How can I demonstrate that a sequence of Bernoulli Random Variables Xn with parameter $\frac1{2n^2}$ converges almost surely to some random variable X?
I know that $X_n$ takes the value $1$ with Probability $\frac1{2n^2}$ and $0$ with probability $1-\frac1{2n^2}$.
Thank you in advance!
[Math] Almost Surely convergence with Bernoulli
borel-cantelli-lemmasconvergence-divergencelimsup-and-liminfprobability theory
Best Answer
Yes. It will converge a.s. to the zero (constant) random variable. This is a simple application of the Borel-Cantelli lemma.
Let $A_n$ be the event that $X_n=1$. Then $\mathrm{Pr}(A_n) = \frac{1}{2n^2}$, which is summable. Hence by the Borel-Cantelli lemma, with probability one, only finitely many of the $A_n$ will occur. Note that independence of $X_n$ is not assumed.
This means that for some $N$, $X_n = 0$ for all $n>N$ a.s., i.e. $X_n \to 0$ a.s.