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

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    $\begingroup$ 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). $\endgroup$
    – drhab
    Jul 3, 2015 at 13:59
  • $\begingroup$ @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? $\endgroup$ Jul 16, 2015 at 9:53
  • $\begingroup$ 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 $P\left(X=c\right)=1$. $\endgroup$
    – drhab
    Jul 16, 2015 at 14:36

3 Answers 3


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.


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 $$P(\omega \in \Omega: |X_n(\omega)-X(\omega)|>\epsilon)\not\to 0.$$

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    $\begingroup$ 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. $\endgroup$ 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.

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

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