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Given $c \in R$, a deterministic probability density $f(x)$ and its cumulative distribution $F(c)$, what can be said about $G(c)$ where:

$G(c)=\int f(x)F\left( x+c\right) dx $

The question specifically is:

A) Whether $G(x)$ is concave or convex in $c$?

B) Whether $G(0)=0$ (i.e. $G(c=0)=0.5$)?

C) The sign of the first and second order derivatives wrt $c$.

Any suggestion will be highly appreciated.

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    you are right, I got it wrong2012-10-15

1 Answers 1

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There is not enough information to answer all of your questions; it depends on the behaviour of $f$. Perhaps useful is the observation that:

$F(x) = \int_{-\infty}^x f(y)\ \mathrm dy, G(c) = \int_{\Bbb R}\int_{-\infty}^{x+c} f(x)f(y)\ \mathrm dy \mathrm dx$

Assuming a sufficiently smooth density $f$ is used, we also have:

$G'(c) = \int_{\Bbb R} f(x)f(x+c) \ \mathrm dx$

Since $f(x) \ge 0$, $G'(c) \ge 0$ for all $c$. Note that it is not even a priori clear that this integral will exist.

Upon differentiating this expression again we get a term $f'(x+c)$ which can behave in arbitrary ways. Noteworthy is that since $0 \le F(x) \le 1$ for all $x$, we must have $0 \le G(c) \le 1$ for all $c \in \Bbb R$.

Now let us look at the domain of the double integration for $G(c)$. We have that it is:

$\{(x,y) \in \Bbb R^2: y \le x + c\} = \{(x,y) \in \Bbb R^2: x \ge y - c\}$

Thus we can rewrite the integral to:

$\int_{\Bbb R} \int_{y - c}^\infty f(x)f(y) \ \mathrm dx \mathrm dy$

but since $x,y$ are dummy variables, this is the same as:

$\int_{\Bbb R} \int_{x - c}^\infty f(x)f(y) \ \mathrm dy \mathrm dx$

If $c = 0$, we obtain:

$2 G(0) = \int_{\Bbb R} \int_{-\infty}^x f(x)f(y) \ \mathrm dy \mathrm dx + \int_{\Bbb R} \int_x^\infty f(x)f(y) \ \mathrm dy \mathrm dx = \int_{\Bbb R} \int_{\Bbb R} f(x)f(y) \ \mathrm dy\mathrm dx = 1$

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    I will be glad to do that, may you tell me how to. I am a new user of this site.2012-10-16