Convergence of random variables.

[trash]

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.