Being a novice to this topic, I am not being able to properly write down a step by step solution to this problem.

For each integer $n$, let $X_n$ be a non negative random variable with finite mean $\mu _n$ . Prove that if $\underset{n \rightarrow > \infty}{\lim}{\mu _n}=0 $ then $X_n \mbox{ converges in probability to }0$.

Update: Please don't down vote if my solution is wrong. Of course you are welcome to edit and correct it...

This is a solution I tried following @whuber’s instructions. I don’t know whether I have gone wrong somewhere or not.

Assume $$ X_n \space doesnot\space converge\space in \space probability \space to\space 0$$ $$\therefore\space there\space exists\space \epsilon \gt 0\space such \space that$$ $$\lim_{n\to\infty} P\left[|X_n-\mu_n|\ge\epsilon\right]\rightarrow 1$$ $$\implies \lim_{n\to\infty} P\left[X_n \ge\mu_n +\epsilon\right]\rightarrow 1$$ $$\implies \lim_{n \to\infty} \frac{\left(E\left[X_n\right]=\mu_n\right)}{\mu_n + \epsilon}\rightarrow 1$$ $ \implies\lim_{n\to\infty} \mu_n\to\mu_n + \epsilon$ which is a contradiction as $\epsilon$ is a number greater than 0. Hence our initial assumption was wrong. Hence proved.

  • 7
    $\begingroup$ For intuition, you might contemplate proving the contrapositive. Suppose $X_n$ did not converge in probability to $0$. Given that these variables are nonnegative, what would that imply about the limit of the means? Intuition is converted into a proof by applying the definitions... . $\endgroup$
    – whuber
    Dec 30, 2015 at 19:31

1 Answer 1


You already have a hint from @whuber on how to solve the problem and I encourage you to work through it.

I am going to offer another approach that is based on an application of a famous inequality relating probabilities and expectations. First, recall the definition of convergence in probability.

We say that a sequence of random variables $X_n$ converges in probability to a random variable $X$ if for every $\epsilon>0$, it holds that $$\lim_{n\to \infty} P\left[|X_n - X| < \epsilon \right] \to 1 $$

We will work with the complement of the above probability. Pick an $\epsilon>0$ and let's try to compute

$$\lim_{n \to \infty} P\left[|X_n - 0| \geq \epsilon \right] = \lim_{n \to \infty} P\left[X_n \geq \epsilon \right]$$

due to the nonnegativity. Applying Markov's inequality, we get

$$\lim_{n \to \infty} P\left[X_n \geq \epsilon \right] \leq \lim_{n \to \infty} \frac{ \mathbb{E}\left[X_n \right]}{\epsilon} \to 0$$

by assumption. Since the probability of the complement goes to zero and $\epsilon$ was arbitrary, the probability of the event $\left\{ X_n < \epsilon \right\}$ goes to one, which is what we wanted to show. Thus, your random variable converges in probability to zero. $\square$


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