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Convergence of random variables.
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[QUOTE="Ray Vickson, post: 4437387, member: 330118"] Your notation is unfortunate, and confuses the issues. The random variables are ##X_n##, but their possible values are just ##x##; that is, we speak of ##F_n(x) = P\{ X_n \leq x \},## etc., where the ##x## has no ##n-##subscript. The point is that convergence in probability of ##Y_n \equiv X_n / \ln(n)## says something about how the functions ##F_n(y) = P\{ Y_n \leq y \}## behave as ##n \to \infty##; here, ##y## does not vary with ##n.## What you have managed to do is more-or-less correctly show convergence in probability. Your argument does NOT show the lack of almost-sure convergence. What you need to do is show that the event [tex] E = \{ \lim_{n \to \infty} Y_n = 0\} [/tex] satisfies ##P(E) < 1.## Alternatively, you can try to show that ##P(E^c) > 0,##, where ##E^c## is the complement of ##E.## What you did was improperly remove the arguments inside the limit when you wrote ##\lim_{n \to \infty} Y_n = 0##; this is meaningless as written, because there are several possible definitions of ##``\lim'', ## (convergence in probability, mean-square convergence, ##L^1## convergence, a.s. convergence, etc) and you have given no indication of which one you intend. [/QUOTE]
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Convergence of random variables.
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