Exchanging $\sup$ and expectation I have a sequence of non-negative random variables $\xi_n\leq b$, and I know that $\Bbb E\xi_n = a_n$ is some decreasing sequence. Is there any chance I can get bounds on $\Bbb E\sup_n \xi_n$ better than just $b$? I guess they may be available in case $\xi$ is a supermartingale, but I am not sure my sequence is. Just to confirm, I don't assume that $\xi$ is iid.
If that makes it easier, instead of supermartingality condition I have a similar one:
$$
  \Bbb E[\xi_{n+1}|\mathscr F_n] \leq \beta\cdot(\varepsilon + \xi_n)
$$
with $\beta < 1$ and small $\varepsilon > 0 $.
 A: I think you can't, I found an example. Set $b \in \mathbb{R}$ and $\beta \in (0,1).$
Take $\xi_n$ independent, such that $\mathbb{P}(\xi_n = b) = 1/2$ and $\mathbb{P}(\xi_n = b - n!)= 1/2$.
[1] (You can change $n!$ for any positive increasing function that satisfies $b(1-\beta) \leq f(n+1)/2 - \beta f(n)$, the factorial is just an abusive solution ).
This sucesion satisfies: 


*

*$\mathbb{E}(\xi_n) = b-(n!/2) = a_n$ decreasing.

*It satisfies you condition, since the r.v. are independent, thus you just need $\mathbb{E}(\xi_{n+1}) \leq \beta \xi_n$, this is why [1] is the requirement for $f$.

*Finally $\sup \xi_n = b \ \ \text{a.e.} $, thus $\mathbb{E}(\sup_n \xi_n) = b$.
A: Let $r=\frac{\beta \varepsilon}{1-\beta}$. Assume $r<b \leq 2r$. We define $\xi_n$ so that it has the following desired properties:


*

*$\xi_n$ are independent

*$P(\xi_n=b)=1/2$.

*$E[\xi_n]=r+\delta_n$, where $\delta_n$ is a decreasing sequence of positive numbers which converges to zero sufficiently rapidly (at least as fast as $\beta^n$) and is strictly less than $b-r$. (Note that in this construction $a_n$ is not given, I am merely choosing $E[\xi_n]$ to be a decreasing sequence.)


Once you've selected $\delta_n$ (which is just a problem related to recurrence relations, with nothing to do with probability), we can actually construct this thing by choosing $P(\xi_n=a)=1/2$ where $a/2+b/2=r+\delta_n$, hence $a=2r+2\delta_n-b$.
For this $\xi_n$, we have $\sup_n \xi_n=b$ a.s., hence $E[\sup_n \xi_n]=b$. Moreover $E[\xi_{n+1} \mid \mathcal{F}_n]=E[\xi_{n+1}]=r+\delta_{n+1}$. Some playing around with recurrence relations shows that this will satisfy the desired inequality for suitable $\delta_n$ (the reason being that $r$ is the fixed point of the recurrence, and that the recurrence is defined by a contraction mapping).
Of course $b \leq 2r$ eventually fails for sufficiently small $\varepsilon$, in which case my construction fails. In this case your result has some hope of being true. Perhaps you could try explicitly assuming $2r<b$, i.e. $\varepsilon<\frac{b(1-\beta)}{2\beta}$?
