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Assume that the following relation holds \begin{align} p\int^1_{\frac{p}{a}} f(\theta)d\theta = \int^p_a f(\theta)d\theta \end{align} where $p$ is some scalar and $a\in[0,1)$. Is there some general rule which confirms the example at hand? Actually I need to know how a factor in front or within the integral can be connected to the interval boundaries.

Edit

Sorry, apparently it shall read \begin{align} p\int^1_{\frac{a}{p}} f(\theta)d\theta = \int^p_a f(\theta)d\theta \end{align} where the lower bound is reversed.

clueless
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1 Answers1

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$$ \int_{a} ^p f(\theta) d \theta = \int_{a/p} ^{p/p} f \left( p \theta \right) p d \theta = p \int_{a/p} ^{1} f \left( p \theta \right) d \theta $$ is the correct equality.

izœc
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  • Anyone care to explain the downvotes? That would be helpful in order for me to revise my answer. – izœc Aug 25 '15 at 22:55
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    This cannot possibly be correct, as the values of the functions on different intervals are independent of each other and can yield completely different integrals. – Dominik Aug 25 '15 at 22:57
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    I believe the correct calculation should be: $$ \int_{a} ^p f(\theta) d \theta = \int_{a/p} ^1 f \left( p \theta' \right) p d \theta' = p \int_{a/p} ^1 f \left( p \theta \right) d \theta $$ – Loreno Heer Aug 25 '15 at 22:59
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    Thanks @LorenoHeer and Dominik for pointing out my sloppy errors - looks like I was trying to write it up and should have taken more time to actually check what I was writing before I submitted. – izœc Aug 25 '15 at 23:07
  • Cheers @LorenoHeer. Is this result standard? Do you mind pointing to references since I'm from econ. Say $f$ is given by $f(\theta)=1-\theta$. What's exactly $f(p\theta)$? – clueless Aug 25 '15 at 23:12
  • @clueless, $f(p \theta) = 1 - p \theta$. This result just comes from a "u" - substitution, or change of variables, in the integrand. – izœc Aug 25 '15 at 23:13
  • @clueless If $f(\theta) = 1 - \theta$ then $f(p \theta) = 1- p \theta$. – Loreno Heer Aug 25 '15 at 23:13
  • I reckon that I can go in the reverse direction by $p\int^1_{a/p} f(\theta)d\theta=\int^p_a f(\theta/p)d\theta$ ?! – clueless Aug 26 '15 at 08:30
  • @clueless exactly. – izœc Aug 26 '15 at 08:50