Convergence in probability but not almost surely nor L^p

**jjacobs** · April 13th 2012, 05:55 AM

Hi,

I'm trying to find a single example of a sequence of random variables X_n such that the sequence converges to random variable X in probability, but not almost surely nor in L^p for any p. Does anyone know on any simple examples, and how to prove the above?

Regards,

John.

**emakarov** · April 13th 2012, 06:54 AM

Let the probability space be [0, 1] with the standard measure. The idea is to have $X_n = n$ on an interval (or set) of measure 1/n and 0 everywhere else. Moreover, the support of $X_n$ (i.e., $\{\omega\in[0,1]: X_n(\omega)\ne0\}$ ) should shift visiting every point over and over again so that every $\omega$ is mapped to zero and non-zero arbitrarily far in the sequence. Then $X_n$ converges to 0 in probability because the measure of the support tends to zero. On the other hand, $X_n(\omega)$ does not converge for any $\omega$ , so there is no almost sure convergence. Similarly, the measure of each $X_n$ is 1, so there is no convergence to 0 in $L^p$ .

Can you give a precise definition of such $X_n$ ?

**jjacobs** · April 13th 2012, 01:36 PM

Originally Posted by emakarov

Let the probability space be [0, 1] with the standard measure. The idea is to have $X_n = n$ on an interval (or set) of measure 1/n and 0 everywhere else. Moreover, the support of $X_n$ (i.e., $\{\omega\in[0,1]: X_n(\omega)\ne0\}$ ) should shift visiting every point over and over again so that every $\omega$ is mapped to zero and non-zero arbitrarily far in the sequence. Then $X_n$ converges to 0 in probability because the measure of the support tends to zero. On the other hand, $X_n(\omega)$ does not converge for any $\omega$ , so there is no almost sure convergence. Similarly, the measure of each $X_n$ is 1, so there is no convergence to 0 in $L^p$ .

Can you give a precise definition of such $X_n$ ?

I see. So if I were to define the sequence of random variables $X_n$ by $P(X_n=n) = \frac{1}{n}$ and $P(X_n=0) = 1 - \frac{1}{n}$ , I can understand how such a sequence would converge to 0 in probability, but would not converge almost surely.But i'm just having a little trouble understanding why it does not converge in $L^p$ . Is it due to the fact that $E(X_n^r) = \frac{n^r}{n} \rightarrow \infty$ as $n \rightarrow \infty$ ?

**emakarov** · April 13th 2012, 02:03 PM

Originally Posted by jjacobs

I see. So if I were to define the sequence of random variables $X_n$ by $P(X_n=n) = \frac{1}{n}$ and $P(X_n=0) = 1 - \frac{1}{n}$ , I can understand how such a sequence would converge to 0 in probability, but would not converge almost surely.

Yes, but note that you can't define, for example,

$X_n(\omega)=\begin{cases}1&\omega\le1/n\\0&1/n<\omega\le1\end{cases}$

Originally Posted by jjacobs

But i'm just having a little trouble understanding why it does not converge in $L^p$ . Is it due to the fact that $E(X_n^r) = \frac{n^r}{n} \rightarrow \infty$ as $n \rightarrow \infty$ ?

Yes, at least when r > 1.

**jjacobs** · April 14th 2012, 07:05 AM

That's brilliant, I think I'm understanding it a bit better now. So to ensure that the sequence does not converge in $L^r$ for all $r > 0$ (thus including $r = 1$ ), am I right in saying I could define the sequence of random variables $X_n$ by $P(X_n=n) = \frac{1}{n^2}$ and $P(X_n=0) = 1 - \frac{1}{n^2}$ and the results that the sequence converges in probability but not almost surely would still hold, but the sequence would also not converge in $L^r$ for all $r > 0$ ?

**emakarov** · April 14th 2012, 10:00 AM

I don't see how changing the support measure from $1/n$ to $1/n^2$ helps ensure non-convergence in $L^r$ for all r > 0. In fact, such sequence would not converge only for r >= 2.

Originally Posted by jjacobs

I could define the sequence of random variables $X_n$ by $P(X_n=n) = \frac{1}{n^2}$ and $P(X_n=0) = 1 - \frac{1}{n^2}$ and the results that the sequence converges in probability but not almost surely would still hold

I want to stress again that the conditions $P(X_n=n) = \frac{1}{n^2}$ and $P(X_n=0) = 1 - \frac{1}{n^2}$ do not define $X_n$ . There are many sequences that satisfy these conditions. Further, some of them do converge almost surely. Also, if the support measure of $X_n$ is $1/n$ , some sequences converge almost surely. In fact, my idea to find a sequence that does not converge almost surely relies on the fact that $\sum_{n=1}^\infty\frac{1}{n}$ diverges. So I am not sure right away if there are sequences with $P(X_n=n) = \frac{1}{n^2}$ and $P(X_n=0) = 1 - \frac{1}{n^2}$ that don't converge almost surely. For another idea, you may want to see Wikipedia's claim that convergence in probability does not imply almost sure convergence and its proof using Borel–Cantelli lemma.

**jjacobs** · April 14th 2012, 04:54 PM

Apologies, what I had intended to write was $P(X_n=n^2) = \frac{1}{n}$ and $P(X_n=0) = 1 - \frac{1}{n}$ , which i believe retains the property of not converging almost surely and also does not converge in $L^1$ .

**emakarov** · April 14th 2012, 05:06 PM

Originally Posted by jjacobs

Apologies, what I had intended to write was $P(X_n=n^2) = \frac{1}{n}$ and $P(X_n=0) = 1 - \frac{1}{n}$ , which i believe retains the property of not converging almost surely and also does not converge in $L^1$ .

Yes, such $X_n$ converges in $L^r$ for r < 1/2. But your conditions do not imply that $X_n$ does not converge almost surely.

**jjacobs** · April 15th 2012, 02:32 AM

The example given on wikipedia of a sequence where $X_n$ assumes the value 1 with probability $\frac{1}{n}$ and zero otherwise can be shown not to converge almost surely by the Borel-Cantelli lemmas. Can this proof not be adapted for the example above, where instead of considering the event that $X_n = 1$ infinitely often, we substitute this for the event $X_n \geq c$ for some constant c? Would this not suffice to show the intended result?

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