Convolution of two binomial distribution The problem and some of my thought are as followed, could you help check if I'm wrong.
Suppose $X∼Bin(n_1,1/2)$ and $Z∼Bin(n_2,p)$, $0<p<1$ being an unknown parameter; $X$ and $Z$ are assumed to be independent. Due to (spatial) aggregation, we can only observe $Y=X+Z$.
Is there always an MLE of $p$?
With the replicate of $Y$, and by the convolution formula of pdf,
$f_Y(y)=\sum_{x=0}^{n_1+n_2}f_X(x)f_Z(y-x)=\sum_{x=0}^{n_1}C_{n_1}^x(1/2)^{n_1}\times C_{n_2}^x p^x(1-p)^{n_2-x}$ something like this.
And then, we construct the likelihood function, find its argmax solve for p. I think it is kind of obvious, only the calculation is annoying.
But what does the question really mean? Is ther any case that a MLE will not exist?
And are there consistent estimators of p based on Y alone?
 A: Let's look at the likelihood function for small values of $n_1$ and $n_2$ (see the R code at the end):

This isn't a mathematical proof, but from these graphs we can fairly confidently conjecture that there is always a unique MLE except when 


*

*$n_2=0$, or 

*$n_2=1$ and $n_1$ is odd with $y=(n_1+1)/2$.


As for whether there is a consistent estimator of $p$ based on $Y$, the answer will depend on the asymptotics that you assume for $n_1$ and $n_2$. It should be intuitively obvious that if $n_2$ does not grow fast enough, relative to $n_1$, then the "noise" from $X$ will drown out the "signal" in $Z$, making consistent estimation impossible. But as long as $n_2$ does grow fast enough, relative to $n_1$, then consistent estimation will be possible.
To see how this works, instead of the MLE we can first look at a simpler, method-of-moments estimator. We have
$$E(Y) = E(X) + E(Z) = \frac{n_1}{2} + n_2p$$
So if we set
$$ \hat p = \frac{Y - \frac{n_1}2}{n_2}$$
then $\hat p$ is an unbiased estimator of $p$ (assuming $n_2>0$). We can then compute the variance
$$ \text{Var}(\hat p) = \frac{n_1}{4n_2^2} + \frac{p(1-p)}{n_2}$$
The variance then converges to zero as long as $n_2\to\infty$ and $n_1/n_2^2 \to 0$. In other words, $n_1$ must grow less quickly than $n_2^2$. Under these conditions, the method-of-moments estimator $\hat p$ is a consistent estimator of $p$, and so the MLE (being asymptotically optimal) must also be consistent under these same conditions.
R code for generating the chart:
library(tidyverse)

like_fun = Vectorize(function(n1, n2, y, p){
  z = 0:min(y,n2);
  sum(choose(n2,z)*p^z*(1-p)^(n2-z)*choose(n1,y-z)*2^(-n1))
})

df = expand.grid(n1 = 0:4, n2 = 0:4, y = 0:8, p = seq(0, 1, .01)) %>% 
  filter(y <= n1 + n2) %>%
  mutate(L = like_fun(n1, n2, y, p),
         n1 = paste("n1 = ", n1),
         n2 = paste("n2 = ", n2))

df %>% 
  ggplot(aes(x = p, y = L, color = factor(y))) +
  facet_grid(n2 ~ n1) +
  geom_line(aes(group = y)) +
  labs(x = 'p',
       y = 'Likelihood',
       color = 'y') +
  theme(axis.text.x = element_text(angle = 90))

