If you assume that

\( X_i \sim \mbox{Log-}\mathcal{N}(\mu_i, \sigma^2_i), i = 1, 2, ..., n \)

and they are mutually independent,

According to wikipedia

http://en.wikipedia.org/wiki/Log-normal_distribution
http://en.wikipedia.org/wiki/Skewness
The definition of the skewness

\( Skew[X_i] = E\left[\left(\frac {X_i - E[X_i]} {SD[X_i]}\right)^3\right]

= (e^{\sigma_i^2} + 2)\sqrt{e^{\sigma_i^2} - 1}\)

for log-normally distributed random variables.

In general, you may expand the skewness of the sum of independent random

variables like this: Consider

\( E\left[\left(\sum_{i=1}^n X_i

- E\left[\sum_{i=1}^nX_i\right]\right)^3\right] \)

\( = E\left[\left(\sum_{i=1}^n (X_i - E[X_i])\right)^3\right] \)

\( = \sum_{i=1}^n E[(X_i - E[X_i])^3] \)

\(

+ 3\sum_{i\neq j}^n E[(X_i - E[X_i])^2(X_j - E[X_j])]

+ 6\sum_{i\neq j\neq k}^n E[(X_i - E[X_i])(X_j - E[X_j])(X_k - E[X_k])]\)

By independence, the terms in the second line all vanish.

Thus,

\( Skew\left[\sum_{i=1}^n X_i\right] =

\frac {\sum_{i=1}^n E[(X_i - E[X_i])^3]}

{(\sum_{i=1}^n Var[X_i])^{\frac {3} {2}} }

= \frac {\sum_{i=1}^n Var[X_i]^{\frac {3} {2}}Skew[X_i]}

{(\sum_{i=1}^n Var[X_i])^{\frac {3} {2}} }\)

Also note

\( Var[X_i] = (e^{\sigma_i^2} - 1)e^{2\mu_i + \sigma_i^2} \)

Now you can just plug in the formula.

PS: In the nice case that all \( X_i \) have the identical distribution,

you may further simplify the above expression into

\( Skew\left[\sum_{i=1}^n X_i\right] = \frac {Skew[X_1]} {\sqrt{n}}\)