# Sequence $\{X_n\}$ of r.v. converging in probability but $\mathbb{P}(\sup_n |X_n|=+\infty) >0$

Does there exist a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and a sequence of real random variables $\{X_n\}_{n\in \mathbb{N}}$ converging in probability, but such that $\mathbb{P}(\{\omega\in \Omega \ |\ \sup_n |X_n| = + \infty \}) > 0$ ?

I managed to prove that if $X_n \to X$ almost surely then the set $\{\omega\in \Omega \ |\ \sup_n |X_n| = + \infty \}$ is negligible, but I found difficulties in finding a counterexample when $X_n \rightarrow X$ in probability.

• Try $(X_n)$ independent with $P(X_n=n)=1/n=1-P(X_n=0)$. – Did Aug 28 '16 at 10:20
• In this case $X_n \to 0$ in probability (and $X_n$ doesn't converge to 0 a.s.) but how can I see that $\{\sup_n |X_n| = + \infty\}$ is not negligible? – Warlock of Firetop Mountain Aug 28 '16 at 10:34
• Borel-Cantelli gives you $P(\sup X_n=+\infty)=1$ (which is exactly the same reason why $X_n$ does not converge to zero...). – Did Aug 28 '16 at 10:35
• Thank you now I've understood. $\forall M>0$ $\sum_n \mathbb{P}\{X_n > M\} = +\infty$ and then for Borel-Cantelli $\mathbb{P}\{\lim\sup X_n > M\} = 1$. – Warlock of Firetop Mountain Aug 28 '16 at 10:43

## 1 Answer

In fact, a counter-example exists in any probability space containing a sequence of non-negative random variables $\left(X_n\right)_{n\geqslant 1}$ such that $\mathbb E\left[X_n\right]\to 0$ but the almost sure convergence of $\left(X_n\right)_{n\geqslant 1}$ to $0$ does not hold. We can also assume that $\mathbb E\left[X_n\right]\gt 0$ for all $n$.

If such a sequence exists , define $$R_n:=\left(\mathbb E\left[X_n\right] \right)^{-1/2}, \quad Y_n:= R_nX_n.$$ Then the sequence $\left(Y_n\right)_{n\geqslant 1}$ converges to $0$ in $\mathbb L^1$ hence in probability and if $\omega$ is such that $\left(X_n\left(\omega\right)\right)_{n\geqslant 1}$ does not converge to $0$, then $\sup_{n\geqslant 1}Y_n\left(\omega\right)=+\infty$. The set of $\omega$'s such that $\left(X_n\left(\omega\right)\right)_{n\geqslant 1}$ does not converge to $0$ has a positive probability hence so have $\left\{\omega\in \Omega \ |\ \sup_n |Y_n| = + \infty \right\}$.