Bayesian inference on a sum of iid random variables with known distribution Let $X_1$, $X_2$, ..., $X_n$ be iid RV's following a mixture distribution of two lognormals such that the pdf of each $X_i$ is $f_{mix}(x)=pf_1(x) + (1-p)f_2(x)$ where $f_1(x)$ and $f_2(x)$ are lognormal pdfs with parameters $\mu_1,\sigma$ and $\mu_2,\sigma$, respectively.
Define $S_i$ as a sum of 10 $X$s, e.g. $S_1 = X_1 +..+ X_{10}, \ \ S_2 = X_{11} +..+X_{20},...$
I am given only $S_1,S_2,\dots,S_{n/10}$.
How can I infer the mixing proportion parameter $p$ here?
That means, I want to know what the proportion of the two lognormals in my mixture is among the 10 samples is (but I only have the sum measurement).
Note that the sum of lognormals is not lognormally distributed.
My problem could be related to this one, but I am not sure:
Bayesian inference on a sum of iid real-valued random variables
update:
assume that I have many samples of $S_i$. If that helps we can assume that the $\mu$s are well separated, making the mixture nicely bimodal.
Generally in mixtures it is that the mixing proportions sum to 1. 
This is also why I thought about the Dirichlet process. I have no restrictions on the values of $p$ otherwise.
 A: Each $X_i$ comes from either one of the two lognormals with probabilities $p$ and $1-p$.  Let $Z_j$ be the number of $X_i$'s in the sum $S_j=\sum_{i=10j-9}^{10j}X_i$ that comes from the first lognormal.  Clearly $Z_j \sim \mbox{bin}(10,p)$.  Conditional on $Z_j$, each $S_j$ is a sum of $Z_j$ lognormals with parameters $\mu_1,\sigma^2$ and $10-Z_j$ lognormals with parameters $\mu_2,\sigma^2$, with pdf 
$$
f_{S|Z=z}(s;\mu_1,\mu_2,\sigma^2), \tag{1}
$$
given by a complicated convolution integral.  The unconditional distribution of each $S_j$ is the 11-component mixture
$$
f_S(s;\mu_1,\mu_2,\sigma^2,p)=\sum_{z=0}^{10}{10 \choose z}p^z(1-p)^{10-z}f_{S|Z=z}(s;\mu_1,\mu_2,\sigma^2). \tag{2}
$$
If $\sigma^2$ is sufficiently small or $\mu_1$ and $\mu_2$ sufficiently different this is going to be a 11-modal distribution with modes located near $10e^{\mu_1+\sigma^2},9e^{\mu_1+\sigma^2}+e^{\mu_2+\sigma^2},\dots,10e^{\mu_2+\sigma^2}$.  This suggest that all five parameters in principle are identifiable given enough data.
Perhaps you can approximate the convolution in (1) by a single moment-matched lognormal as discussed here, use (2) to compute the likelihood and then compute approximate maximum likelihood estimates by maximising the resulting log likelihood numerically.  Or you could do approximate Bayesian inference using this approximate likelihood function.  This option would allow using informative priors on some of the parameters which might be necessary in practice if there is too much overlap between each component of (2).
A: A simulation based answer is to treat the $X_i$'s as latent variables and include these in a global MCMC sampler. At iteration $t$, it could proceed as follows


*

*Simulate the $X_i^t$'s given the $S_j$'s and the current value of the parameters ${\theta}^t$

*Given the $X_i^t$'s and their log-Normal mixture distribution, simulate the next value of the parameters, $\theta^{t+1}$


Step 2 is straightforward in that this is equivalent to simulate the posterior of a Normal sample. Step 1 can be decomposed in a sequence of Gibbs steps where each $X_i$ is generated conditional on the $8$ other $X_k$'s and the corresponding $S_j$. Meaning any MCMC move targeting the distribution
$$\{pf_1(x_i;\mu_1,\sigma_1)+(1-p)f_2(x_i;\mu_2,\sigma_2)\}\times
\{pf_1(s_j-x_{10(j-1)+1}-\ldots-x_{10j};\mu_1,\sigma_1)+(1-p)f_2(s_j-x_{10(j-1)+1}-\ldots-x_{10j};\mu_2,\sigma_2)\}$$
A: The parameter $p$ is not identifiable. This can be seen by writing
$f_{mix}=p\cdot \mathcal{LN}(\mu_1,\sigma) + (1-p)\cdot \mathcal{LN}(\mu_2,\sigma)$ 
as 
$f_{mix}=\mathcal{LN}(\nu_1,\sigma) +\mathcal{LN}(\nu_2,\sigma)$ 
with $\nu_1=\log p + \mu_1$ and $\nu_2=\log (1-p) + \mu_2$
Then we can always choose $\mu_1=\nu_1-\log p$ and $\mu_2=\nu_2-\log (1-p)$. Then 
$$p(f_{mix}|\mu_1,\mu_2,\sigma,p)=p(f_{mix}|\nu_1,\nu_2,\sigma,p)=p(f_{mix}|\nu_1,\nu_2,\sigma)$$
 and hence 
$$p(S_n|\nu_1,\nu_2,\sigma,p)=p(S_n|\nu_1,\nu_2,\sigma)$$
In other words the data likelihood is independent of $p$ given $\mu_1$ and $\mu_2$.
