# Convergence of random variables.

## Homework Statement

Given a sequence of independent random variables ${X_n}$, each one with distribution Exp(1). Show that $Y_n = \displaystyle\frac{X_n}{\log(n)}$ with $n \geq 2$ converges to 0 in probability but it doesn't coverges almost surely to 0.

## Homework Equations

Density for each $X_n$ given by $f(x_n) = e^{-x_n}$ if $x_n \geq 0$ and 0 otherwise.

## The Attempt at a Solution

Since $\displaystyle\frac{e^{-x_n}}{\log(n)}$ tends to 0 as $n \rightarrow +\infty$, given $\epsilon > 0$, then there's a $N>0$ such that if $n>N$ we have $\displaystyle\frac{e^{-x_n}}{\log(n)} < \epsilon$. This implies that $\displaystyle\lim_{n \to{+}\infty}{} P\{ |Y_n| < \epsilon \} = 1$.

Now, what about almost surely convergence?.
I have to prove that $P \{\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0 \} \neq 1$ but it seems to me that since $\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0$ will follow that $P \{\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0 \} = 1$ .

Ray Vickson
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## Homework Statement

Given a sequence of independent random variables ${X_n}$, each one with distribution Exp(1). Show that $Y_n = \displaystyle\frac{X_n}{\log(n)}$ with $n \geq 2$ converges to 0 in probability but it doesn't coverges almost surely to 0.

## Homework Equations

Density for each $X_n$ given by $f(x_n) = e^{-x_n}$ if $x_n \geq 0$ and 0 otherwise.

## The Attempt at a Solution

Since $\displaystyle\frac{e^{-x_n}}{\log(n)}$ tends to 0 as $n \rightarrow +\infty$, given $\epsilon > 0$, then there's a $N>0$ such that if $n>N$ we have $\displaystyle\frac{e^{-x_n}}{\log(n)} < \epsilon$. This implies that $\displaystyle\lim_{n \to{+}\infty}{} P\{ |Y_n| < \epsilon \} = 1$.

Now, what about almost surely convergence?.
I have to prove that $P \{\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0 \} \neq 1$ but it seems to me that since $\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0$ will follow that $P \{\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0 \} = 1$ .

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
$$E = \{ \lim_{n \to \infty} Y_n = 0\}$$ 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.