Show minimal sufficient statistic is not complete in normal distribution

Let $$Z_i$$ for $$1 \leq i \leq n$$ be a sample from the $$N(ap, bp(1-p))$$ density, where $$a \gt 0, b \gt 0$$ are known but $$p \in (0,1)$$ is an unknown parameter.

I have shown that $$T = (\sum^n_{i = 1} Z_i, \sum^n_{i = 1} Z^2_i)$$ is a two-dimensional minimal sufficient statistic for the unknown parameter $$p$$. I want to show that $$T$$ is not complete.

We have

$$E_p[\sum^n_{i = 1} Z_i] = nap$$,

$$E_p[\sum^n_{i = 1} Z^2_i] = nbp(1-p) + na^2p^2$$,

$$E_p[(\sum^n_{i = 1} Z_i)^{2}] = nbp(1-p) + n^2a^2p^2$$.

I have tried to find with no success a function $$g(T)$$ such that $$E_p[g(T)] = 0$$ for all $$p$$ but $$P_p(g(T) = 0) \neq 1$$.

Any hints will be appreciated.

• Welcome to cv, Oscar24680 :-). In your problem, you are dealing with a subfamily of normal distributions with only one parameter $p$ instead of the usual two parameters $\mu,\sigma^2$. On Wikipedia's page about completeness you can find a related example - maybe you can take that as inspiration.
– Ute
Commented Jul 27, 2023 at 10:39
• Another hint: also use the constants $a$ and $b$ when you build your function $g(T)$ as a function of the components $T_1=\sum_i Z_i$ and $T_2=\sum_i Z_i^2$.
– Ute
Commented Jul 27, 2023 at 11:17
• Hint 3: Consider $p(1-p) = p - p^2$.
– Ute
Commented Jul 27, 2023 at 11:20
• @Ute thanks for the hints. If we let $c_1 = na^2 - nb$ and $c_2 = n^2a^2 - nb$ then $g(T) = -(\frac{1}{c_1} - \frac{1}{c_2})\frac{b}{a}T_1 + \frac{1}{c_1}T_2 - \frac{1}{c_1}T^2_1$ works. Commented Jul 27, 2023 at 12:11
• Great! change the factor of $T_1^2$ to $\frac{1}{c_2}$, and we have equivalent solutions; I find $g(T)=(n-1)ab T_1 -c_2T_2 + c_1 T_1^2$ (you can double check - maybe I'm wrong). If you want, you can write your own answer (worked out a bit) and accept it. Could be that this gives you some weird badge here on cv ;-)
– Ute
Commented Jul 27, 2023 at 14:09

Thanks to the hints of @Ute we can observe that:

$$E_p[\sum^n_i Z_i] = nap$$

$$E_p[\sum^n_i Z^2_i] = nbp + (na^2 - nb)p^2$$

$$E_p[(\sum^n_i Z_i)^2] = nbp + (n^2a^2 - nb)p^2$$

Let $$c_1 = na^2 - nb$$, $$c_2 = n^2a^2 - nb$$. Then:

$$\frac{1}{c_1}E_p[\sum^n_i Z^2_i] - \frac{1}{c_2}E_p[(\sum^n_i Z_i)^2] = (\frac{1}{c_1} - \frac{1}{c_2})nbp$$.

Note that $$-(\frac{1}{c_1} - \frac{1}{c_2})\frac{b}{a}E_p[\sum^n_i Z_i] = -(\frac{1}{c_1} - \frac{1}{c_2})nbp$$.

Let $$T = (T_1, T_2) = (\sum^n_i Z_i, \sum^n_i Z^2_i)$$.

Let $$g(T) = g(T_1, T_2) = -(\frac{1}{c_1} - \frac{1}{c_2})\frac{b}{a}T_1 + \frac{1}{c_1}T_2 - \frac{1}{c_2}T^2_1$$. This function (and any non-zero scalar multiple of it) will work. Indeed,

$$E_p(g(T)) = E_p(-(\frac{1}{c_1} - \frac{1}{c_2})\frac{b}{a}T_1 + \frac{1}{c_1}T_2 - \frac{1}{c_2}T^2_1) = -(\frac{1}{c_1} - \frac{1}{c_2})\frac{b}{a}E_p(T_1) + \frac{1}{c_1}E_p(T_2) - \frac{1}{c_2}E_p(T^2_1) = 0, \forall p\in(0, 1).$$

However, $$g(T)$$ is not identically zero with probability one. So the minimal sufficient statistic $$T$$ is not complete.

• Hi Oscar24680, these are all correct calculations, great :-). You need to explain $T_1$ and $T_2$ - you and I understand it, but otherwise only people who have read the comments, and comments get deleted after a while. The you can still turn this in an even better answer, by making it super useful for a new user who has the same question comes by. They will see the solution, but would they understand why, and what is going on?
– Ute
Commented Jul 27, 2023 at 19:04
• Hi @Ute, I just edited the answer. Do you think it's better now? Thanks Commented Jul 27, 2023 at 19:12
• Yes, $T_1$ and $T_2$ are explained. But what is the idea that you had and that helped, the principle that you applied, so to say. I know solutions in books or lectures often come out of the blue, and you had already stated some of the things in your question. But in fact, you need to find g such that $g(T_1, T_2)$ has mean 0, but is not trivially = 0, that is there is a positive probability that $g(T) >0$. Then trick 1 is to search among linear combinations of $T_1$, $T_2$ or any functions of these for which you can calculate the mean. Because then you can apply linearity of the mean.
– Ute
Commented Jul 27, 2023 at 19:21
• That is how you started (in your question already). You took these three candidates $T_1, T_2, T_1^2$ and establish the connection to the unknown parameter $p$. It turns out that the expressions of the means of these three statistics include $p$ and $p^2$. For $E_p g(T)=0$ to hold for all $p$, you then need to make sure to find coefficients of $T_1, T_2, T_1^2$ such that the coefficients of $p$ and the coefficients of $p^2$ both add up to 0, / you solved a system of linear equations / whatever the method you used.
– Ute
Commented Jul 27, 2023 at 19:25
• :-) I'll vote up. Btw, can you accept it? Or is there no checkmark, when you write your own answer?
– Ute
Commented Jul 27, 2023 at 19:55