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Let $(X_n)_{n\ge 1}$ be a sequence of dependent non negative random variables , where $X_n$ has density w.r.t. Lebesgue on $[0,n]$ and $\mathbb{E}(X_n) < \infty$.

We know that $X_n$ converges weakly to $X$ which has density w.r.t. Lebesgue on $[0,\infty]$ and $\mathbb{E}(X) < \infty$.

Question: $X_n/\log(n)$ converges almost surely to $0$?

My attempt: Applying Slutzsky's theorem I can say that $X_n/\log(n)$ converges in probability to $0$, but from it I cannot say anything a.s.

Furthermore, I know that convergence of marginal distributions does not say anything about a.s. convergence.

However, here $1/log(n)$ is a deterministic sequence going to $0$ and so the intuition is that in order to have a negative answer we need $X_n(\omega)$ diverges on a subset of $\Omega$ (the space where $X_n$ are defined) of positive probability, which seems against the hypotesis of $X_n$ converges weakly to $X$.

Thanks for the help!

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Unfortunately we cannot conclude $X_n/\log(n) \to 0$ a.s.

Take $$Y_n = 1_{\left[k2^{-m},(k+1)2^{-m}\right]} \quad\quad\text{for } n = 2^m + k$$ which is a well known example for weak but no almost sure convergence.

Now define $$X_n := \log(n)Y_n$$ then $X_n$ fulfills your properties with $X \equiv 0$ but $$\frac{X_n}{\log_n} = Y_n$$ by construction so there is no a.s. convergence.

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  • $\begingroup$ Thank you very much for the help! Now I'm (almost ;) ) sure that we cannot conclude $X_n/\log(n) \rightarrow 0$ a.s. I think that the idea is that $\limsup X_n$ can go to $\infty$ even if $X_n$ converges weakly to $X$. In your example $X_n$ does not fulfills the assumptions since $X$ has not density w.r.t. Lebesgue, but it was still very usefull to me to understand the problem. Thanks! $\endgroup$ – Gio Mar 6 '20 at 13:56

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