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I'm searching for an upper bound for a function like

$$ E\left[ \max_{x \in R} \left( \frac{ \sum_{i=1}^n K(\frac{x-X_i}{f(x,X_1, \dots X_n)}) \cdot Y_i } { \sum_{i=1}^n K(\frac{x-X_i}{f(x,X_1, \dots X_n)})} - \frac{ \sum_{i=1}^n K'(\frac{x-X_i}{f(x,X_1, \dots X_n)}) \cdot Y_i } { \sum_{i=1}^n K'(\frac{x-X_i}{f(x,X_1, \dots X_n)})} \right) \right], $$

where $R \subset \mathbb{R}$, $K$ and $K'$ are kernel-functions (i.e. a non-negative real-valued integrable function) , $X_1, \ldots, X_n$ are iid Random Variables over a probability space $\Omega$ and $Y_1, \ldots Y_n$ are iid Random Variables over a probability space $\Omega$ and $f$ is a $ \mathbb{R}\times \Omega$ measurable function.

$\frac{0}{0} = 0$

$Y_1, X_1, f$ are bounded.

I'm hoping for some kind of exponential inequality, maybe much more general one, but couldn't find one...

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  • $\begingroup$ Please provide more information about the distribution of the random variables and the types of kernel functions being considered. An explanation of the notation is needed, too: please don't make us guess what all these letters mean, even if you think you are using a standard notation for this kind of problem. $\endgroup$ – whuber Apr 7 '15 at 14:19
  • $\begingroup$ Hey, thanks alot for the reply. I just added some requirements of the rv's and added the explanation of the notation. Sorry for the sloppiness... $\endgroup$ – Michael Apr 7 '15 at 17:20
  • $\begingroup$ Thanks. Just one more little thing: both denominators vary with the index $i$, but of course the iid assumptions imply the expectations won't change with $i$. I am wondering, though, whether you intended these denominators to be inside the summations. These expressions bother me because without additional assumptions, it looks like those denominators could easily be zero, suggesting it's hopeless to look for any kind of universal upper bound. $\endgroup$ – whuber Apr 7 '15 at 17:27
  • $\begingroup$ Thank you very much. I really made a stupid mistake in the denominator. $\endgroup$ – Michael Apr 8 '15 at 12:35
  • $\begingroup$ Now it's looking more like a kernel estimator! $\endgroup$ – whuber Apr 8 '15 at 15:27

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