How do you calculate the expected value of mixed lognormal distribution? Suppose $X=\log(Y)$ can be modeled by a mixture of two normal distributions with proportion $p$ of $X_1$ and proportion $1-p$ of $X_2$, where $X_1\sim\mathcal N(U_1, \sigma^2_1)$ and $X_2\sim\mathcal N(U_2, \sigma^2_2)$.
How do you calculate $E(Y)$; i.e., $E(\exp(X))$ where $X$ is a mixture of two normals?
 A: I'll try to give an answer for the mixture case. Let's formalize the set-up. We consider a random variable $X$ and an indicator random variable $I$, with $P[I=1] = 1-P[I=2] = p$, independent of $X$. Furthermore, for the mixture we have that the law of $X$ given that $I=1$ is the law of $X_1$, which is Gaussian with mean $U_1$ and variance $\sigma^2_1$; and, if $I=2$, the law is that of $X_2$ with law $N(U_2,\sigma^2_2)$.
Then, for $Y=\exp(X)$ we can calculate the expectation as
$$
\begin{align}
E[Y] &= E[\exp(X)] = p E[\exp(X)|I=1] + (1-p) E[\exp(X)|I=2] \\
     &= p E[\exp(X_1)] + (1-p) E[\exp(X_2)] \\
     &= p \exp(U_1+ \sigma^2_1/2) + (1-p) \exp(U_2+ \sigma^2_2/2) 
\end{align}
$$
by using the expectation of a log-normal.
A: For instance, if you are dealing with a two component mixture, the first moment is calculated as follows:
$\mu_{\rm mixture}=\pi_{1} \cdot \mu_{1} +  \pi_{2} \cdot \mu_{2}$,
where $\pi_{i}\ (i=1,2)$ stands for the components' weights and $\mu_{i}\ (i=1,2)$ for the means. 
