# Mixture of power-law distribution

I have two large sets A and B of (integer) numbers, both obtained with two different (and unknown) probability distributions $\rho_A$ and $\rho_B$. A (also large) set C contains a proportion $p$ of elements sampled from $\rho_A$ and a proportion $1-p$ sampled from $\rho_B$, hence $\rho_C = p\ \rho_A + (1-p)\ \rho_B$.

I have access to the three sets A, B and C and I want to estimate $p$.

The catch is, the distributions $\rho_A$ and $\rho_B$ seem to have a power-law tail. So normalization is impossible, I can't compute the mean... I don't know the exact form of the distributions either.

How can I extract the value of the mixture coefficient $p$ from my three populations $A$, $B$ and $C$ ?

• I work with a similar problem. Can the integers be thought of as ratios or as something similar to a count of times x exceeds y? I ask because this may give you a clue as to the cause of the underlying distribution. – Dave Harris Dec 28 '17 at 11:16
• @DaveHarris It's not the case, they correspond to the number of clones in a cell population to be precise – Jeannette Dec 28 '17 at 11:39
• You definitely have a power law. I will post an answer later. I am in a restricted location and have limited access. – Dave Harris Dec 28 '17 at 13:01
• Sorry, still no computer. Does the amount of food impact the mutation rate? Also, if you thought of the count of mutations as $x_{t+1}=\beta{x_t}+\epsilon_{t+1}$, even though $x_{t+1}$ and $x_t$ are not independent, are the errors independent? – Dave Harris Dec 29 '17 at 2:09
• Your question is unclear: if the two distributions $\rho_A$ and $\rho_B$ are completely determined, the coefficient $p$ can be estimated by an EM algorithm. If the two distributions $\rho_A$ and $\rho_B$ are themselves parameterised, a modification of EM will work. – Xi'an Dec 29 '17 at 9:09