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Is it possible to find a lower bound of this integral? $\displaystyle\int^A_0 (A-x)p(x)\ dx$. Here $p(x)$ is some probability distribution with known mean and standard deviation and $A$ is a constant.

I was trying to simplify this as $A\displaystyle\int^A_0 p(x)\ dx - \displaystyle\int^A_0 xp(x)\ dx $. The lower bound on the first integral can be found using Markov's inequality but how to find the upper bound of the second integral? Also, will this bound be tight?

Dip
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1 Answers1

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You can use Holder's inequality to bound the second one. We should have $$\int_{0}^{A} xf(x)dx\le (\int^{A}_{0}x^{2}dx)^{1/2}(\int^{A}_{0}f(x)dx)^{1/2}=(\frac{A^{3}}{3})^{1/2}(\int^{A}_{0}f(x)^{2}dx)^{1/2}$$ The maximum achieves if and only if $f(x)$ is a constant function. So we have a strict upper bound (if the standard deviation is not zero) this way. I guess a better upper bound exists; but I do not know how.

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Bombyx mori
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