Intereting Posts

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Consider the relation R given by divisibility on positive integers that is xRy <-> x|y

The computed moments of log normal distribution can be found here. How to compute them?

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If $X$ is lognormal, then $Y = \log X$ is normal. So consider $${\rm E}[X^k] = {\rm E}[e^{kY}] = \int_{y=-\infty}^\infty e^{ky} \frac{1}{\sqrt{2\pi}\sigma} e^{-(y-\mu)^2/(2\sigma^2)} \, dy. $$ Now observe that $$\begin{align*} ky – \frac{(y-\mu)^2}{2\sigma^2} &= – \frac{-2k\sigma^2 y + y^2 – 2\mu y + \mu^2}{2\sigma^2} \\ &= -\frac{1}{2\sigma^2}\left(y^2 – 2(\mu + k\sigma^2)y + (\mu + k \sigma^2)^2 + \mu^2 – (\mu + k \sigma^2)^2\right) \\ &= -\frac{\left(y – (\mu+k\sigma^2)\right)^2}{2\sigma^2} + \frac{k(2\mu + k \sigma^2)}{2}. \end{align*}$$ Thus the $k^{\rm th}$ raw moment is simply $${\rm E}[X^k] = e^{k(2\mu + k\sigma^2)/2} \int_{y=-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-(y – \mu’)^2/(2\sigma^2)} \, dy,$$ where $\mu’ = \mu + k \sigma^2$. But this latter integral is equal to 1, being the integral of a normal density with mean $\mu’$ and variance $\sigma^2$. So ${\rm E}[X^k] = e^{k(2\mu + k\sigma^2)/2}$. The variance of $X$ is then easily calculated from ${\rm Var}[X] = {\rm E}[X^2] – {\rm E}[X]^2$.

In fact, the expression for the $k^{\rm th}$ raw moment of $X$ that we derived is actually also the moment generating function of $Y = \log X$.

**Addendum.** A somewhat different computation can be made from the observation that $$\frac{Y – \mu}{\sigma} = Z \sim \operatorname{Normal}(0,1),$$ so $$\operatorname{E}[X^k] = \operatorname{E}[e^{kY}] = \operatorname{E}[e^{k(\sigma Z + \mu)}] = e^{k \mu + (k \sigma)^2/2} \operatorname{E}[e^{(k\sigma) Z – (k\sigma)^2/2}].$$ Then $$\operatorname{E}[e^{(k \sigma) Z – (k \sigma)^2/2}] = \int_{z=-\infty}^\infty \frac{e^{-z^2/2 + (k \sigma) z – (k \sigma)^2/2}}{\sqrt{2\pi}} \, dz = \int_{z=-\infty}^\infty \frac{e^{-(z-k\sigma)^2/2}}{\sqrt{2\pi}} \, dz = 1,$$ and the result is proven.

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