I am reading on martingale theory from "Probability: a graduate course" and where the following theorem is proved (in the proof, theorem $9.1$ refers to Doob's maximal inequality):

enter image description here

Do you have any idea how he obtained inequality $(1),$ especially how he obtained $\{\max_{1 \leq k \leq n} c_kX_{k}^+>\lambda\}$ in the integral in inequality $(1)$ ?


1 Answer 1


Because of the following two things:

  1. $\{c_k\}$ is not increasing so the integrand is less or equal than zero.
  2. $Y_k \geq c_k X^+_k$ so $\{\max_{0 \leq k \leq n} c_k X^+_k > \lambda\} \subset \{\max_{0 \leq k \leq n} Y_k > \lambda\}$.
  • $\begingroup$ Do you think there is a typo while passing to inequality $(1)$ ? Instead of $c_j-c_{j-1}$ it should be $c_{j}-c_{j+1}$ $\endgroup$
    – john
    Commented Dec 15, 2021 at 21:21
  • $\begingroup$ Yes, it's a typo. Above where they define $Y_k$, it says $c_{j+1}$. The funny thing is that my answer was based on that. I missed the typo!, But it's clearly a typo. Also the summation starts in $j=0,$ so we would have a $c_{-1},$ which doesn't exist. $\endgroup$
    – UBM
    Commented Dec 15, 2021 at 23:30
  • $\begingroup$ In this case, his argument is incorrect $\endgroup$
    – john
    Commented Dec 16, 2021 at 3:15
  • $\begingroup$ what part of his argument is incorrect? $\endgroup$
    – UBM
    Commented Dec 16, 2021 at 11:37
  • $\begingroup$ the passage to inequality $(1)$, especially that points 1. and 2. do not hold in this case (integral is positive) $\endgroup$
    – john
    Commented Dec 16, 2021 at 21:49

You must log in to answer this question.

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