# Example of convergence in distribution but not in probability

While I was looking for an example of a sequence of random variables which converges in distribution, but doesn't converge in probability, I have read that it should be enough to consider a sequence of independent and identically distributed non-degenerate random variables. I don't understand why... Can someone explain (or correct if it isn't right)? Thank you

• They all have the same distribution (identically distributed) so there is convergence in distribution. But if they are not degenerate then there is no convergence in probability (independent). Jul 3, 2015 at 13:59
• @drhab Sorry for answering so late, let's take X and Y indep. and identically distributed and $X_n = X \forall n \in \mathbb{N}$. It's clear that there is convergence in distribution. If there is convergence in probability, then it must be $P(X=Y)=1$. I want to show that in this case there is $c \in \mathbb{R}$ such that $P(X=c)=P(Y=c)=1$. I think I'm able to do this if X and Y are discrete (suppose $P(X=c)<1 \forall c \in \mathbb{R}$, then $P(X=c)=(P(X=c))^2=0 \forall c$, absurd, where the last equality follows from indep.). Could you help me for the general case? Jul 16, 2015 at 9:53
• If $X,Y$ are independent with $P\left(X=Y\right)=1$ and $F$ denotes their common CDF then$F\left(x\right)=P\left(X\leq x\right)=P\left(X\leq x\wedge Y\leq x\right)=P\left(X\leq x\right)P\left(Y\leq x\right)=F\left(x\right)^{2}$. This implies that $F\left(x\right)\in\left\{ 0,1\right\}$ for each $x\in\mathbb{R}$. Then the characteristics of $F$ as a CDF (right-continuous, non-decreasing, non-constant and taking values in $[0,1]$) tell us that $F=1_{[c,\infty)}$ for some constant $c$. This implies that $P\left(X=c\right)=1$. Jul 16, 2015 at 14:36

Choose the probability space $([0,1],\mathscr{B},m)$. ($\mathscr{B}$ consists of all Borel sets of $[0,1]$, $m$ is the Lebesgue measure.)

Let $X_{2n}(\omega)=\omega$, $X_{2n-1}(\omega)=1-\omega$.

Show that $X_{2n}$ and $X_{2n-1}$ have the same distribution. (Uniform distribution.)

Show that $\{X_n\}$ does not converge in probability.

• the most simple exam ever seen Mar 3, 2022 at 19:06
• Wow, I'm THAT stupid. Aug 14, 2022 at 5:43
• I don't really understand this. Can you unpack further? Jan 18, 2023 at 20:15
• @Learningstatsbyexample I've posted the proof in my answer. math.stackexchange.com/a/4652815/481713 Mar 5, 2023 at 22:11
• @paperskilltrees I've posted the proof in my answer. math.stackexchange.com/a/4652815/481713 Mar 5, 2023 at 22:11

Here is an example of a sequence of random variables $$\{X_n\}_{n\geq1}$$ defined on $$\Omega$$ which converges in distribution to $$X:\Omega\to\mathbb{R}$$ but does not converge in probability to $$X$$.

Consider the abstract space $$\Omega=\{0,1\}$$ and

$$\begin{cases} X_n(0)=0\\ X_n(1)=1\\ X(1)=0\\ X(0)=1 \end{cases}$$ all with probability $$\frac{1}{2}$$.

Then we have $$X_n\xrightarrow[]{\text{dist}}X$$ because $$F_n=F$$, where $$F_n$$ and $$F$$ are the distribution functions of $$X_n$$ and $$X$$ respectively.

However, $$X_n\xrightarrow[]{\text{prob}}X$$ does not hold since we have $$|X_n(\omega)-X(\omega)|=1$$, for $$\omega\in\Omega$$. Thus $$P(\omega \in \Omega: |X_n(\omega)-X(\omega)|>\epsilon)\not\to 0.$$

• So just pick 2 different random variables with the same distribution. Amazing I struggled so hard when the answer is so simple, I keep getting this idea that any type of convergence must mean the $X_n$ must look more and more similar to $X$. My iq must be sub 50. Aug 14, 2022 at 5:59

Let $$X_0 = \text{Uniform}[0,1]$$,

$$X_{2n} = X_0$$, $$X_{2n+1} = 1 - X_0$$

$$X_n$$ has the uniform distribution on $$[0,1]$$, so they converge in distribution.

Suppose $$X_n$$ converges in probability to some random variable $$X$$, then \begin{align} P(|X_n - X| > \epsilon) + P(|X_{n+1} - X| > \epsilon) &\ge P(|X_n - X| > \epsilon \text{ or } |X_{n+1} - X| > \epsilon)\\ &= P(\max(|X_n - X|, |X_{n+1} - X|) > \epsilon)\\ &= P(\max(|X_n - X|, |1 - X_{n} - X|) > \epsilon)\\ &\ge P(\max(|X_n - X|, |1 - X_{n} - X|) > \epsilon \text{ and } |X-\frac{1}{2}| > \frac{1}{4}) \end{align}

Given that $$|X - \frac{1}{2}| > \frac{1}{4}$$, it's impossible for both $$X_n$$ and $$1 - X_n$$ to be within $$\epsilon$$ of $$X$$. Therefore $$P(|X_n - X| > \epsilon) + P(|X_{n+1} - X| > \epsilon)$$ cannot possibly go to zero.

• This is an addendum to Eclipse Sun's answer showing that $(X_n)$ does not converge in probability (Eclipse Sun skipped the details). Mar 5, 2023 at 22:26