# Same Example for Two Counter Examples

I'm learning some probability theory and I've come across the following:

For an example of a sequence of random vairables that converges in the mean square sense but not almost surely: We set $$P(X_n=1) =1/n \qquad P(X_n=0)=1-1/n$$ and $$X_n\to X$$ in mean square but not almost sure. The not almost sure follows from the second Borel Cantelli Lemma. I got the examples from here

However, in the book Counter Examples in Probability, they give the following example for a sequence that converges almost surely but not completely $$Y_n(w) = \left\{\begin{array}{ll} 1 & 0\leq w< 1/n\\ 0 & 1/n \leq w

Don't $$X_n$$ and $$Y_n$$ define identical distributions? How does $$X_n$$ converse almost surely but $$Y_n$$ doesn't?

• $X_n$ is only partially defined. To see why that is, consider how you might compute (say) $\Pr((X_n, X_{n+1})=(1,1)).$ What would it be?
– whuber
May 11 '19 at 14:03
• What do I need to define them further? I could say they are independent. Are the $Y_n$ independent?
– Rdrr
May 11 '19 at 15:01
There is a subtle difference between the $$X$$'s and the $$Y$$'s. In the example where the $$X$$'s are defined, they are constructed to be independent. This independence is essential in proving why there is no almost sure convergence. OTOH the $$Y$$'s are defined to ensure pointwise convergence occurs almost everywhere.
It is true that the $$X$$'s and $$Y$$'s have the same marginal distribution, i.e., each $$X_n$$ has the same distribution as the corresponding $$Y_n$$. But we've just seen that it's impossible to decide almost sure convergence by the marginal distributions alone.