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$\int_0^1 g(x) dx = 1$

$\int_0^1 xg(x) dx = \beta $

$\int_0^1 x^2g(x) dx = (\beta)^2 $

I was assigned this homework problem and I don't know how to solve it. I let $ g(x) = 1 $ and let $ g(x) = nx^{n-1} $ , but that got me nowhere. I am stumped.

Flinn Bella
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1 Answers1

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Use the fact that $\int_0^1 g(x) \, dx=1$ to get \begin{align*} \int_0^1 2x \beta g(x) \, dx& =2\beta^2\\ \int_0^1 x^2 g(x) \, dx&=\beta^2 \end{align*} So we have $$\int_0^1 (x-\beta)^2 g(x) \, dx=0.$$ We have this integral as $0$ and $(x-\beta)^2g(x)$ is a continuous positive function, so...?

Anurag A
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  • @FlinnBella if $f$ is continuous and $f \geq 0$ on $[a,b]$, then $\int_a^b f(x) , dx \geq 0$ (think geometrically in terms of areas etc.). So the integral can be zero only when the function is identically equal to zero. In our case this tells us that $(x-\beta)^2g(x)$ must be the zero function. But that forces $g$ to be the zero function which contradicts the positivity. – Anurag A Sep 01 '20 at 04:48
  • @FlinnBella No!! $x$ varies in the interval $[0,1]$ so $x$ cannot be a fixed number. This simply means there is NO positive continuous function $g$ that can satisfy these requirements. If instead of positive we were given that $g \geq 0$, then we could have concluded that $g=0$. – Anurag A Sep 01 '20 at 04:55