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I'm quite new to the field so please bare with me.

Problem: Let ξ be a random variable distributed according to a log-normal distribution with parameters μ and $σ^2$, i.e. log(ξ) is normally distributed with mean μ and variance $σ^2$. Show that the kth moment of ξ is given by $E[ξ^k]=e^{kμ+(k^2σ^2)/2}$

Question: I suppose for this problem we can use the pdf of log-normal distribution and by the definition of the nth moment we need to take an integral $\int_{-\infty}^{\infty}x^nf(x)dx$ where f(x) is our pdf. However what is our x?

P.S.: I am confused by the log(ξ), it seems that we needed to kinda switch the representation of x to log(x) and then use the pdf of normal distribution to obtain a solution for random variable y, which corresponds to y=log(x). Once we get that we can simply find x=exp(y). Or is it just a different approach and we can boldly use log-normal distribution?

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First of all observe that the moment generating function of a Gaussian random variable $X$ is: $$ M(t)=\mathbb{E}(e^{tX})=e^{\mu t+\frac 12\sigma^2 t^2}. $$ Given that $\log\xi$ is normally distributed, we can see: $$ \mathbb E(\xi^n)=\mathbb E(e^{n\log\xi})=M(n)=e^{\mu n+\frac 12\sigma^2 n^2}. $$


To explicitly answer your question, to find $n$-th moment of a random variable $\xi$, means to find $\mathbb E(\xi^n)$. Therefore if $\xi$ is a log-normal RV, the $n$-th moment is as follows: $$ \mathbb E(\xi^n)=\int_0^\infty \xi^n\frac{1}{\xi\sqrt{2\pi\sigma^2}}e^{-\frac{(\log\xi-\mu)^2}{2\sigma^2}}d\xi. $$

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  • $\begingroup$ I really don't see where all your formulas come from. For the last one: it's not what definition says, apart from pdf there should also be ξ^n inside the integral. en.wikipedia.org/wiki/Moment_(mathematics) $\endgroup$ – Denys S. May 15 '14 at 22:31
  • $\begingroup$ Regarding the $\xi^n$, you are right. That was a typo! $\endgroup$ – Arash May 15 '14 at 22:47
  • $\begingroup$ Now which one exactly you do not see? Because the MGF of Gaussian is known, anyway you can write it down. On the other hand that last one is also direct features of expected value. Let me know $\endgroup$ – Arash May 15 '14 at 22:48
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A "moment" of Q is the integral $\int x\; dq$. This means that you integrate the coordinate on some axis (eg radius), over the elements of Q. The n'th moment is $\int x^n \; dq$.

Volume, for example, is 'moment of surface'. This is because surface can be treated as a normal vector by area. If you bound a surface by a loop, you can freely vary the surface, because the implication of $\int x^0 \;dq=0 $ is that the volume does not depend on the position of the observer's coordinate. This means that for any surface bounded by a loop, the vector area is identical regardless of the shape of the surface.

Likewise, one can see that a dipole moment is simply the $p = \int x \; dq$ gives a fixed vector if the total charge is zero.

Momentum is the rate of change of moment (of "mass" = weight), since while the actual moment of mass varies by the observer, the rate of change does not depend on coordinate.

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  • $\begingroup$ Thank you for putting the intuition behind math! $\endgroup$ – Denys S. May 15 '14 at 22:33

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