# Challenging Problem

• Aug 15th 2009, 02:54 AM
ILoveDaVinci
Challenging Problem
Hello math minds,
This is my first post. I was over at physics forums (under a different name) but have quickly migrated to this site after finding it (Rofl).

I would appreciate it if someone could walk me through this, if like me, you find yourself with a lot of free time.

Let $\displaystyle X$ be a random variable with probability density function

$\displaystyle f(x) = (x \sigma \sqrt{2\pi})^{-1}) (1 + \varepsilon \ sin (2 \pi n \frac{(log(x) - \mu)}{\sigma})) exp (- \frac{1}{2 \sigma^2} (log(x) - \mu)^2) \ \ \, x>0$
and $\displaystyle f(x) = 0 \ \ \, x \leq 0$ (a piecewise function) where $\displaystyle \mu \in R$ $\displaystyle , \sigma > 0$ $\displaystyle , \varepsilon \in [0,1), \ and \ n \in N$ are parametres (N and R denote natural and real numbers). Note that when $\displaystyle \varepsilon = 0, f(x)$ is the pdf corresponding to a long-normal distribution.

(1) Show that $\displaystyle \int_{- \infty}^{\infty} f(u) du = 1$

(2) Show that the moments $\displaystyle E[X^{\alpha}]$for all $\displaystyle \alpha \in \ R$ do not depend on $\displaystyle \varepsilon$ or n.

(3) Show that $\displaystyle M(t) = E[e^{tX}]$ is infinite for any t > 0

Good Luck
• Aug 15th 2009, 09:53 AM
kobylkinks
first is obvious
Integral in 1 is just a sum of integral over log-normal density and integral which is reduced (by introducing new variable =( log(x)-mu )/ sigma ) to integral over odd function which equals to zero