 Two-Gaussian mixture example

Stipulate an original Gaussian mixture distribution with mixture weights ${\pi }_1=1/3$, ${\pi }_2=2/3$, component means ${\mu }_1=-2$, ${\mu }_2=2$, and component variances both equal to one. These are chosen statistical parameters, with no observed sample or fitting. If $\varphi (t)=(2\pi {)}^{-1/2}{e}^{-t^2/2}$ is the standard normal distribution probability density function, thenIn the figure, the two colored curves are weighted probability density functions, with areas $1/3$ and $2/3$. Adding their heights at every $x$ gives the black curve. Neither colored curve stops at a dividing line.

The law of total expectation and law of total variance give the total mean and variance:All of the extra variance $32/9$ comes from variation between component means. Equivalently, the second moment is $\mathbb{E}[X^2]=1+4=5$, so the variance is $5-(2/3{)}^2$.

A mixture distribution is different from a deterministic linear combination of random variables. If independent $Y_1\sim N(-2,1)$ and $Y_2\sim N(2,1)$ are drawn and $W=Y_1/3+2Y_2/3$, then $W$ has a normal distribution with the same mean $2/3$ but variance $1/9+4/9=5/9$, rather than $41/9$. Every draw of $W$ uses both random variables. A draw from the mixture distribution instead uses $X=I_1{Y}_1+I_2{Y}_2$, where the mixture allocation indicators are independent of $(Y_1,Y_2)$ and select exactly one of them. The expression for $f$ adds densities, not simultaneous random outcomes.