2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.