Nonparametric mixture estimation

Question

Let's assume that we have two samples $\{X_i\}_{i=1..N}$ and $\{Y_i\}_{i=1..M}$ corresponding to random variables $X$ and $Y$. Let there also be a sample $\{Z_i\}_{i=1..K}$ corresponding to random variable $Z$. Assume that $Z$ is unknown mixture of $X$ and $Y$, i.e. $f_Z = \alpha f_X + (1-\alpha) f_Y$.

Is there any way to estimate the alpha coefficient without resorting to parametric hypotheses about $X$ and $Y$? Since we have their samples, it seems possible with the help of kernel density estimation.

P.S. Articles about the nonparametric EM method that I found immediately go into the wilds. But the current case seems to be simple.

Answer 1

Maybe I have missed some nuance of your Question; the answer seems almost too simple. Sorry if I'm wasting your time with this.

The brief Wikipedia article on mixture distributions has some relevant material.

Because $f_Z = \alpha f_X + (1-\alpha)f_Y$ you have $$E(Z) = \alpha E(X) + (1-\alpha) E(Y),$$ and then $\alpha = E(Z-Y)/E(X-Y).$

You can use means of the samples you have to estimate the required expectations, and thus $\alpha.$

Here is a brief simulation of 100,000 $Z_i$s that are a mixture with $\alpha = 1/3$ of standard normal and standard uniform. [See this page for more elegant and more general methods of simulation.]

m = 10^5
x = rnorm(m);  y = runif(m)
MAT = cbind(x,y)
id = sample(1:2, m, rep=T, p=c(1,2))
z = numeric(m)
for(i in 1:m) {
 z[i] = MAT[i,id[i]] }
mean(z)
[1] 0.3333354  # aprx 1/3 = 0(1/3) + (1/2)(2/3)
hist(z, prob=T, br=50, col="skyblue2")
 curve((1/3)*dnorm(x)+(2/3)*dunif(x), add=T, col="red", n=10001)

mean(z-y)/mean(x-y)
[1] 0.3314798  # aprx 1/3 = alpha