I know that the following implications are true:

$$\text{Almost sure convergence} \Rightarrow \text{ Convergence in probability } \Leftarrow \text{ Convergence in }L^p $$


$$\text{Convergence in distribution}$$

I am looking for some (preferably easy) counterexamples for the converses of these implications.


1 Answer 1

  1. Convergence in probability does not imply convergence almost surely: Consider the sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ (endowed with Lebesgue measure $\lambda$) defined by $$\begin{align*} X_1(\omega) &:= 1_{\big(\frac{1}{2},1 \big]}(\omega) \\ X_2(\omega) &:= 1_{\big(0, \frac{1}{2}\big]}(\omega) \\ X_3(\omega) &:= 1_{\big(\frac{3}{4},1 \big]}(\omega) \\ X_4(\omega) &:= 1_{\big(\frac{1}{2},\frac{3}{4} \big]}(\omega)\\ &\vdots \end{align*}$$ Then $X_n$ does not convergence almost surely (since for any $\omega \in (0,1]$ and $N \in \mathbb{N}$ there exist $m,n \geq N$ such that $X_n(\omega)=1$ and $X_m(\omega)=0$). On the other hand, since $$\mathbb{P}(|X_n|>0) \to 0 \qquad \text{as} \, \, n \to \infty,$$ it follows easily that $X_n$ converges in probability to $0$.
  2. Convergence in distribution does not imply convergence in probability: Take any two random variables $X$ and $Y$ such that $X \neq Y$ almost surely but $X=Y$ in distribution. Then the sequence $$X_n := X, \qquad n \in \mathbb{N}$$ converges in distribution to $Y$. On the other hand, we have $$\mathbb{P}(|X_n-Y|>\epsilon) = \mathbb{P}(|X-Y|>\epsilon) >0$$ for $\epsilon>0$ sufficiently small, i.e. $X_n$ does not converge in probability to $Y$.
  3. Convergence in probability does not imply convergence in $L^p$ I: Consider the probability space $((0,1],\mathcal{B}((0,1]),\lambda|_{(0,1]})$ and define $$X_n(\omega) := \frac{1}{\omega} 1_{\big(0, \frac{1}{n}\big]}(\omega).$$ It is not difficult to see that $X_n \to 0$ almost surely; hence in particular $X_n \to 0$ in probability. As $X_n \notin L^1$, convergence in $L^1$ does not hold. Note that $L^1$-convergence fails because the random variables are not integrable.
  4. Convergence in probability does not imply convergence in $L^p$ II: Consider the probability space $((0,1],\mathcal{B}((0,1]),\lambda|_{(0,1]})$ and define $$X_n(\omega) := n 1_{\big(0, \frac{1}{n}\big]}(\omega).$$ Then $$\mathbb{P}(|X_n|>\epsilon) = \frac{1}{n} \to 0 \qquad \text{as} \, \, n \to \infty$$ for any $\epsilon \in (0,1)$. This shows that $X_n \to 0$ in probability. Since $$\mathbb{E}X_n = n \cdot \frac{1}{n} = 1$$ the sequence does not converge to $0$ in $L^1$. Note that $L^1$-convergence fails although the random variables are integrable. (Just as a side remark: This example shows that convergence in probability does also not imply convergence in $L^p_{\text{loc}}$.)
  • 1
    $\begingroup$ The example in 2 is a bit weird as convergence in distribution is not directly related to random variables. Perhaps another example is $X_n=(-1)^n X$ where $X$ is a symmetric non trivial random variable. $\endgroup$
    – Kolmo
    Jun 21, 2015 at 9:40
  • $\begingroup$ @Kolmo Yeah, you are right, that's also a nice counterexample. :) $\endgroup$
    – saz
    Jun 21, 2015 at 9:42
  • $\begingroup$ @Kolmo, nothing weird. But it is not the best example, as the sequence does converge in probability (although to a different r.v.). You example is better in this respect: it shows that a convergent in distribution sequence may be divergent in probability. $\endgroup$
    – zhoraster
    Sep 13, 2015 at 6:38
  • $\begingroup$ I have question in Example 4: Does the sequence {X_n} converge almost surely to zero in this example? I guess not, but how can I verify that? $\endgroup$
    – user67724
    Oct 21, 2018 at 17:09
  • $\begingroup$ Let me add to my question regarding Example 4: Suppose I define X_n = \sqrt{n} I{(0, 1/n]}. Does X_n converge almost surely to zero? $\endgroup$
    – user67724
    Oct 21, 2018 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.