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?


2 Answers 2


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


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