Is the independence of this sequence of random variables not implicitly given when we define their probability distributions?

Question

In this post, the user asks whether the following random variable converges to $0$ almost surely:

$X_n = \begin{cases} 0, & \text{with probability 1 - $\frac{1}{2n}$,} \\ n, & \text{with probability $\frac{1}{2n}$} \end{cases}$

In the comments and answers to the question, it is said that it depends on whether $X_n$ is independent or not. But it seems to me that the $X_n$ are independent because since as we have specified their probability distributions, we always know the probability of getting a $0$ or an $n$ for any particular $n$-th event, and it doesn't matter what has happened for previous (or future) events.

One of the answers to the post outlines two cases where we can have, or not have, almost sure convergence:

• Case 1) Doesn't converge: The $X_n$ are independent then the fact that $X_n = n$ infinitely often with probability one follows from the second Borel-Cantelli lemma
• Case 2) Does converge: Let $U \sim$ uniform$(0, 1)$ and set $X_n = n$ if $U < 1 / 2n$. Then our sequence satisfies the conditions of the problem and $X_n \to 0$ almost surely.

Can someone show me how to prove these statements using the definition of independent events $P(A \cup B) = P(A)P(B)$? I'm hoping that if I can see it written out clearly that I can pinpoint where my intuition is wrong.

Answer 1

So, you're somewhat correct that there is a colloquial tendency to imply that a sequence like that is independent without stating it explicitly, but it's definitely not a guaranteed thing. Let's take a look at the cases.

The first case is independent by definition so no work there.

Take the second case. We want to know if $X_n$ and $X_k$ are independent. Without loss of generality, say $n<k$.

Independence here implies $P(X_n=n, X_k=k)=P(X_n=n)P(X_k=k)$. In order for $X_n=n$ and $X_k=k$, we need $U<1/2k$ (since $n<k$ this implies $U<1/2n$).

We know that $P(U<1/2k)=1/2k$. So $P(X_n=n, X_k=k)=1/2k$. We also know $P(X_n=n)=1/2n$ and $P(X_n=k)=1/2k$.

So $P(X_n=n, X_k=k)=1/2k$ and $P(X_n=n)P(X_k=k)=\frac{1}{4nk}$ which means $P(X_n=n, X_k=k)\neq P(X_n=n)P(X_k=k)$ and thus $X_n$ and $X_k$ are not independent.

The reason they're not independent is that they're both drawing from the same random variable ($U$), so when we write out these probabilities, whether $X_n=1$ is being determined by the same underlying variable for whether $X_k=1$.

The distributions given for $X_n$ are just when we look at it by itself (marginal distributions as Xi'an said). But that doesn't mean it isn't correlated with other variables. To give a dumb example, if $X$ is $1$ when I'm eating pizza and $0$ otherwise and $Y$ is $1$ if I'm happy and $0$ otherwise, we can write those out as $X=1$ with probability $0.1$ and $Y=1$ with probability $0.5$, but as a pizza lover, I can tell you they definitely aren't independent .