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.

$E(p_n(X))=\sum_{k=1}^{n}a_kE(X^k)=\sum_{k=1}^{n}a_kE(e^{Zk})=\sum_{k=1}^{n}a_k e^{\frac{1}{2}k^2}$

Not sure if this is helpful, but thought it would be worth sharing.


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