# Mean and kinda almost sure convergence relation

Considering:

Almost sure convergence as $$X_n \xrightarrow[]{\text{a.s}} X \Leftrightarrow P(\lim_{n \rightarrow \infty}X_n=X)=1$$ and

Mean $$L_p$$-convergence as $$X_n \xrightarrow[]{\text{mean}} X \Leftrightarrow \mathbb{E}|X_n-X|^p \rightarrow 0$$

Are there any examples of such Random variables that they converge in terms of mean and do not converge in terms of 'almost sure' convergence and vice-versa - converging a.s (in terms of definition listed above) and do not converging in terms of mean.

May you help me with this one? (Doing this as a part of my types of RV convergence implication tree proof building - I've used infamous Riss model and similar during proofing, however don't have any ideas for this two - I want to prove that mean (so-called $$L_p$$) convergence do not necessarily imply a.s convergence and vice versa).

• Surely your definition of almost sure convergence should have the probability equalling $1$? Jan 28 at 21:32
• math.stackexchange.com/questions/3959062/…
– Ian
Jan 28 at 21:33
• @jlammy oddly yes.. Jan 28 at 21:49
• What do you mean by "oddly yes"? The thing that you wrote, with $0$ on the RHS, is literally the complete opposite of convergence. I'm sure it's just a typo in your text -- it should be $1$ on the RHS. Jan 28 at 21:52
• @jlammy maybe it is not exactly a.s convergence, I'm living not in the USA/England/etc and we call it here a.s convergence) Jan 28 at 21:56

Consider $$X_n=\begin{cases}n^3 & \text{with probability }n^{-2} \\ 0 & \text{with probability }1-n^{-2}.\end{cases}$$ Then $$X_n\to0$$ almost surely: indeed, $$\sum\mathbb P(\lvert X_n\rvert>\varepsilon)\leq\sum n^{-2}$$ which converges, so we can apply Borel-Cantelli. But $$\mathbb E[X_n]=n\to\infty$$.
Now consider independent rvs $$Y_n=\begin{cases}1 & \text{with probability }n^{-1} \\ 0 & \text{with probability }1-n^{-1}.\end{cases}$$ Then $$\mathbb E[Y_n]=1/n\to0$$, but $$Y_n$$ doesn't converge a.s. Indeed, as $$\sum\mathbb P(Y_n=1)=+\infty$$ and the events $$\{Y_n=1\}$$ are independent, the second Borel-Cantelli lemma gives that $$\mathbb P\left(\limsup_{n\to\infty}\{Y_n=1\}\right)=1,$$ so we can't have $$Y_n\to0$$ almost surely.
• @9cloudalpha First example converges a.s. to $0$, but doesn't converge in mean; second example converges in mean to $0$, but doesn't converge a.s. Jan 29 at 0:11
• And $\Omega$ for $X$ is $[0; +\infty]$? Jan 29 at 7:25