suppose we have a uniform distribution such that $u\in \left[0,\:\frac{1}{\phi }\right]$ from which a single observation is $x\left(0\right)$.
We want to prove that an unbiased estimator $\phi'\:=\:h\left(x\left(0\right)\right)$ would be equal to $\int _0^{\frac{1}{\phi }}\:h\left(v\right)dv = 1$
So here's my work below:
I think we should minimize the error in order to obtain the best estimate. That's $e\:=\:E\left\{\left(\phi -\phi '\right)^2\right\} = 0 $
In this case, $e\:=\:E\left\{\left(\phi -h\left(x\left(0\right)\right)\right)^2\right\}\:=\:\int _0^{\frac{1}{\phi }}\:\left(\phi \:-h\left(x\left(0\right)\right)\right)^2p\left(\phi \right)d\phi = 0$
distributing and carrying forward, we get the following: $e\:=\:\int _0^{\frac{1}{\phi }}\:\phi ^2p\left(\phi \right)d\phi -2\int _0^{\frac{1}{\phi \:}}\:\phi \:h\left(x\left(0\right)\right)p\left(\phi \:\right)d\phi \:+\int _0^{\frac{1}{\phi \:\:}}\:h^2\left(x\left(0\right)\right)p\left(\phi \:\right)d\phi \:\:=\:0$
since uniform distribution, we know that $p\left(\phi \right)\:=\:\phi $ Applying the substitution, we get
$e\:=\:\int _0^{\frac{1}{\phi }}\:\phi d\phi \:-\:2\int _0^{\frac{1}{\phi \:}}\:\phi \:^2h\left(x\left(0\right)\right)\:d\phi \:\:+\:\int _0^{\frac{1}{\phi \:\:}}\:\phi \:h^2\left(x\left(0\right)\right)\:d\phi \:=0\:$
I don't know where to go from here. Would someone help me with this please? Thank you.