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Suppose that $X_1,X_2...$ is a sequence of non-negative real random variables. I have that $\mathbb{E}(X_i^2) \to 0$ as $i \to +\infty$, therefore my sequence converges at least in distribution to the random vairbale that is identically zero.

Does it converge in proba ? almost surely ?

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    $\begingroup$ • convergence in distribution to a constant implies convergence in probability, it does not imply convergence almost surely, see en.wikipedia.org/wiki/… $\endgroup$ Commented Jan 29, 2021 at 14:47
  • $\begingroup$ It is true that convergence in distribution to a constant implies convergence in probability, but IMO it is easier in this case to use that $L^2$ convergence implies convergence in probability. Carlo Beenakker's link is rather exhaustive, and as described by Will Sawin, non-negativity does not help. $\endgroup$
    – Pierre PC
    Commented Jan 29, 2021 at 15:47
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    $\begingroup$ Thanks a lot for the link, i always forgot these. This is my new favorite bookmark ;) $\endgroup$
    – lrnv
    Commented Jan 29, 2021 at 15:49

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Yes in probability: The definitions of convergence in distribution to a constant random variable and convergence in probability to a constant random variable are the same.

No almost surely: If we let $\alpha$ be randomly distributed between $[0,1]$ and we let $X_n$ be $1$ if $$\alpha + \sum_{i=1}^{n-1} (1/i) \mod 1 > 1 -1/n $$ and $0$ otherwise then $X_n$ has a probability $1/n$ of being $1$ and a probability $1-1/n$ of being $0$, but with probability $1$, $X_n$ is $1$ infinitely often (at the points where $\alpha + \sum_{i=1}^n (1/i)$ passes an integer).

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  • $\begingroup$ Thanks for the counter-example. $\endgroup$
    – lrnv
    Commented Jan 29, 2021 at 15:48

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