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I already know how to get the moments if $X$ is log-normally distributed. But what happens when $X$ is being shifted: $Y=aX+b$, $a>0$ and $b>0$. How to compute the moments of $Y$?

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You have $Y=aX+b$, but the multiplication by $a$ still leaves you with a lognormal, not really changing anything. If $X\sim \text{logN}(\mu,\sigma^2)$ then $aX\sim \text{logN}(\mu+\log(a),\sigma^2)$, so if you know how to compute moments for a lognormal you can do it for $aX$ as easily as for $X$.

So for $X\sim \text{logN}(\mu,\sigma)$ the problem reduces to working out moments for a simple shift, $Y=X+\gamma$.

You have an answer for raw moments, but I presuming that apart from the mean you want central moments.

So we already have $E(Y) = E(X)+\gamma$ for the mean.

The central moments (and functions of them, like skewness or kurtosis) are all unaffected by the location-shift.

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We have

$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$

so

$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k$.

The rest remains on what is $X$ (i.e. what are its $\mathbb{E}X^k$ moments). For log-normal distribution we have

$\mathbb{E}X^k=e^{k\mu+k^2\sigma^2/2}$.

Thus

$\mathbb{E}Y^n=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k e^{k\mu+k^2\sigma^2/2}$.

I don't immediately see whether this has a closed-form (someone might supplement this).

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  • $\begingroup$ Thank you Tamas, You made me happy! What if n is being replaced with real number t ? $\endgroup$ – Min Pien Jun 10 '15 at 8:54
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    $\begingroup$ @MinPien: I don't think you can bypass integration in this case. Realizing simplifications, such as the very good idea suggested by Glen_b, may be very helpful. $\endgroup$ – Tamas Ferenci Jun 12 '15 at 13:24

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