4.6 Deviation: Markov & Chebyshev Bounds

Suppose \(X\) is a nonnegative integer random variable for which \(E[X]\)=5. You know nothing else about \(X\).

Complete the following statements so that they are guaranteed to be true.

1. The probability that \(X=5\) is
   Exercise 1

     positive 

     less than 1 

     at least \(\frac{1}{2}\) 

     at most \(\frac{1}{2}\) 

     could be 0 

     1 
   Two cases disprove all the others: \(X\) is always 5 or \(X\) is 0 or 10 with equal likelihood.


2. With positive probability, \(X\) is
   Exercise 2

     equal to 0 

     at most \(\frac{5}{2}\) 

     less than 5 

     at most 5 

     equal to 5 

     greater than 5 

     at least 10  
   If \(X\) were always >5, then we would have \(E[X] >\)5. So, it must be at most 5 with positive probability.

   The rest of the options (except equal to 5) are disproved by the following counterexample: Let \(X\) be the constant random variable equal to 5. Then \(E[X]\)=5, and the other statements are obviously false. Equal to 5 is disproved by the random variable that takes values 0 or 10, each with probability \(\frac{1}{2}\).



3. \(E[X^2]\)
   Exercise 3

     is 25 

     is not 25 

     is at most 25 

     is at most \(E[X]\) 

     could be 0 

     could be 100 
   \(X\) could be 20 with probability \(\frac{1}{4}\) and 0 with probability \(\frac{3}{4}\). Then its expectation would be 5, but the expectation of its square would be \(100=\frac{3}{4}\cdot 0^2 + \frac{1}{4}\cdot 20^2\). In fact, by setting \(X\) to be very large with low probability, we can make its expectation arbitrarily large. We can rule out all the other answers:

   • \(X\) could be 0 with probability \(\frac{1}{2}\) and 10 with probability \(\frac{1}{2}\). Then the expectation of \(X^2\) would be 50. So, the 1st, the third, and the fourth answers are wrong.
   • \(X\) could be 5 with probability 1. Then the expectation of \(X^2\) would be 25. So the second answer is wrong.
   • \(E[X^2]\) is at least \(E[X^2]\), since \(Var[X]=E[X^2]-E[X]^2\) and Variance is nonnegative, so we know it is not 0.


4. The probability \(\Pr[X \geq 1000]\) is
   Exercise 4

     zero 

     positive 

     at most \(\frac{1}{1000}\) 

     at most \(\frac{1}{500}\) 

     at most \(\frac{1}{200}\) 

     at most \(\Pr[X \geq 5\cdot1000]\) 
   A direct application of Markov's Theorem. Since \(X\) is a nonnegative variable, we know that \[\Pr [X \geq a] \leq \frac{E[X]}{a}\] for any \(a\). In this case, we have \[\Pr [X \geq 1000] \leq \frac{5}{1000} = \frac{1}{200}\] On the other hand, \(X\) could be 1000 with probability \(\frac{1}{200}\) and 0 otherwise, in which case the smaller bounds would not hold. Of course if \(X\) is always 5, this probability is 0, so it's not necessarily positive.
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