# Show that the law of $P(X_n=0)=\frac{1}{n^\alpha}=1-P(X_n=1)$ converges weakly?

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?

1. 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.

1. 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?

Convergence in probability to a constant is equivalent to convergence in distribution (i.e. weak convergence) to the constant, so your proof that $$X_n\overset{p}{\rightarrow}1$$ already implies $$X_n\overset{d}{\rightarrow}1$$ for all $$\alpha>0$$. One minor note in your proof is I would specify $$\epsilon\in (0,1)$$ to justify $$P(|X_n-1|>\epsilon)=P(X_n=0),$$

because in the case $$\epsilon\geq 1$$, the LHS is already zero.

Now, as you have noted, you can use Borel-Cantelli lemma to look at a.s. convergence. Take any $$\epsilon \in (0,1)$$ and note that

$$\sum_n P(|X_n-1|>\epsilon )=\sum_n P(X_n=0)=\sum_n n^{-\alpha}.$$

Consider two cases:

1. $$\alpha>1$$: Since the series converges to $$\zeta(a)<\infty$$, the first Borel-Cantelli lemma tells us $$X_n\overset{a.s.}{\rightarrow}1.$$

2. $$\alpha\in(0,1]$$: Since the series diverges and the $$X_n$$ are independent, the second Borel-Cantelli lemma tells us $$X_n\not \overset{a.s.}{\rightarrow}1.$$

Weak convergence of $$X_n$$ can obtained by direct computations: Suppose $$f$$ is a bounded continuous function on $$\mathbb{R}$$. Then $$E[f(X_n)]=\frac{1}{n^a}f(0)+\Big(1-\frac{1}{n^a}\Big)f(1)\xrightarrow{n\rightarrow\infty}f(1)$$ This is what the law of $$X_n$$ converges weakly to the atomic measure $$\delta_1$$. From this, we also get that $$X_n$$ converges to $$1$$ in probability (it is well known that for any $$a\in\mathbb{R}$$, $$X_n\Longrightarrow a$$ as $$n\rightarrow\infty$$ iff $$X_n\rightarrow a$$ in probability).

As for almost surely convergence, the direct part of Borel Cantelli shows that if $$a>1$$, then $$A_n$$ converges to $$1$$ almost surely ($$\sum_nP[A_n]=\sum_n\frac{1}{n^a}<\infty$$). The indirect part of the Borel-Cntelli theorem (as the OP argue in his posting) implies that if the $$X_n$$ are independent, then convergence almost surely fails.

Observation: If independence is not assumed, then we may get convergence almost surely to $$1$$ as the following example shows:$$(\Omega,\mathscr{F},P)=([0,1],\mathscr{B}([0,1]),\lambda)$$ where $$\mathscr{B}([0,1])$$ is the collection of Borel sets in $$[0,1]$$ and $$\lambda$$is Lebesgue measure restricted to $$\mathscr{B}([0,1])$$. $$X_n(t)=\mathbb{1}_{[0,1-n^{-a}]}(t)$$ satisfies $$P[X_n=0]=n^{-a}=1-P[X_n=1]$$. Clearly $$X_n\xrightarrow{n\rightarrow\infty}1$$ almost surely