# Approximating the first moment of $h(x)$ where $x$ ~${\rm log\,normal}(\mu, \sigma)$ [closed]

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?

• 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 $\infty$. Oct 30, 2018 at 18:20
• The change of variable is $y = ln(x)$, so the new variable $-\infty <y < \infty$.
– AIM
Oct 30, 2018 at 19:07
• 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.
– whuber
Jun 25, 2021 at 19:03