MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my research (I am not a mathematician by training, but I frequently use mathematics in my research), I often come across "approximation" problems in the following form:

Let $X$ and $Y$ be two possibly dependent continuous non-negative real-valued random variables. One can assume all the nice things about the PDFs of $X$ and $Y$. The situation is now as follows: Usually, I can prove that an estimate of the form $X \leq Y \leq \beta X$ for some $\beta > 1$ holds. The ultimate goal is to find an upper bound on the integral $I = \int_{a}^{a(1+\epsilon)} f_Y(y)\mathrm{d}y$, where $a>0$ and $\epsilon>0$ is "small." Here, the assumption is that the PDF of $Y$ is "extremely hard to evaluate" and one hopes to find an upper bound using the PDF of $X$ (which is much easier to evaluate let's say) only.

One can go by writing $I = F_Y(a(1+\epsilon)) - F_Y(a)$ and bound each term via the CDF of $X$. Suppose that by doing so, we get the upper bound $I_{\mathrm{bad}}$ on $I$. The problem is that obviously, $I_{\mathrm{bad}}$ will not decay to $0$ as $\epsilon \rightarrow 0$ while $I$ does. I was wondering whether there is another trick of the trade that will give a more sensical bound for small $\epsilon$. I am not (necessarily) looking for an answer that will hold for any distribution; even ideas or methods regarding a specific class of distributions would be greatly appreciated. Thanks for your time.

share|cite|improve this question
It doesn't seem like you have enough to go on. As far as I understand you're trying to bound the probability that $Y$ belongs to an interval $[a,b]$ based only on the fact that $X \le Y \le \beta X$. If there are no other assumptions then the upper bound "probability that $[X,\beta X]$ intersects $[a,b]$" is the best possible and this will converge to the probability that $X \le a \le \beta X$ if we let $b \to a$. – Pablo Lessa Apr 4 '13 at 20:12
Dear Pablo, thank you for your reply. That is what I was suspecting as well. I guess I will have to get my hands dirty once again.. – Michael Brown Apr 4 '13 at 21:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.