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Uniform distribution

We consider a random variable $X$ defined on a probability space $\left(\Omega,\mathcal{F},P\right)$. This random variable represents a number of visitors in a site. We make the following assumption :-

The probability distribution of $X$ is the uniform distribution $\mathcal{U}(0,1)$.

We also make the following hypothesis :-

The number of people visiting the site is assumed to be independent of the total number of daily visitors.

Is the following proposition true.

$$\frac{\int_0^1 x^4 P(dx)}{\int_0^1 x P(dx)}=1$$

Intuitively, the result seems to be true but I can’t find an easy way to prove this result. If anyone has a solution, I will be glad to see it.

A:

1) Yes, because $\mathcal U(0,1)$ has density $$f_X(x) = \frac 1{1+\int_1^x\frac 1t\,\mathrm dt}=\frac{1+x}2$$
2) Yes, to see this note that, for any integrable random variable $\xi$ we have $$\int \xi\frac 1x\,\mathrm dx = \ln\xi + C$$ where $C$ does not depend on $\xi$ and is finite. Apply this to $\xi=\chi_A$ where $A=\{0\le x\le 1\}$.

Use of an intramedullary pin in open-wedge high tibial osteotomy for the treatment of medial osteoarthritis: a 4-year follow-up.
To analyze the long-term results of an intramedullary (IM) pin in open-wedge high tibial oste
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