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i'd like to prove the following inequality: $$ (\int_{0}^{\infty}e^{-\alpha ((A+t)^b-A^b)} dt )^2 \geq \int_{0}^{\infty}t e^{-\alpha ((A+t)^b-A^b)} dt $$

where $\alpha \geq 0$ (scale parameter), $b \geq 1$ (shape parameter) and A positive reel number (age).

In fact, what i im trying to prove is that, a random variable following a weibull distribution with age A, has a coefficient of variation smaller than 1.

I tried to get a closed forme of these 2 integrals, where the first one is (first moment):

$$ m_1=\int_{0}^{\infty}e^{-\alpha ((A+t)^b-A^b)} dt = \frac{\alpha^{-1/b}}{b}e^{\alpha A^b} \Gamma(\frac{1}{b},\alpha A^b)=\frac{A}{b}U(1,1+\frac{1}{b},\alpha A^b) $$ where $\Gamma$ the upper incomplete gamma function and U the confluent Hypergeometric function;

and the second one(half of the 2nd moment): $$\frac{m_2}{2}=\int_{0}^{\infty}t e^{-\alpha ((A+t)^b-A^b)} dt =\frac{\alpha^{-2/b}}{b}e^{\alpha A^b} \Gamma(\frac{2}{b},\alpha A^b)-A \cdot m_1 $$

However, i still don't see how can we prove this inequality.

Cn anyone give some help or references?

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Since the two exponents are identical, we can proceed in a very rough naive but interesting way:

Let's instead prove the following:

$$\left(\int_0^{+\infty} e^{-at}\ dt\right)^2 \geq \int_0^{+\infty} t e^{-at}\ dt$$

Now the integrals are pretty trivial, and we conclude:

$$\frac{1}{a^2} \geq \frac{1}{a^2}$$

If instead of $e^{-at}$ we chose $e^{-at^2}$ thence we'd have

$$\left(\frac{\sqrt{\pi}}{2\sqrt{a}}\right)^2 \geq \frac{1}{2a}$$

$$\frac{\pi}{4a} \geq \frac{1}{2a}$$

Which is of course true.

By a simil method we can extend it to your integral, hence the inequality holds true.

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  • $\begingroup$ Thanks for your answer Von Neumann. However i still d'ont see how can we generalize the results to a more complicated situation $\endgroup$ – Xingheng Liu Feb 13 '18 at 8:34

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