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?

  • $\begingroup$ 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, 2018 at 18:20
  • $\begingroup$ The change of variable is $y = ln(x)$, so the new variable $-\infty <y < \infty$. $\endgroup$
    – AIM
    Oct 30, 2018 at 19:07
  • $\begingroup$ Since $h$ is not stipulated, it doesn't matter what distribution you assume for $X.$ That's because, letting $F$ be the distribution function of $X$ (which is invertible) and letting $G$ be any other invertible continuous distribution function, $$h(X)=h(F^{-1}G(G^{-1}((F(X)))))=(h\circ F^{-1}\circ G)(Y)$$ where $Y$ has $G$ for its distribution and now $h\circ F^{-1}\circ G$ plays the role of $h.$ That makes this question too broad to be answerable. $\endgroup$
    – whuber
    Jun 25, 2021 at 19:03


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