Minimal Sufficient Statistic for Bivariate Binomial

Question

Find a minimal sufficient statistic for $p$ where $Y\sim\mathsf{Binom}(n,p)$ and $Z\sim\mathsf{Binom}\left(n,p^2\right)$ are independent random variables. Determine if this statistic is complete. If it's not, find a counterexample.

My try:

We have that $\mathbf T=(\sum Y_i+\sum Z_i,\sum Z_i)$ is sufficient for $p$ since

$$\begin{align*} \mathbb P(\mathbf Y=\mathbf y, \mathbf Z=\mathbf z) &\overset{\text{ind}}{=}\mathbb P(\mathbf Y=\mathbf y)\mathbb P(\mathbf Z=\mathbf z)\\\\ &=\prod_{i=1}^n {n\choose y_i}p^{y_i}(1-p)^{n-y_i}{n\choose z_i}p^{2z_i}(1-p^2)^{n-z_i}\\\\ &=\left[\prod_{i=1}^n {n\choose y_i}{n\choose z_i}\right]p^{\sum y_i}(1-p)^{n^2-\sum y_i}p^{2\sum z_i}(1-p^2)^{n^2-\sum z_i}\\\\ &=\left[\prod_{i=1}^n {n\choose y_i}{n\choose z_i}\right]\exp\left[\sum y_i\log\left(\frac{p}{1-p}\right)+\sum z_i\log\left(\frac{p^2}{1-p^2}\right)+B(p)\right]\\\\ &=\left[\prod_{i=1}^n {n\choose y_i}{n\choose z_i}\right]\exp\left[\sum y_i\log\left(\frac{p}{1-p}\right)+\sum z_i\left[\log\left(\frac{p}{1-p}\right)+\log\left(\frac{p}{1+p}\right)\right]+B(p)\right]\\\\ &=\left[\prod_{i=1}^n {n\choose y_i}{n\choose z_i}\right]\exp\left[\left(\sum y_i+z_i\right)\log\left(\frac{p}{1-p}\right)+\sum z_i\log\left(\frac{p}{1+p}\right)+ B(p)\right] \end{align*}$$

However this is not of full rank since $\log\left(\frac{p}{1-p}\right)$ and $\log\left(\frac{p}{1+p}\right)$ depend on one another so we cannot immediately conclude that $\mathbf T$ is minimal sufficient. I next considered the "ratio method": For any two possible sample points $(\mathbf{y}^{(1)},\mathbf z^{(1)})$ and $(\mathbf y^{(2)},\mathbf z^{(2)})$ we have

$$\frac{f(\mathbf y^{(1)},\mathbf z^{(1)}\mid p)}{f(\mathbf y^{(2)},\mathbf z^{(2)}\mid p)}\underset{p}{\propto}\exp\left[\left(\sum y^{(1)}_i+z^{(1)}_i-\sum y^{(2)}_i+z^{(2)}_i\right)\log\left(\frac{p}{1-p}\right)+\left(\sum z^{(1)}_i-\sum z^{(2)}_i\right)\log\left(\frac{p}{1+p}\right)\right]$$

but it's not clear to me that this ratio doesn't depend on $p$ if and only if $\mathbf T\left(\mathbf{y}^{(1)},\mathbf z^{(1)}\right)=\mathbf T\left(\mathbf{y}^{(2)},\mathbf z^{(2)}\right)$. Assuming what I have done thus far is correct, how can I proceed to conclude that $\mathbf T$ is minimal sufficient?

As for deciding whether or not $\mathbf T$ is complete, perhaps the following counterexample is viable? Let $g(T)=\mathbf T_1-\mathbf T_2-n^2p$ so that

$$\mathbb E_p(g(\mathbf T))=\mathbb E_p\left(\mathbf T_1-\mathbf T_2-n^2p\right)=0$$

but

$$\mathbb P_p(g(\mathbf T)=0)=\mathbb P_p\left(\mathbf T_1-\mathbf T_2-n^2p=0\right)\neq 1$$

for all $p$.

Answer 1

Quoting from Lester Mackey's notes [see Corollary 6.16 in Lehmann and Casella, 1999, for more details):

For any minimal $s$-dimensional exponential family the minimal natural statistic$$\left(\sum_iT_1(X_i),...,\sum_iT_s(X_i)\right)$$ is a minimal sufficient statistic.

and from Chuan Goh's (this is also Theorem 6.22 in Lehmann and Casella, 1999, for more details):

For any full rank exponential family with minimal sufficient statistic $$\left(\sum_iT_1(X_i),...,\sum_iT_s(X_i)\right)$$ this statistic is complete.