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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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Maybe you should restate the title of your question, as one might expect the integral you are trying to bound to be "infinite"... –  Sh4pe Oct 26 '12 at 12:42
    
Good idea, thanks! –  Donagh Oct 26 '12 at 12:51

1 Answer 1

up vote 0 down vote accepted

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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Thanks, that's really interesting! Measure theory isn't exactly my area, so I'm finding your notations a bit alien. Using the functions you defined, if I were to apply the second mean value theorem, I would get the following: $\int_{0}^{\infty} F(x) (g_{1}(x) - g_{2}(x)) dx = F(0) \int_{0}^{y}(g_{1}(x) - g_{2}(x)) dx + F(\infty) \int_{y}^{\infty}(g_{1}(x) - g_{2}(x)) dx$ –  Donagh Oct 26 '12 at 13:54
    
Apologies, I hit enter too soon. My previous comment should continue: Assuming that $F(0) = 0$ and $F(\infty) = 1$, this simplifies to: $I = \int_{y}^{\infty}(g_{1}(x) - g_{2}(x)) dx$ and so: $I \leq max \left( \int_{y}^{\infty}(g_{1}(x) - g_{2}(x)) dx | y \geq 0\right)$. Is this equivalent to your argument? –  Donagh Oct 26 '12 at 14:01
    
The formulas are equivalent, because $\int\limits_0^{+\infty}(g_1(x)-g_2(x))\,\mathrm dx=0$. –  Did Oct 26 '12 at 14:03
    
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! –  Donagh Oct 26 '12 at 14:09

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