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

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


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

for all $p$.

  • $\begingroup$ $P(Y=y, Z=z) = \binom{n}{y}p^y(1-p)^{n-y}\binom{n}{z}p^{2z}(1-p^2)^{n-z}$ $\endgroup$
    – Onyambu
    Commented Nov 20, 2020 at 20:04
  • $\begingroup$ I'm considering the case where we don't just have a single observation. $\endgroup$
    – Remy
    Commented Nov 20, 2020 at 20:09
  • 2
    $\begingroup$ Perhaps it would help to simplify the likelihood by expressing $$\log\left(\frac{p^2}{1-p^2}\right)=\log\left(\frac{p}{1-p}\right) + \log\left(\frac{p}{1+p}\right).$$ That makes it relatively easy to demonstrate minimal sufficiency. $\endgroup$
    – whuber
    Commented Nov 20, 2020 at 20:10
  • $\begingroup$ Can't believe I didn't recognize that! I'll try again and perhaps answer my own question. $\endgroup$
    – Remy
    Commented Nov 20, 2020 at 20:14
  • $\begingroup$ I have encountered the same issue and I'm not sure if it can be simplified further. $\endgroup$
    – Remy
    Commented Nov 20, 2020 at 20:36

1 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.

  • $\begingroup$ The definition I'm seeing of a minimal exponential family is as follows: "an exponential family is referred to as minimal if there are no linear constraints among the components of the parameter vector nor are there linear constraints among the components of the sufficient statistic." Does that mean I should use $\sum y_i$ and $\sum z_i$ as the sufficient statistics rather than $\sum y_i+z_i$ and $\sum z_i$ since the latter sufficient statistics depend on one another? In that case $\log\left(\frac{p}{1-p}\right)$ and $\log\left(\frac{p^2}{1-p^2}\right)$ would be the natural parameters. $\endgroup$
    – Remy
    Commented Nov 23, 2020 at 19:43
  • $\begingroup$ A bijective transform of a (minimal) sufficient statistic is a (minimal) sufficient statistic. $\endgroup$
    – Xi'an
    Commented Nov 23, 2020 at 20:15
  • $\begingroup$ @Xi'an For the second result, does the dimension of the exponential family have to match the number of statistics we have? Or is it sufficient that the parameter spaces has non-empty interior? $\endgroup$
    – user82261
    Commented Jan 9, 2021 at 1:00

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