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Halbort
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Halbort
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Maximizing the $\alpha$-moment of a distributution

Given $\alpha$ and constant $\mu$,

$$\begin{array}{ll} \text{maximize} & \displaystyle\int_0^\infty p(x)x^\alpha \,\mathrm d x\\ \text{subject to} & \displaystyle\int_0^\infty p(x)\,\mathrm d x = 1\\ & \displaystyle\int_0^\infty p(x)x \, \mathrm d x = \mu\end{array}$$

I had previously posted this problem on math stack exchange. However, on second thought, I think this may be a more involved problem than I previously thought.


Here is one Idea I had. Let $M(t)$ be the moment generating function of $p(x)$. We know for integers $n$ that:

$$\frac{d^n}{dt^n}M(t) \bigg|_{t=0} = \int_0^\infty p(x)x^n \,\mathrm d x$$. I think I could extend this so:

$$\frac{d^\alpha}{dt^\alpha}M(t) \bigg|_{t=0} = \int_0^\infty p(x)x^\alpha \,\mathrm d x$$

Now our problem boils down to this. Find the moment generating function $M(t)$ that has the maximum fractional derivative $\frac{d^\alpha}{dt^\alpha}M(t) |_{t=0}$ such that $M(0) = 1$ and $M'(0) = \mu$. I do not know how to proceed further. I know little about fractional derivatives.