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Let $X_1, \cdots X_n \stackrel{\text{iid}}{\sim} N(\alpha \sigma, \sigma^2)$, where $\alpha$ is known, and $\sigma > 0$ is unknown. Show that the family of distributions of $$T(\mathbf{X})=(\sum X_i, \sum X_i^2)$$ is not complete.

My work:

I am getting that this family is complete with the following work.

\begin{align*}E_\sigma[g(T(X))]&=\int_{-\infty}^\infty g(T(x))\frac{1}{\sqrt{2 \pi}\sigma}\exp(\frac{-1}{2\sigma^2}(x-\alpha \sigma)^2)dx\\ &=\frac{1}{\sqrt{2\pi}\sigma}\exp(\frac{-\alpha^2}{2})\int_{-\infty}^\infty g(T(x))\exp(\frac{-x^2}{2\sigma^2} + \frac{x\alpha}{\sigma})dx\end{align*}

For this to be $0$, $\int_{-\infty}^\infty g(T(x))dx=0$, since the exponential terms can never be equal to 0. Does this imply that the family of distributions of $T(X_1,\cdots,X_n)$ is complete?

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    $\begingroup$ What enabled you to replace "$g(T)$" by "$g(X)$" in the first equation?? Indeed, given that the distribution of $X$ is $n$-variate, how are we to make any sense of "$\mathrm{d}x$"? $\endgroup$
    – whuber
    Feb 6, 2020 at 4:40
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    $\begingroup$ $T(X)$ explicitly is two dimensional: it has two components. $\endgroup$
    – whuber
    Feb 6, 2020 at 4:52
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    $\begingroup$ When thinking about this, I found myself looking for simple functions of the components of $T$ that had easy-to-compute expectations. Assuming $\alpha$ known, I was able to find distinct functions of them that had the same expectations no matter what value $\sigma$ might have; and thereby could construct a nontrivial $g$ with zero expectation for all $\sigma.$ $\endgroup$
    – whuber
    Feb 6, 2020 at 5:00
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    $\begingroup$ I'm afraid both expectations rely on $\sigma:$ it's right there in the formulas. $\endgroup$
    – whuber
    Feb 6, 2020 at 5:13
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    $\begingroup$ Ansqered at stats.stackexchange.com/q/353431/119261 $\endgroup$ Feb 6, 2020 at 6:19

1 Answer 1

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The argument is incorrect: it is not because $$\int_{-\infty}^\infty g(T(x))\exp(\frac{-x^2}{2\sigma^2} + \frac{x\alpha}{\sigma})\text{d}x=0$$that $g\circ T$ is necessarily zero. (The argument does not even use the specific functional form of $T$.) Furthermore, as pointed out by @whuber, the integral in your approach should be on $\mathbb R^n$ rather than $\mathbb R$.

As suggested by @whuber, the standard line of attack is to find a function of $T(X)$ that is independent from $\sigma$. What could help in this regard is to rewrite the observations as $X_i\sim\sigma Y_i$, where $Y_i\sim N(\alpha,1)$, and to notice that $$T(X)\sim(\sigma\sum_i Y_i,\sigma^2\sum_i Y_i^2)$$ to guess a transform of $T(X)$ that does not depend on $\sigma$. (Hint: $\sigma^2=(\sigma)^2$.)

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  • $\begingroup$ I must admit that I do not know how to leverage your hint in finding a transformation of $T(X)$ that does not rely on $\sigma$. I feel like $\frac{T(X)}{\sigma}$ is not correct. Can you not just divide both components by $\sigma$ or $\sigma^2$? $\endgroup$
    – Ron Snow
    Feb 7, 2020 at 1:52
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    $\begingroup$ The transform of $T(X)$ cannot depend on $\sigma$ because this is not longer a statistic. Have a further look at your course notes and textbook to get a better grasp of the notions of statitics, sufficient statistics, and complete statistics. Check the examples provided in class. $\endgroup$
    – Xi'an
    Feb 7, 2020 at 6:55
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    $\begingroup$ Stronger hint: what ratio involving the components of $T$ will cancel out the powers of $\sigma$? (There are many answers; choose a simple one.) $\endgroup$
    – whuber
    Feb 7, 2020 at 18:04
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    $\begingroup$ It's the right idea, but are you sure about your algebra? $\endgroup$
    – whuber
    Feb 12, 2020 at 3:42
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    $\begingroup$ The point is to be independent from $\sigma$ not to compute the value. $\endgroup$
    – Xi'an
    Feb 13, 2020 at 5:34

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