Convergence of random variables.

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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 $P \{\displaystyle\lim_{n \to{+}\infty}{} {Y_n} = 0 \} = 1$ .

 

[Ray Vickson]

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.