Consider the sequence of random variable given by distribution : $$\mathbb{P}(X_n=1)=1-\frac{1}{n},$$ $$\mathbb{P}(X_{n}=0)=\frac{1}{n}$$ and $Y_n=X_n \cdot Y$ for random variable Y. Does the $Y_{n}$ converge almost surely?

My attempts: $Y_n$ converges in probability to $Y$. What is more ${\{\omega : Y_n\nrightarrow Y\} }=\{\omega : X(\omega)\neq0 \wedge \forall n_0 \exists n>n_0 X_n(\omega)=0\} $ Now, for instance, when $\mathbb{P}(Y=0)=1$, we have almost sure convergence. Does the answer depend on the variable $Y$? I was also trying to show something about $\mathbb{P}(\forall n_0 \exists n>n_0 X_n(\omega)=0)$, but without independence of $X_n$ I have nothing.

  • $\begingroup$ Indeed, both situations are possible. Sure you were not told that $P(Y=0)\ne1$ and/or that $(X_n)$ is independent? $\endgroup$ – Did Jun 14 '15 at 19:29
  • $\begingroup$ I was told that $(X_n)$ don't have to be independent and nothing about $Y$. $\endgroup$ – user231964 Jun 14 '15 at 19:34
  • $\begingroup$ Then both situations are possible. $\endgroup$ – Did Jun 14 '15 at 19:36
  • $\begingroup$ But how to show that if $(X_n)$ is independent and $P(Y=0)\neq1$, then the sequence does not converge? Is it true? :) $\endgroup$ – user231964 Jun 14 '15 at 19:41
  • $\begingroup$ Yes it is true and it follows directly from Borel-Cantelli lemma. $\endgroup$ – Did Jun 14 '15 at 19:42

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