4
$\begingroup$

Consider the Hilbert space $L^2_w$ with scalar product $\langle f,g\rangle_w =\int_0^\infty f(x)g(x)w(x)dx$ where the weight $w$ is the density function of a log-normal distribution $$ w(x)=\frac{1}{\sqrt{2\pi}\sigma x}e^{-\frac{(\ln(x)-\mu)^2}{2\sigma^2}},$$ for some $\mu\in\mathbb{R}$ and $\sigma>0$. The log-normal distribution has finite moments of any order but it is well known that it is not determined by its moments and as a consequence we have that the polynomials do not lie dense in $L^2_w$. Are there any known results that describe subspaces of functions that lie in the colsure of the polynomials in this space? In particular I am interested to know whether or not the square root function lies in the polynomial closure?

$\endgroup$
0
$\begingroup$

By squaring this is the same question as is $X$ in the span of $1, X^2, X^4,...$. I will produce a function that has the property that $E(FX^{2j}) = 0, E(FX) \ne 0$. Set $Z = log(X)$ which is a normal r.v. Regarding $F$ as a function of Z, $E(FX^{2j}) = 0$ is equivalent to $E(F(Z+2j) ) = 0$ since $X^{2j} e^{ - 2j^2} $ is the likelihood ratio for $N(2j, 1)$ with respect to $N(0,1)$. Similiarly, $E(FX^{j}) \ne 0$ is equivalent to $E(F(Z+1) ) \ne 0$. Take $F(x) = 1, 0< x < 2, -1 2 < x < 4$ and extend to be periodic with period 4. Then I claim $F(Z) = F(log(X)) $ is as claimed. The function is antisymmetric so $E(F(Z)) = 0$ but also, $F(x-2j)$ is antisymmetric, so $E(F(Z-2j)) = 0$ and therefore $E(FX^{2j}) = 0$ . But $F(x - 1)$ is not antisymmetric and will not have 0 expectation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.