Simplify $int F(x+a)dF(x)$

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Any hints on how to simplify



$$int_a^b F(x+c)dF(x),$$



where $a,b,c in mathbbR^+$, $F(x)$ is a CDF of random variable $X$ and its PDF $f(x)$ is continuously differentiable over the support $[-infty, infty].$



I played around with integration by parts, but did not achieve too much.







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    This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
    – Kavi Rama Murthy
    Aug 22 at 7:22















up vote
3
down vote

favorite












Any hints on how to simplify



$$int_a^b F(x+c)dF(x),$$



where $a,b,c in mathbbR^+$, $F(x)$ is a CDF of random variable $X$ and its PDF $f(x)$ is continuously differentiable over the support $[-infty, infty].$



I played around with integration by parts, but did not achieve too much.







share|cite|improve this question
















  • 1




    This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
    – Kavi Rama Murthy
    Aug 22 at 7:22













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Any hints on how to simplify



$$int_a^b F(x+c)dF(x),$$



where $a,b,c in mathbbR^+$, $F(x)$ is a CDF of random variable $X$ and its PDF $f(x)$ is continuously differentiable over the support $[-infty, infty].$



I played around with integration by parts, but did not achieve too much.







share|cite|improve this question












Any hints on how to simplify



$$int_a^b F(x+c)dF(x),$$



where $a,b,c in mathbbR^+$, $F(x)$ is a CDF of random variable $X$ and its PDF $f(x)$ is continuously differentiable over the support $[-infty, infty].$



I played around with integration by parts, but did not achieve too much.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 22 at 6:43









Green.H

1,057216




1,057216







  • 1




    This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
    – Kavi Rama Murthy
    Aug 22 at 7:22













  • 1




    This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
    – Kavi Rama Murthy
    Aug 22 at 7:22








1




1




This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
– Kavi Rama Murthy
Aug 22 at 7:22





This is the so-called convolution of $F$ with itself evaluated at $-c$. There is no general formula for this.
– Kavi Rama Murthy
Aug 22 at 7:22
















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