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I know that the following implications are true:

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

$$\Downarrow$$

$$\text{Convergence in distribution}$$

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

BCLC
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Mathlearner
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1 Answers1

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  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 $L^p_{\text{loc}}$.)
saz
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    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. – Kolmo Jun 21 '15 at 09:40
  • @Kolmo Yeah, you are right, that's also a nice counterexample. :) – saz Jun 21 '15 at 09:42
  • @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. – zhoraster Sep 13 '15 at 06:38
  • 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? – user67724 Oct 21 '18 at 17:09
  • 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? – user67724 Oct 21 '18 at 17:16
  • @user67724 Yes, they do converge to zero almost surely (with respect to Lebesgue measure). More generally, if $X_n = a_n 1_{(0,1/n]}$ for some sequence $(a_n){n \in \mathbb{N}} \subseteq \mathbb{R}$, then $X_n \to 0$ almost surely. The proof goes as follows: For any $\omega \in (0,1)$ we have $\omega > \frac{1}{n}$ for $n \geq N(\omega)$ sufficiently large. Hence, $1{(0,1/n]}(\omega)=0$ for any such $n$, and so $X_n(\omega)=0$ for all $n \geq N$. This proves $X_n(\omega) \to 0$ as $n \to \infty$. – saz Oct 21 '18 at 18:53
  • @saz. Thank you. Your proof also gives an example of when a.s. convergence does not imply L1 convergence. – user67724 Oct 23 '18 at 00:00
  • @saz: Suppose {a_n} goes to infinity at a rate n^k, k>1, then does X_n = a_n I{(0,1/n]} still converge almost surely to 0 (w.r.t. Lebesgue measure)? My doubt is that even though X_n(\omega) goes to zero, it is multiplied by a term that goes to infinity at a faster rate. Can you please clarify? Thanks. – user67724 Oct 23 '18 at 12:42
  • @user67724 The point is simply that $1_{(0,1/n)}(\omega)$ **equals $0$ for $n$ large enough. You can multiply zero with whatever constant you like, and itstill equals zero. Hence, $X_n(\omega)=0$ for $n \geq N(\omega)$ large enough. – saz Oct 23 '18 at 12:43