# Convergence almost surely

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!

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.
• 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! – Gio Mar 6 '20 at 13:56