Probability - almost sure convergence

[Gregg]

Homework Statement

We have $\mathbb{P}(X_n = 1) = p_n$ and $P(X_n=0) = 1-p_n$ the question is about almost sure convergence. i.e. does $X_n \overset{a.s.}{\longrightarrow} 0$ if $p_n = 1/n$?

Homework Equations

$X_n \overset{a.s.}{\longrightarrow } X$ if $\mathbb{P}( \omega \in \Omega : X_n(\omega) \to X(\omega) \text{ as } n\to \infty) = 1$

The Attempt at a Solution

I don't think I understand this properly. Looking at my attempt I've tried a quick $\epsilon -\delta$ setting $\epsilon = 2/N$ and having $|1/n| < \epsilon$ for $n>N$

I don't think this is what it's asking. Can I say that $X(\omega) = 0$ "clearly" and then that $\mathbb{P}( \omega \in \Omega : |X_n(\omega) - X(\omega)| > \epsilon \text{ i.o. }) = 0$ ?

Where i.o. means infinitely often.

 

[mighty2000]

thank you for your question. Let me try to clarify the concept of almost sure convergence and how it applies to the given situation.

Firstly, the statement $X_n \overset{a.s.}{\longrightarrow} 0$ means that the sequence of random variables $X_n$ converges to the constant value 0 with probability 1. This means that for almost all outcomes (i.e. for almost all possible values of the random variable), the sequence will eventually settle down to 0. In other words, the probability that the sequence does not converge to 0 is 0.

In order to prove that $X_n \overset{a.s.}{\longrightarrow} 0$, we need to show that the probability of the sequence not converging to 0 is 0. In other words, we need to show that $\mathbb{P}(\omega \in \Omega : X_n(\omega) \nrightarrow 0) = 0$. To do this, we can use the definition of almost sure convergence that you have provided: $X_n \overset{a.s.}{\longrightarrow } X$ if $\mathbb{P}( \omega \in \Omega : X_n(\omega) \to X(\omega) \text{ as } n\to \infty) = 1$.

In this case, we have $X(\omega) = 0$ for all outcomes $\omega$, since we are trying to prove that the sequence converges to 0. Therefore, we can rewrite the definition as: $X_n \overset{a.s.}{\longrightarrow } 0$ if $\mathbb{P}( \omega \in \Omega : X_n(\omega) \to 0 \text{ as } n\to \infty) = 1$.

Now, let's look at the given probabilities: $\mathbb{P}(X_n = 1) = p_n$ and $P(X_n=0) = 1-p_n$. We can see that as $n \to \infty$, $p_n = 1/n$. This means that as $n \to \infty$, the probability of $X_n$ taking the value 1 decreases to 0, while the probability of it taking the value 0 increases to 1