# Approximating the first moment of h(x) where $x$~Lognomal($\mu, \sigma$)

What is the best way to approximate $$E(h(X))$$, where $$X$$ ~ Lognomal($$\mu, \sigma$$).

So far, I can think of Monte Carlo Methods and Gaussian Hermite quadrature as below: \begin{align} E(h(X)) &= \int_{0}^{\infty} h(x) \frac 1 x \cdot \frac 1 {\sigma\sqrt{2\pi\,}} \exp\left( -\frac{(\ln x-\mu)^2}{2\sigma^2} \right) dx \\[8pt] \end{align} using a change of variable $$x = e^y$$: \begin{align} &= \int_{-\infty}^{\infty} h(e^y) \frac 1 {e^y} \cdot \frac 1 {\sigma\sqrt{2\pi\,}} \exp\left( -\frac{(y -\mu)^2}{2\sigma^2} \right) e^y dy \\ &= \int_{-\infty}^{\infty} h(e^y) \cdot \frac 1 {\sigma\sqrt{2\pi\,}} \exp\left( -\frac{(y -\mu)^2}{2\sigma^2} \right)dy \end{align} having $$h(e^y) = g(y)$$ $$= \int_{-\infty}^{\infty} g(y) \cdot \frac 1 {\sigma\sqrt{2\pi\,}} \exp\left( -\frac{(y -\mu)^2}{2\sigma^2} \right) dy.$$ Using the Gauss-Hermite quadrature from this link in Wikipedia: \begin{align} \int_{-\infty}^{\infty} g(y) \cdot \frac 1 {\sigma\sqrt{2\pi\,}} \exp\left( -\frac{(y -\mu)^2}{2\sigma^2} \right) dy &\approx \frac{1}{\sqrt{\pi}} \sum_{i=1}^n w_i g(\sqrt{2} \sigma x_i + \mu) \\ &= \frac{1}{\sqrt{\pi}} \sum_{i=1}^n w_i h(e^{(\sqrt{2} \sigma x_i + \mu)}). \end{align} Is what I am doing here fine? Or this would produce approximation errors?

Not sure what conditions function $$h(X)$$ satisfies, but let's assume that it can be approximated by a polynomial $$p_n(X)=\sum_{k=1}^{n}a_kX^k$$ of degree $$n$$.
It's not hard to calculate the expectation of $$p_n(X)$$ using normal MGF. Random variable $$Z=\ln(X)$$ is normally distributed by definition. It is well know, that $$E(e^{Zt})=e^{\frac{1}{2}t^2}$$. Hence, we can obtain.
$$E(p_n(X))=\sum_{k=1}^{n}a_kE(X^k)=\sum_{k=1}^{n}a_kE(e^{Zk})=\sum_{k=1}^{n}a_k e^{\frac{1}{2}k^2}$$