Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it possible to find a sequence of r.v.'s $\{X_n\}$ that converge in probability but not almost surely nor in $L^p$?

Does anyone know of a single example that illustrates this point? I've been trying to think of one for ages and I've not come up with anything. I'm sure such a sequence must exist, but can't think what it could be.

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

Perhaps a more probabilistic example:

For each $n$, let $Z_n$ be uniformly distributed on the set $\{1, 2, \dots, n\}$. For $1 \le k \le n$, let $$X_{n,k} = \begin{cases} n^2, & \text{if $Z_n = k$}\\ 0, & \text{otherwise}.\end{cases}$$ Then consider the sequence $\{X_{1,1}, X_{2,1}, X_{2,2}, \dots, X_{n,1}, \dots, X_{n,n}, \dots\}$ which steps through the "rows" of this triangular array. It converges to 0 in probability since for any $\epsilon < 1$, we have $P(|X_{n,k}| > \epsilon) = P(Z_n = k) = 1/n$ which goes to $0$ as $n \to \infty$. It does not converge almost surely, since the sequence will always have infinitely many values greater than 1 (one in each row) but all the rest are zero. Finally, we have $E|X_{n,k}|^p = \frac{1}{n} n^{2p} = n^{2p-1} \to \infty$ for all $p \ge 1$, so the sequence does not converge in $L^p$ either.

share|improve this answer
add comment

Let $P$ be the Lebesgue measure on $[0,1]$ and $$f_{n,k}(x) = \begin{cases} 4^n & x\in \left[\frac{k-1}{2^n}, \frac{k}{2^n}\right] \\ 0 & \text{otherwise} \end{cases}$$

for $k\in \{1, \dots, 2^n\}$.

Then $$\|f_{n,k}\|_1 =\frac{4^{n}}{2^{n}} = 2^n$$ $f_{n,k}$ doesn't converge pointwise (going through the sequence $(n,k)$ as $(1,1), \;(2,1),\; (2,2),\; (3,1),\; \dots$) and $$P[f_{n,k} > \epsilon] = 2^{-n}\to 0$$ for all $0<\epsilon<1$.

share|improve this answer
add comment

$X_n$ independent such that $P(X_n=n)=\frac{1}{n}$ and $P(X_n=0)=1-\frac{1}{n}$.

Then $\forall \varepsilon>0$, $P(|X_n|>\varepsilon)=\frac{1}{n} \to 0$ but by second Borel Cantelli $P(\limsup |X_n|>\varepsilon)=1$ and $E|X_n|^p=n^{p-1}$ that doesn't converge to 0 $\forall p\geq1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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