1
$\begingroup$

I'm working through Lamperti's 1960 paper "Criteria for the recurrence or transience of stochastic process. I" (J. Math. Anal. Appl. 1(3–4), 314–330. DOI: 10.1016/0022-247x(60)90005-6) as part of my master's thesis work, as assigned by my advisor.

The paper is open access and can be found here: https://www.sciencedirect.com/science/article/pii/0022247X60900056?via%3Dihub

In the proof (starting on page 14), I've encountered what seems to be an error in bounding the third term in equation (A.6) on page 15:

$\int_{x^{\alpha}}^{+\infty} f(x+y) dF(y; x, x_{i})$

Here, $f(t) = \log\log(t+e)$. It appears the author bounds $\log\log(x+y+e)$ by $(x+y)^{d_2 - d_1}$ for all large $x$ and $y \geq x^{\alpha}$, where $2 < d_2 < d_1 \leq 3$.

However, this seems impossible because $\log\log(x+y+e)$ increases to infinity as $x+y$ grows, while $(x+y)^{d_2 - d_1}$ decreases to 0 since $d_2 < d_1$.

Am I missing something, or is this an error in the proof? If this bounding is indeed incorrect, how might it be fixed? I'd appreciate any insights into this step of the proof or potential corrections.

Also, if anyone happens to know where the criteria in this paper are used, can you tell me in the comments below? This can helps me understand the implications and the importance of these criteria, and provide very valuable help for my subsequent research direction.

$\endgroup$
1
  • $\begingroup$ When $y \leq x^{\alpha}$, we use taylor-lagrange formula for $f(x+y) - f(x)$ like what's done in the paper. When $y > x^{\alpha}$, we could use $f(x+y) - f(x) \leq \log(1+y) < y^{\beta}$. So the theorem remains correct, there is no problem. $\endgroup$
    – ZENG
    Commented Jun 25 at 13:14

1 Answer 1

1
$\begingroup$

It looks to be a typo: In the second display on p. 329, use $(x+y)^{d_1-d_2}$ instead of $(x+y^{d_2-d_1}$. (In the discussion, $2<d_2<d_1$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .