7
$\begingroup$

Let $\xi_1,\xi_2\cdots$ be nonnegative, independent random variables with $\xi_n$ having distribution $\lambda_n \exp(-\lambda_n x)dx$, $x\geq0$ and $\lambda_n>0$. Assume $\sum_{n=1}^\infty \lambda_n^{-1}=\sum_{n=1}^\infty E(\xi_n)=\infty$, show that $\sum_{n=1}^\infty \xi_n = \infty$ a.s.

This is a homework problem, I'd found the same problem here( $X_n \sim \text{Exponential}(\lambda_n)$, independent, $\sum 1/\lambda_n = \infty$, then, $\sum X_n=\infty$ a.s.), but I don't have characteristic function as tool yet, I don't know how otherwise to check $P\{\sum_{n=1}^\infty\xi_n = \infty\} \neq 0$ after using zero- one law.

BTW, I will accept characteristic function method though if no other clean proof is found(I don't know how step 3 in the above link gives contradiction after obtaining the characteristic function.)

$\endgroup$
3
  • 2
    $\begingroup$ I find this in Ash's probability and measure theory, problem 6.2.6(b), p247. He gave a hint to consider $\exp \{-\sum_{j=1}^{n}\xi_j\}$. $\endgroup$
    – vita nova
    Oct 27 '17 at 8:50
  • 3
    $\begingroup$ Following nova vita's comment, notice that $$ \mathbb{E}\left[\exp\left\{ -\sum_{j=1}^{\infty} \xi_j \right\}\right] = \prod_{j=1}^{\infty}\frac{\lambda_j}{1+\lambda_j} = \exp\left\{ - \sum_{j=1}^{\infty} \log\left( 1+ \frac{1}{\lambda_j} \right) \right\} = 0. $$ Since we are taking expectation to a non-negative random variable, this immediately yields $\sum_{j=1}^{\infty} \xi_j = \infty$ almost surely. $\endgroup$ Oct 27 '17 at 8:58
  • $\begingroup$ It's a pretty nice solution way better than my expectation! $\endgroup$
    – CYC
    Oct 27 '17 at 13:24
3
$\begingroup$

A cheap trick is to use domination of a symmetric variable. Define $$X_n=\begin{cases} 0&\text{ if $\xi_n\leq\lambda_n^{-1}\ln(2)$}\\ \lambda_n^{-1}\ln(2)&\text{ otherwise.} \end{cases}$$

Since $\lambda_n^{-1}\ln(2)$ is the median of $\xi_n,$ each variable $X_n$ is symmetrically distributed about its mean: it has probability $\tfrac 1 2$ of being zero and probability $\tfrac 1 2$ of being $\lambda_n^{-1}\ln(2).$

Since $\sum_{n=1}^\infty\lambda_n^{-1}$ diverges, there is a sequence $0=n_1<n_2<\dots$ with $\sum_{n=n_k+1}^{n_{k+1}}\lambda_n^{-1}\ln(2)\geq 2$ for each $k.$ Since a sum of independent symmetric variables is symmetric, the random variables $Y_k=\sum_{n=n_k+1}^{n_{k+1}}X_i$ are symmetric, and the mean of $Y_k$ is at least $1$ so $\mathbb P[Y_k\geq 1]\geq \tfrac 1 2.$ The event that $Y_k\geq 1$ occurs for infinitely many $k$ has probability $1$ by the Borel-Cantelli lemma. And in that event, $\sum_{n=1}^\infty \xi_n\geq \sum_{k=1}^\infty Y_k$ must be infinite.

$\endgroup$
1
  • $\begingroup$ Very clever approach, thanks for posting ! $\endgroup$ Oct 27 '17 at 9:48
2
$\begingroup$

Applying Kolmogorov's three-series theorem immediately gives a proof.

Putting jokes aside, here is a rather crude solution. I am sure there will be a shorter and neater solution, but my brain has almost stopped working...


Write $\beta_n = 1/\lambda_n$ for simplicity and write $T_n = \lambda_n \xi_n$. It is easy to check that $(T_n)$ are i.i.d. and have the common distribution $\operatorname{Exp}(1)$. Also we write $S_n = \sum_{k=1}^{n} \beta_k$.

  • Assume first that $\beta_n / S_n \not\to 0$. Then there exist $\epsilon > 0$ and $(n_k)$ such that $\beta_{n_k} \geq \epsilon S_{n_k}$ for all $k$. Then

    $$ \sum_{j=1}^{n_k} \xi_j = \sum_{j=1}^{n_k} \beta_j T_j \geq \epsilon S_{n_k} T_{n_k} $$

    and since $\mathbb{P}(T_{n_k} \geq 1 \text{ i.o.}) = 1$ by the second Borel-Cantelli's lemma, we have $\sum_{j=1}^{n_k} \xi_j \uparrow \infty$ with probability one. This proves the desired claim.

  • Now assume that $\beta_n / S_n \to 0$. On the one hand, for $\epsilon \in (0, 1)$ Chebyshev's inequality yields

    $$ \mathbb{P}\left( \sum_{j=1}^{n} \xi_j \leq \epsilon S_n \right) = \mathbb{P}\left( \sum_{j=1}^{n} \beta_j(1 - T_j) \geq (1-\epsilon) S_n \right) \leq \frac{1}{(1-\epsilon)^2 S_n^2} \sum_{j=1}^{n} \beta_j^2. $$

    On the other hand, by the Stolz-Cesaro theorem applied to $\frac{\beta_n^2}{S_n^2 - S_{n-1}^2} = \frac{\beta_n}{S_n + S_{n-1}} \to 0$, it follows that $\frac{1}{S_n^2} \sum_{j=1}^{n} \beta_j^2 \to 0$. Then by extracting a subsequence $(n_k)$ so that

    $$ \sum_{k=1}^{\infty} \frac{1}{S_{n_k}^2} \sum_{j=1}^{n_k} \beta_j^2 < \infty $$

    is satisfied, Borel-Cantelli's lemma tells that

    $$ \mathbb{P}\left( \sum_{j=1}^{n_k} \xi_j > \epsilon S_{n_k} \text{ eventually} \right) = 1 - \mathbb{P}\left( \sum_{j=1}^{n_k} \xi_j \leq \epsilon S_{n_k} \text{ i.o.} \right) = 1. $$

    Therefore the claim follows also in this case.

$\endgroup$
2
  • $\begingroup$ Why $P(T_{n_k}\geq 1\,\,i.o.) =1 $? $\endgroup$
    – CYC
    Oct 27 '17 at 13:24
  • $\begingroup$ @CYC, Notice that $(T_{n_k})_{k=1}^{\infty}$ are i.i.d. and satisfies $\sum_{k=1}^{\infty} \mathbb{P}(T_{n_k} \geq 1) = \sum_{k=1}^{\infty} e^{-1} = \infty$. So the statement follows from the second Borel-Cantelli's lemma. $\endgroup$ Oct 27 '17 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.