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If a sequence of non-negative random variables $X_1, X_2, \dots$ converges almost surely to a random variable $X$, that is $X_n \xrightarrow{a.s} X$ or equivalently $P(\lim\limits_{n\to\infty}X_n=X)=1$, then what can we say about convergence of $1/X_n$?

I believe since $P(\lim\limits_{n\to\infty}1/X_n=1/X) = P(\lim\limits_{n\to\infty}X_n=X)=1$, we conlude $1/X_n \xrightarrow{a.s}1/X$. Is this complete?

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  • $\begingroup$ X_n^{-1}=1/X_n? Or what? $\endgroup$ – Martín Vacas Vignolo May 16 '16 at 0:23
  • $\begingroup$ @vvnitram yes, I also edited it. $\endgroup$ – Susan_Math123 May 16 '16 at 0:25
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    $\begingroup$ is necessary that P(X_n=0)=0, and the same with X... But if this condition is satisfying, the proof is right $\endgroup$ – Martín Vacas Vignolo May 16 '16 at 0:27
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Since $X_n \to X$ a.s, there's a set $A$, $P(A)=1$ such that for every $\omega\in A$, $X_n(\omega) \to X(\omega)$. Therefore (calculus) on $A \cap \{\omega:X(\omega)\ne 0\}$ we have $\frac{1}{X_n(\omega)}\to \frac{1}{X(\omega)}$.

Summary:

  1. Special case. If $X\ne 0$ a.s., then indeed $\frac{1}{X_n}\to \frac{1}{X}$ a.s.
  2. General. $\frac{1}{X_n} \to \frac{1}{X}$ a.s. on $\{\omega:X(\omega)\ne 0\}$ (that is for $\omega$ in that set, possibly except for a subset of measure zero, the limit holds). Note however, that without assuming anything $\{\omega:X(\omega)\ne 0\}$ may be itself of measure zero in which case, the statement is vacuous.
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