Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

a busy cat

How to do this problem, I am really confused. Also, what is the definition of Markov's inequality?

share|cite|improve this question

Markov inequality is the integrated version of the almost sure inequality $a\mathbf 1_{X\geqslant a}\leqslant X$ hence the equality case happens if and only if $a\mathbf 1_{X\geqslant a}=X$ almost surely, that is, $X\in\{0,a\}$ almost surely.

share|cite|improve this answer

Let $X$ be a non-negative random variable with a finite expectation $E(X)$. Markov's Inequality states that in that case, for any positive real number $a$, we have $$\Pr(X\ge a)\le \frac{E(X)}{a}.$$

In order to understand what that means, take an exponentially distributed random variable with density function $\frac{1}{10}e^{-x/10}$ for $x\ge 0$, and density $0$ elsewhere. Then the mean of $X$ is $10$.

Take $a=100$. Markov's Inequality says that $$\Pr(X\ge 100)\le \frac{E(X)}{100}=\frac{10}{100}.$$

It is straightforward to show that for this exponential, we have $\Pr(X\ge 100)=e^{-100/10}$. This is a number that is very close to $0$.

Note that the Markov Inequality did not lie. The probability that $X\ge 100$ really is $\le \frac{1}{10}$. But $e^{-10}$ is very much less than $\frac{1}{10}$. So in this case the Markov Inequality produced a correct but lousy bound for the "tail probability" of the exponential.

Could we produce a general but better bound. Yhe purpose of the exercise is to show that in a sense we cannot.

That is the price we pay for having a theorem that applies to all random variables that have a finite expectation. The bound produced by the Markov Inequality is often not at all sharp.

The question asks us, given a positive integer $a$, to produce a random variable $X=X_a$ such that $\Pr(X\ge a)=\frac{E(X)}{a}$.

We can produce a very boring random variable that will do the job. Let $X=a$ with probability $1$. Then $\Pr(X\ge a)=1$. You should show that in this case, $\frac{E(X)}{a}=1$. That will show that it is perfectly possible to have $\Pr(X\ge a)$ exactly equal to $\frac{E(X)}{a}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.