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Let $X_1.....X_n$ be independent Normal($\theta, \theta^2$) for $\theta >0$. Find a minimal sufficient statistic. Is it complete?

I have a minimal sufficient statistic, $T=(\sum_{1}^{n}X_i,\sum_{1}^{n}X_i^2$), but cannot show it is complete. I suspect I can find a counterexample to the completeness condition, but I have no intelligent strategy other than guessing random functions with expectation $0$. A sketch or even starting point would be helpful.

Edit: Duplicate of Not complete but minimal sufficient statistic

But I would still like some intuition as to how one would come up with such a function.

Y.D.X.
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Muselive
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  • Does this answer your question? Not Complete Statistic in Normal Distribution – NCh Feb 26 '20 at 04:48
  • Only in the strictest sense. It appears as if this function comes out of thin air. How does one guess this function in say, an exam setting?Edit; to be clear I mean the functional $2x_1^2 - (n-1)x_2$ – Muselive Feb 26 '20 at 04:51
  • You should suppose that this two-variate statistics is incomplete for one-variate parameter. So you need to find a nonzero function $g$ of two variables with zero expectation $\mathbb Eg(T)=0$ for all $\theta$. How to find it? Look at expected value of $\sum_{i=1}^n X_i$ (preferably the second moment since we need smth in terms of $\theta^2$, not $\theta$), at expected value of $\sum_{i=1}^n X_i^2$ and try to combine zero from it. Nothing comes out of thin air. – NCh Feb 26 '20 at 05:00
  • https://math.stackexchange.com/q/2936487/321264 – StubbornAtom May 14 '20 at 07:23

1 Answers1

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You should suppose that this two-dimensional statistics is incomplete for one-dimensional parameter. So you need to find a function $g$ of two variables with zero expectation $\mathbb Eg(T)=0$ for all $\theta$ and s.t. $\mathbb P(g(T)=0)\neq 1$. How to find it? Let us look at some moments of $\sum_{i=1}^n X_i$ and $\sum_{i=1}^n X_i^2$ and try to combine zero from it.

$$\mathbb E\left(\sum_{i=1}^n X_i\right)=n\theta$$ $$\mathbb E\left(\biggl(\sum_{i=1}^n X_i\biggr)^2\right)=\text{Var}\left(\sum_{i=1}^n X_i\right)+n^2\theta^2 = n\theta^2+n^2\theta^2=\theta^2n(n+1)\tag{1}$$ $$\mathbb E\left(\sum_{i=1}^n X_i^2\right)=2n\theta^2\tag{2}$$

It is easy to see that if we divide (1) by $n+1$ and subtract (2) divided by $2$ we obtain zero for any $\theta$. So the function $g(x,y)$ can be $\frac{x^2}{n+1}-\frac{y}2$, $$ g(T) = \frac{\bigl(\sum_{i=1}^n X_i\bigr)^2}{n+1} - \frac{\sum_{i=1}^n X_i^2}{2} $$ or smth without fractions, like as $g(x,y)=2x^2-(n+1)y$ etc.

NCh
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  • Here how you know that $\mathbb P(g(T)=0)\neq 1$? – Alexis Sandoval Feb 23 '21 at 07:22
  • @AlexisSandoval Calculate $g(T)$ and obtain some non-trivial quadratic function of independent normal r.v.'s. It has absolutely continuous distribution, so the probability of reach any value is zero. – NCh Feb 25 '21 at 03:06