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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?

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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.