Skewness of a sum of independent lognormal variables?

TheNewbie

New Member
Hey,

I have a set of lognormally distributed independent variables X1,...,Xn each with different E[Xk] and Var[Xk] for k=1,2...n. (So each Xk has its own my_k and sigma_k^2)

As far as I understand, if you calculate the sum of these variables, say:
S=Sum(Xk) , then this S will also be lognormally distributed with
my_s=sum(my_k) and sigma_s^2=sum(my_k), right?

Now for the real question, what about the calculation of the 3rd moment, the so called skewness, is this skewness also additive so that the skewness of S is the sum of all the skewnessses for the independent variables X1,...,Xn?

Dragan

Super Moderator
Hey,

.....Now for the real question, what about the calculation of the 3rd moment, the so called skewness, is this skewness also additive so that the skewness of S is the sum of all the skewnessses for the independent variables X1,...,Xn?
...
No, it is not additive....because of the Central Limit Theorem.

The distribution of the sum of independent variables will become more "normal" like....i.e. the skew will get closer to zero.

TheNewbie

New Member
Thanks! Ok, I was suspecting something like that...

...So, does anyone know how to proceed with the calculations? I can't find a decent source so a lil' push in the right direction would be much appreciated!

TheNewbie

New Member
Uh oh, Dason you're right. Oh crap, looks like this problem was more massive than I thought...

BGM

TS Contributor
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

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

TheNewbie

New Member
Holy mother of god, I probably can not thank you enough for taking the time to explain this. This was EXACTLY what I needed!

Dragan

Super Moderator
PS: In the nice case that all $$X_i$$ have the identical distribution,
you may further simplify the above expression into

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

BGM: You just jogged my memory on this topic. If I remember correctly, I think we can make a broader statement. That is, let $$\gamma _{k}$$ denote the k-th standardized cumulant. It turns out that:

$$\frac{\gamma _{3}}{n^{\frac{1}{2}}}$$; $$\frac{\gamma _{4}}{n^{2}}$$; $$\frac{\gamma _{5}}{n^{\frac{3}{2}}}$$; $$\frac{\gamma _{6}}{n^{2}}$$ …

where $$\gamma _{3}$$ is the usual measure of skew and $$\gamma _{4}$$ is kurtosis.