I am a bit loss with the convergence in probability and the absolute value.

Let $X_n$ be a random variable defined in $\mathbb{R}$ with $\lim_{n \rightarrow \infty} E[X_n] = a$ and $V[X_n] = O(n^{-1})$. Then, we know that by applying Chebyshev's inequality

$$P(|X_n - a| > e) < O(n^{-1})$$

which means that $X_n$ converges in probability to $a$.

Now imagine that I can show that $E|X_n - a| = o(1)$, using Markov's inequality we can I conclude that $X_n$ converges in probability to $a$.

But what about if I am only able to show that

$$|E[X_n] - a| = o(1)? $$


1 Answer 1


$|E[X_n] - a| = o(1)$ tells you very little about the behaviour of the distribution of $X_n$ apart from its mean.

For example

  • if independent $Y_j \sim \mathcal N\left(\frac{1}{j^2}, j^2\right)$ and $X_n=\sum\limits_{j=1}^n Y_j$
  • then you have $\left|E[X_n] - \frac{\pi^2}6\right| \le \frac1n = o(1)$, meaning $E[X_n]$ converges to $\frac{\pi^2}6$
  • but $X_n$ does not converge to a constant, as it has a normal distribution with increasing variance of $\frac16n(n+1)(2n+1)$
  • $\begingroup$ Doesn't the question rule out this example by specifying the variance of $X_n$ vanishes at a rate of $1/n$? Indeed, what distinction are you and the OP drawing between $E[X_n]$ limiting to $a$ and having "$o(1)$" behavior relative to $a$?? $\endgroup$
    – whuber
    Mar 31, 2023 at 14:03
  • 1
    $\begingroup$ @whuber I read the question with "Now imagine ..." resetting all the assumptions (it does not need a converging variance) and the question I was responding to starting "But what about ..." resetting them again. If you keep the assumption of variance reducing to zero, then this in effect takes you back to the start as I see no substantial difference between $E[X_n]$ limiting to $a$ and having $o(1)$ behaviour relative to $a$ $\endgroup$
    – Henry
    Mar 31, 2023 at 15:12
  • $\begingroup$ Thank you for the explanations -- that helps connect your answer to the questions. (+1) $\endgroup$
    – whuber
    Mar 31, 2023 at 19:03

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