nov 14 '17 at 2:42

Do you remember the one-line proof of Markov's inequality ? It readily adapts:

$$a\mathbb 1_{X>a}\leq X$$ still holds true. Taking expectations on both sides, $aP(X>a)\leq E(X)$, hence $$P(X>a)\leq \frac{E(X)}a$$

  • 1
    $\begingroup$ Of course, this also follows directly from the original inequality, and the trivial fact that $P(X > a) \le P(X \ge a)$. $\endgroup$ – Ilmari Karonen Nov 12 '17 at 16:33

The bounds are correct, all comes from the equation you wrote
$P(X\geq a)-P(X=a)\leq \frac{E(X)}{a}-P(X=a)$

And clearly in the continuous case $P(X=a)=0$


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