The question is suppose that $\mu_n$ is the law of independent random variables $\{X_n\}$, where
\begin{equation*}
P(X_n=0)=\frac{1}{n^\alpha}=1-P(X_n=1)
\end{equation*}
where $\alpha \in (0,\infty)$.
For what value of $\alpha$ will weak convergence, almost sure convergence and convergence in probability holds true?
- Convergence in Probability:
Let $\{X_n\}$ be a sequence of independent random variables defined as
\begin{align*}
\begin{cases}
X_n = 0 & w.p~~\frac{1}{n^\alpha}\\
X_n=1 & w.p~1-\frac{1}{n^\alpha}
\end{cases}
\end{align*}
Then, $\lim_{n\rightarrow \infty}P(|X_n-1|>\epsilon)=\lim_{n\rightarrow \infty}P(X_n=0)=\lim_{n \rightarrow \infty}\frac{1}{n^\alpha}=0$.
This implies that $X_n \rightarrow 1$ in probability.
- Almost sure convergence:
Let $A_n$ be the event that $\{X_n=0\}$. Then, $A_n's$ are independent and $\sum_{n=1}^{\infty} P(A_n)=\infty$ for $\alpha \in(0,1]$. Then by Borel-Cantelli lemma 2w.p. 1 infinitely many $A_n's $ will occur.
I am not sure what to say about when $\alpha >1$ and weak convergence?
Can anyone give some idea about it?