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I'm trying to work out an upper bound for the following problem, but I'm making very little progress. Hopefully, someone will be able to make a suggestion.

The integral I'm attempting to bound is:

$I = \int_{0}^{\infty} f(x) \left( g(x) - \hat{g}(x) \right) dx$

Here, $f(x)$ is a cumulative distribution function, and so is monotonically increasing on $[0, \infty]$, $g(x)$ is a probability density function, such that $\int_{0}^{\infty} g(x) dx = 1$, and $\hat{g}(x)$ is an approximation to $g(x)$, so as $\hat{g}(x) \rightarrow g(x)$, $I \rightarrow 0$.

I would like to derive some bound $I^{*}$, so that $I^{*} \geq I$ (or $I^{*} > I$), i.e. an error bound on the effect of the mismatch between $g(x)$ and $\hat{g}(x)$. I've looked at some general integral inequalities (Cauchy-Schwarz, Holder, Minkowski, etc.), but with no luck so far. So, my question is this: based on the properties of $f(x)$, $g(x)$ and $\hat{g}(x)$ outlined above, are there any further techniques I can use to upper bound the integral? To be really demanding, I'd love a form along the lines of $I^{*} = c - k \int_{0}^{\infty} (g(x) - \hat{g}(x)) dx$, where $c$ and $k$ are constant with respect to $x$, but any tips on how to tackle this would be great.

I can go into more detail on the exact functions I'm using, if necessary, but I thought it best to keep it general for now.

Thanks,

Donagh

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Let me reformulate the question, with somewhat more orthodox notations. One is given a CDF $F$ and two PDF $g_1$ and $g_2$, and one is interested in bounding the integral $ I=\int_0^{+\infty}F(x)\,(g_1(x)-g_2(x))\,\mathrm dx. $ Note that $I=\int\limits_0^{+\infty}(G_2(x)-G_1(x))\,\mathrm d\mu(x)$ where $F$ is the CDF of the measure $\mu$, that is, for every $x\geqslant0$, $F(x)=\mu([0,x])$.

Hence the only upper bound of $I$, valid for every probability measure $\mu$, is $I\leqslant\max(G_2-G_1)$, that is, $ I\leqslant\max\left\{\int_0^x(g_2(t)-g_1(t))\,\mathrm dt\,{\Large\mid}\,x\geqslant0\right\}. $

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    I figured as much, but my brain goes a bit fuzzy when I see measure theoretic arguments. Thanks for the clarification, and for all your help - it's going to be a good Friday after all!2012-10-26