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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 $y = ln(x)$: \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?

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  • $\begingroup$ Minor comment about your question: When you do change of variables, the integration limits should also change, i.e., it should be from $0$ to $\infty$ instead of $-\infty$ to $\infty$. $\endgroup$ – Maxtron Oct 30 '18 at 18:20
  • $\begingroup$ The change of variable is $y = ln(x)$, so the new variable $-\infty <y < \infty$. $\endgroup$ – BIM Oct 30 '18 at 19:07

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