Problem

Let $\{X_n\}$ be a sequence of independent random variables, $X_n\sim exp(q_n)$.

1. suppose $\sum \frac{1}{q_n}<\infty$, show that $\sum X_n<\infty$ a.s.
2. suppose $\sum \frac{1}{q_n}=\infty$, show that $\sum X_n=\infty$ a.s.

Solution

solve only for 2. here, and as a background, $\{\sum X_n\text{ exists}\}$ is a tail event (as long as $\{X_n\}$ being independent), so from Kolmogorov’s 0-1 law, $\sum X_n$ either $=\infty$ a.s. or $< \infty$ a.s.

• method 1: Kolmogorov’s three series theorem

  compute for some $A$: 1️⃣ $\sum \mathbb{P}[|X_n|\geqslant A]$ 2️⃣$\sum\mathbb{E}[Y_n]$ 3️⃣ $\sum \operatorname{Var}(Y_n)$, where $Y_n=X_n\mathbb{\boldsymbol 1}_{\{|X_n|\leqslant A\}}$.

  this works well especially when $\{X_n\}$ is (uniformly) bounded.

• method 2: moment generating function

  if $\sum X_n<\infty$ a.s. then for all $s>0$,

  $$ 0<\mathbb{E}[e^{-s\sum X_n}]=\prod_{n=1}^\infty\frac{q_n}{s+q_n} $$

  taking $-\ln(\cdot)$ for both sides, i.e. $\sum\ln\frac{q_n}{s+q_n}=\sum\ln\left(1-\frac{s}{s+q_n}\right)<\infty$, which is equivalent to $\sum\frac{s}{s+q_n}<\infty$. This implies $\sum \frac{1}{q_n}<\infty$, since $s>0$ is arbitrary.

• method 3: index function

  consider $Y_n=\frac{1}{q_n}{\bold 1}_{\{X_n>\frac{1}{q_n}\}}$, we have nice properties:

  • $\{Y_n\}$ independent, again, from Kolmogorov’s 0-1 law, $\mathbb{P}[\sum Y_n=\infty]=0$ or $1$
  • $X_n\geqslant Y_n\geqslant 0$
  • $\mathbb{E}[Y_n]=\frac{1}{q_n}e^{-q_n\frac{1}{q_n}}=\frac{1}{q_n}$ , as a result, $\mathbb{E}[\sum Y_n]=\sum \frac{1}{q_n}=\infty$

  thus $\mathbb{P}[\sum Y_n=\infty]=0$ fails, and $\sum X_n\geqslant \sum Y_n=\infty$ a.s.

• method 4(bad): Chernoff/Chebyshev+Borel-Cantelli

  this is a (sometimes) useful method, usually works well for proving $\sum <\infty$ a.s.

  a.s. convergence $\iff \mathbb{P}[\cap_M\cup_N\cap_n\{\text{blabla}_M\}]=1$ $\iff$$\mathbb{P}[\cap_N\cup_n\{\text{blabla}_t\}^c]=0$, and the last one can be deducted from Borel-Cantelli lemma, via $\sum \mathbb{P}[\{\text{blabla}_t\}^c]<\infty$

  • sometimes $\{\text{blabla}_t\}$ is like $\{|S_n|>t\}$, we use Chebyshev inequality $\mathbb{P}[~]\leqslant t\mathbb{E}[~]$
  • sometimes $\{\text{blabla}_t\}$ is like $\{|S_n|<t\}$, we use Chernoff type argument:

  $$ \mathbb{P}[|S_n|<t]=\mathbb{P}[e^{-\lambda_n|S_n|}>e^{-\lambda_n t}]\leqslant e^{\lambda_n t}\mathbb{E}[e^{-\lambda_n|S_n|}] $$

  this may not work well in general, since finally we arrive at $\sum e^{-a_n}$ with only $a_n\to\infty$, which is not enough. a possible refinement is to modify $\lambda_n$.