# Upper bounds for an integral with an infinite upper limit

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
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, $$I\leqslant\max\left\{\int_0^x(g_2(t)-g_1(t))\,\mathrm dt\,{\Large\mid}\,x\geqslant0\right\}.$$
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