2
$\begingroup$

Say, we have the sequence of random variables defined on $\Omega=[0,1]$ with uniform distribution:

$$X_n(\omega) := \begin{cases} \omega, & \text{if $n$ is odd} \\ 1-\omega, & \text{if $n$ is even} \\ \end{cases}$$

And we'd like to investigate if the sequence converges

  • almost surely
  • in probability
  • in distribution

To show that it doesn't converge a.s., we could take any $\omega \in \Omega \setminus \{ \frac{1}{2}\}$ and show that $X_n(\omega)$ isn't Cauchy.

It does converge in distribution since they all have the same CDF.

My question is how to show from the definition that the sequence doesn't converge in probability? It seems that we'd need to show that $$\text{For all r.v. }X \ \ \ \exists \varepsilon>0 \ \ \text{such that} \ \ P(|X_n-X|>\varepsilon)\nrightarrow 0$$ But how would we do that?

$\endgroup$
  • $\begingroup$ Hint: A sequence of random variables $X_n$ converges in probability to $X$ if and only if every subsequence has a further subsequence which converges to $X$ almost everywhere. $\endgroup$ – Chris Janjigian Feb 17 '14 at 0:45
2
$\begingroup$

There are probably prettier ways, but this seems to work:

Take $\varepsilon = \frac18$ and suppose $P(|X_n - X|\gt \frac18) \lt \frac1{16}$ for all $n \gt 2N-1$

then $P(\omega \lt \frac14 \text{ and } |X_{2N}(\omega) - X(\omega)|\gt \frac18) \lt \frac1{16}$

so $P(\omega \lt \frac14 \text{ and } |X_{2N+1}(\omega) - X(\omega)|\gt \frac18) \gt \frac14-\frac1{16} = \frac{3}{16} \gt \frac1{16}$, contradicting the supposition.

So $P(|X_n-X|>\frac18)\nrightarrow 0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.