I'm having trouble understanding the answer to one of the mit opencourseware Statistic For Applications homeworks

The question: (question 1.): [0] The answer: [1]

The question asks to show this random variable converges in probability:

$P[X_n=1/n] = 1-1/n^2$

$P[X_n=n] = 1/n^2$

The answer first calculates the expected value:


Then it says "On the other hand:

$\lim_{n \to inf} P[|X_n|>\epsilon] = \lim_{n \to inf} P[X_n>\epsilon] <= (\lim_{n \to inf}E[X_n])/\epsilon = \lim_{n \to inf}(2/n-1/n^3)/\epsilon = 0$

...hence Hence $X_n$ converges in probability"

The part I don't understand is this:

$\lim_{n \to inf} P[X_n>\epsilon] <= (\lim_{n \to inf}E[X_n])/\epsilon$

Where did that come from? I've been looking online to see if I can find any examples that does something similar but I have yet to see any. As far as I can tell the expected value shouldn't have this relationship to the probability of the value of $X_n$. Is this a common tick to use to solve these problems? Can someone give me some intuition as to how this "<=" makes sense?

[0] https://github.com/hoangnguyen7699/StatisticsForApplication_solution/blob/master/MIT18_650F16_PSet1.pdf

[1] https://github.com/hoangnguyen7699/StatisticsForApplication_solution/blob/master/PS1/ProblemSet1.pdf

  • 1
    $\begingroup$ Markov's inequality. $\endgroup$ Commented Apr 14 at 16:51
  • 1
    $\begingroup$ Oh. look at that. Thank you! $\endgroup$
    – tom tom
    Commented Apr 15 at 11:18
  • 1
    $\begingroup$ sorry for the "concise" response, I was on my phone. $\endgroup$ Commented Apr 15 at 11:21
  • $\begingroup$ No worries, I really appreciate the help. It completely unstuck me $\endgroup$
    – tom tom
    Commented Apr 15 at 17:30

1 Answer 1


It's Markov's inequality. thanks Matthew Towers.


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