1

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?

Xi'an
  • 90,397
  • 9
  • 157
  • 575
Jen Snow
  • 1,595
  • 2
  • 18
  • 2
    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 "$\mathrm{d}x$"? – whuber Feb 06 '20 at 04:40
  • @whuber I must admit that this notation is quite confusing to me. Also, I do not know how to make sense of your second sentence. Do you mean that this approach, when integrating over $x$ is not appropriate since $T(\mathbf{X})$ is of dimension $n$? – Jen Snow Feb 06 '20 at 04:51
  • 2
    $T(X)$ explicitly is *two* dimensional: it has two components. – whuber Feb 06 '20 at 04:52
  • Should I try a different approach? I know that if the function of a sufficient statistic is ancillary, then the sufficient statistic is not complete. So, should I try to find a function of $T(\mathbf{X})$ that is ancillary? Since I am working with the normal distribution, I suppose I can try to do a scale family. – Jen Snow Feb 06 '20 at 04:55
  • 2
    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 $\sigma.$ – whuber Feb 06 '20 at 05:00
  • @whuber I think your approach may differ from what my professor has been doing, but I am definitely game to try it. So, $E(\frac{\Sigma X_i}{\sigma})=\alpha n$ and $E(\frac{\Sigma x_i^2}{\sigma^2})=n+ \alpha^2$, so neither of these simple functions' expectations rely on $\sigma$. However, can you please fill in the gap of how constructing $g$ with these new, simple functions shows that $T(x)$ is not complete? – Jen Snow Feb 06 '20 at 05:08
  • 2
    I'm afraid *both* expectations rely on $\sigma:$ it's right there in the formulas. – whuber Feb 06 '20 at 05:13
  • 2
    Ansqered at https://stats.stackexchange.com/q/353431/119261 – StubbornAtom Feb 06 '20 at 06:19
  • 1
    Does this answer your question? [On the existence of UMVUE and choice of estimator of $\theta$ in $\mathcal N(\theta,\theta^2)$ population](https://stats.stackexchange.com/questions/353431/on-the-existence-of-umvue-and-choice-of-estimator-of-theta-in-mathcal-n-th) – Xi'an Feb 08 '20 at 07:00

1 Answers1

2

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: $\sigma^2=(\sigma)^2$.)

Xi'an
  • 90,397
  • 9
  • 157
  • 575
  • 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 $\sigma^2$? – Jen Snow Feb 07 '20 at 01:52
  • 2
    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. – Xi'an Feb 07 '20 at 06:55
  • I understand that the transform of $T(X)$ should not depend on $\sigma$. However, I do not know how to use your hint. What should my initial thought be given this amount of work? – Jen Snow Feb 07 '20 at 17:48
  • 2
    Stronger hint: what ratio involving the components of $T$ will cancel out the powers of $\sigma$? (There are many answers; choose a simple one.) – whuber Feb 07 '20 at 18:04
  • @whuber If $Y=\sum x_i^2$ and $X=\sum x_i$, and we let $g(T(\mathbf{X}))=\frac{X^2}{n\alpha^2+1}-\frac{Y}{1+\alpha^2}$, then we get $E_\sigma[g(T(X))]=n\sigma^2-n\sigma^2=0$. Is this adequate? – Jen Snow Feb 12 '20 at 02:15
  • 2
    It's the right idea, but are you sure about your algebra? – whuber Feb 12 '20 at 03:42
  • @whuber I keep checking my algebra and getting the same thing. Which part did I get wrong? $X=\sum^n_{i=1}X_i \sim N(\alpha n \sigma, n\sigma^2)$, so $E(X^2)=Var(X)+E(X)^2=n\sigma(1+n\alpha^2)$ and $E(Y)=nE(X_i^2)=n(Var(X_i)+(E(X_i))^2)=n(\sigma^2+\alpha^2\sigma)$ – Jen Snow Feb 13 '20 at 02:55
  • @whuber or Xi'an - is it actually just as simple as $g(T(X))=X^2-Y$? I think this works – Jen Snow Feb 13 '20 at 04:42
  • I was trying to follow your ratio idea, where $Z_1 = \sigma\sum_i Y_i$ and $Z_2=\sigma^2 \sum_iY_i^2$. Then if $g(T(X))=\frac{X^2}{Y}$, the transform of $T(X)$ does not depend on $\sigma$, but I do not know how to find the expected value of such a transformation. – Jen Snow Feb 13 '20 at 04:53
  • 2
    The point is to be independent from $\sigma$ not to compute the value. – Xi'an Feb 13 '20 at 05:34
  • @Xi'an is that equivalent to the definition here: https://en.wikipedia.org/wiki/Completeness_(statistics)? I was not aware that would be a sufficient (pun) answer to show completeness. That is much easier than trying to find the expected value of $g(T)$. Would you mind explaining how those are equivalent? I appreciate all of your help. – Jen Snow Feb 13 '20 at 06:06
  • @Xi'an Also, do you see anything wrong with the work I presented in earlier comments? – Jen Snow Feb 13 '20 at 06:15
  • An easy argument is that, when the distribution of $g(T)$ is independent from $\sigma$, its expectation is obviously constant when $\sigma$ varies. If $g(\cdot)$ is not a constant function, this means $T$ is not complete. – Xi'an Feb 19 '20 at 18:33