# MIT Statistic For Applications course Question 1

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:

$$E[x_n]=2/n-1/n^3$$

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?

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