Skewness of mixture density

I have the mixture density of two normal distributions:

\begin{align} f(l)=\pi \phi(l;\mu_1,\sigma^2_1)+(1-\pi)\phi(l;\mu_2,\sigma^2_2) \end{align}

The skewness is given by

\begin{align} \gamma_1 = \operatorname{E}\Big[\big(\tfrac{X-\mu}{\sigma}\big)^{\!3}\, \Big] = \frac{\mu_3}{\sigma^3} = \frac{\operatorname{E}\big[(X-\mu)^3\big]}{\ \ \ ( \operatorname{E}\big[ (X-\mu)^2 \big] )^{3/2}} \end{align}

Now my question is, what is the Skewness of a mixture gaussian? How can I derive it? I searched for it, but I could not find anything. A mathematical derivation would be helpful, also I have estimated both densities and I have the estimates for $\mu$ and $\sigma$. I want a formula in what I can insert those values to get the skewness of the mixed density.

Well, may be I didn't really get the problem, but you wrote: "...A mathematical derivation would be helpful, also I have estimated both densities and I have the estimates for $\mu$ and $\sigma$...". Isn't that all you need to compute the skewness. You have the pdf of a mixture. Assuming that you know all the parameter, then just treat it like any other pdf (for example single Gaussian pdf). So in order to find skewness we need to compute first, second and third moments of a mixture. For constant (non-random) weights $\pi, (1-\pi)$, the pdf of the mixture is: $$f(x)=\sum_{i=1}^2\pi_i f_i(x),$$ where $f_i(x)=f(x|\mu_i,\sigma_i)$ So it follows immediately for any moment k:

$$\mu^{(k)} = \mathbb{E}_{f}[x^k] = \sum_i^2{\pi_i \mathbb{E}_{f_i}[x^k]} = \sum_i^2{\pi_i \mu_i^{(k)}}.$$

$\mu^{(k)}$ is the $k$-th moment of $f$ and $\mu_i^{(k)}$ is the $k$-th moment of $f_i$.

CORRECTION.
But we know all moments of the gaussian pdf. Then $$\operatorname{E}_{f}\big[(X-\mu)^3\big]=\operatorname{E}_{f}\big[(X-\operatorname{E}_{f}\big[x\big])^3\big]=\operatorname{E}_{f}\big[x^3\big]-3\operatorname{E}_{f}\big[x\big]\operatorname{E}_{f}\big[x^2\big]+3\operatorname{E}_{f}\big[x\big]^2\operatorname{E}_{f}\big[x \big]-\operatorname{E}_{f}\big[x\big]^3$$ $$\operatorname{E}_{f}\big[(X-\operatorname{E}_{f}\big[x\big])^3\big]=\operatorname{E}_{f}\big[x^3\big]-3\operatorname{E}_{f}\big[x\big]\operatorname{E}_{f}\big[x^2\big]+2\operatorname{E}_{f}\big[x\big]^3=\mu^{(3)}-3\mu^{(1)}\mu^{(2)}+2(\mu^{(1)})^3$$ $$\operatorname{E}_{f}\big[(X-\operatorname{E}_{f}\big[x\big])^3\big]=\sum_i^2{\pi_i \mu_i^{(3)}}-3\sum_i^2{\pi_i \mu_i^{(1)}}\sum_i^2{\pi_i \mu_i^{(2)}}+2\bigg(\sum_i^2{\pi_i \mu_i^{(1)}}\bigg)^3$$ The moments of the gaussian pdf are defined as follows: $$\mu_i^{(1)}=\mu_i$$ $$\mu_i^{(2)}=\mu_i^2+\sigma_i^2$$ $$\mu_i^{(3)}=\mu_i^3+3\mu_i\sigma_i^2$$ So after substituting them and simplifying the equation one will get: $$\operatorname{E}_{f}\big[(X-\operatorname{E}_{f}\big[x\big])^3\big]=(1-\pi ) \pi \left(\mu _1-\mu _2\right) \left((1-2 \pi ) \left(\mu _1-\mu _2\right){}^2+3 \left(\sigma _1^2-\sigma _2^2\right)\right)$$ That was the numerator of your skewness. Working with the denominator in similar way one will get: $$\operatorname{E}\big[ (X-\mu)^2 \big]=\pi \left((1-\pi ) \left(\mu _1-\mu _2\right){}^2+\sigma _1^2-\sigma _2^2\right)+\sigma _2^2$$ As a result the skewness will look like: $$\gamma_1 = \frac{\operatorname{E}\big[(X-\mu)^3\big]}{\ \ \ ( \operatorname{E}\big[ (X-\mu)^2 \big] )^{3/2}} =\frac{(1-\pi ) \pi \left(\mu _1-\mu _2\right) \left((1-2 \pi ) \left(\mu _1-\mu _2\right){}^2+3 \left(\sigma _1^2-\sigma _2^2\right)\right)}{\left(\pi \left((1-\pi ) \left(\mu _1-\mu _2\right){}^2+\sigma _1^2-\sigma _2^2\right)+\sigma _2^2\right){}^{3/2}}.$$ Hope this will help.

• Unfortunately, this defines $\mu^{(k)}$ as a moment, not a central moment hence the sentence "But we know..." does not apply.
– Did
Apr 1 '13 at 18:08
• @Did Thanks, you were right. Hope now it is correct. Apr 3 '13 at 9:37
• @Caran-d'Ache thanks! Apr 4 '13 at 9:57