Find the MLE of $p_{1}-p_{2}$ from binomial distributions

Question

I've been struggling with this exercise for a while, I would appreciate any hint or help :). I'm given two random samples from binomial distributions $X_{1},...,X_{n}\sim bin(n,p_{1})$ and $Y_{1},...,Y_{m}\sim bin(m,p_{2})$ and I'm asked to find the MLE of $p_{1}-p_{2}$.
I'm able to find the MLE for $p_{1}$ or $p_{2}$ but I don't know how to do it for the difference.

Answer 1

Let $X$ and $Y$ be independent random vectors with parametric densities $x\mapsto f_{X}(x;p_{X})$ and $y\mapsto f_{Y}(y;p_{Y})$. Here, I am using $p$ to denote the parameters.

The MLE is \begin{align*} \hat{p}\equiv(\hat{p}_{X},\hat{p}_{Y}) & =\operatorname{argmax}_{p_{X},p_{Y}}f_{X}(x;p_{X})f_{Y}(y;p_{Y})\\ & =\operatorname{argmax}_{p_{X},p_{Y}}\log f_{X}(x;p_{X})+\log f_{Y}(y;p_{Y})\\ & =(\operatorname{argmax}_{p_{X}}\log f_{X}(x;p_{X}),\operatorname{argmax}_{p_{Y}}\log f_{Y}(y;p_{Y}))\\ & =(\operatorname{argmax}_{p_{X}}f_{X}(x;p_{X}),\operatorname{argmax}_{p_{Y}}f_{Y}(y;p_{Y})). \end{align*} That is, the MLE of $p$ is obtained by computing the MLE of $p_X$ and $p_Y$ separately. In your example, $p_X$ and $p_Y$ are the probability of success in two independent Binomial distribution.

By the equivariance of MLEs, the MLE of $g(p)$ is $g(\hat{p})$. In your example, $g(p)=p_{X}-p_{Y}$, and hence the MLE of $g(p)$ is $g(\hat{p})=\hat{p}_{X}-\hat{p}_{Y}$.

Can you finish?