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Consider an arbitrary probability density function defined on the interval $[x_\text{min},1]$ with mean $\mu$ and $x_\text{min} > 0$. What is the smallest $n$-th moment? The precise distribution may not be needed. Concretely, what is $$\min_{f(x)}\int_{x_\text{min}}^1 x^n f(x)dx$$ where $$\int_{x_\text{min}}^1f(x)dx = 1$$ and $$\int_{x_\text{min}}^1 xf(x)dx = \mu$$

The fact that the interval contains $0 \leq x \leq 1$ might be helpful since $$\lim_{n\to\infty} \int_{x_\text{min}}^1 x^n f(x)dx = 0$$

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  • $\begingroup$ I don't think the expression you're after is well defined. What do you mean by taking the minimum of that integral with respect to $f(x)$? Once you integrate, you'll be left with a real number or a function $g(x_{\min})$, usually, so what is there to minimize with respect to $f(x)$ or $x$? $\endgroup$ – Nap D. Lover Jun 29 '17 at 22:05
  • $\begingroup$ $f(x)$ is not known. Effectively, we are minimizing over the space of all functions. I didn't specify any space in particular. $\endgroup$ – A15234B Jun 29 '17 at 22:16
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If $n>1$ the function $x\mapsto x^n$ is convex on $[x_\mathrm{min},1]$ and then by Jensen's inequality we know (switching to probability language) that $E X^n \ge (EX)^n = \mu^n$. In fact, if all the probability mass were concentrated at $\mu$, the minimum $\mu^n$ would be achieved. But if you insist that $f$ be a proper density function, and not the Dirac $\delta(x-\mu)$, the best you can do is get arbitrarily close to $\mu^n$, by (say) taking $f$ to be supported in a very short interval centered at $\mu$.

If $0<n<1$ the function is concave, and a similar argument shows the minimum is attained by concentrating all the mass at $1$ and $x_\mathrm{min}$ in such a way as to make $EX=\mu$. And so on.

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  • $\begingroup$ I can't believe I didn't see it. It's intuitively obvious when you think of multiplying $x^n$ and $f(x)$. Thanks! $\endgroup$ – A15234B Jun 29 '17 at 22:18

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