The question is this: Does there exist an integrable function $f\colon\mathbb R\to\mathbb R$ such that $f$ differs from $0$ on a set of nonzero Lebesgue measure and \begin{equation} \int_{\mathbb R}g_\theta(x)f(x)\,dx=0 \tag{1} \end{equation} for all real $\theta$, where $g_\theta(x):=e^{-(\theta x-1)^2/2}$?

More generally, one may ask this: Does there exist a nonzero tempered distribution $F$ on $\mathbb R$ such that $\langle g_\theta,F\rangle=0$ for all real $\theta$? The latter question can of course be restated as follows: Does there exist a nonzero tempered distribution $\hat F$ on $\mathbb R$ such that $\langle h_\tau,\hat F\rangle=0$ for all real $\tau$, where $h_\tau(t):=e^{i\tau t-\tau^2 t^2/2}$?

One may note that $(1)$ will hold for all $\theta>0$ if $f=I_{(0,\infty)}-c$, where $I_{(0,\infty)}$ is the indicator function of the set $(0,\infty)$ and $c=\frac1{\sqrt{2\pi}}\int_0^\infty e^{-(u-1)^2/2}\,du$. However, the question is whether $(1)$ can hold for (an essentially nonzero $f$ and) all real $\theta$.

  • $\begingroup$ In statisticians' language: Does the family $\displaystyle \left\{ N\left( \frac 1 \theta, \frac 1 {\theta^2} \right) \mid \theta\in\mathbb R\right\}$ of probability distributions admit any non-trivial unbiased estimator of zero? Or in other words, is this family of distributions complete (i.e. it does not admit such an estimator)? $\qquad$ $\endgroup$ Mar 6 '17 at 22:29
  • $\begingroup$ @MichaelHardy : This was indeed the original formulation, and this is why one of the tags was st.statistics. However, the essence of the problem is analytical. My partial answer suggests that much may depend on the behavior of $f(x)$ as $|x|\to\infty$. $\endgroup$ Mar 7 '17 at 1:29
  • $\begingroup$ Why the downvote? Can you please explain? $\endgroup$ Dec 27 '17 at 19:07

Here is an idea of a partial answer; cf. the answer by Christian Remling. Assume that $(1)$ holds for all $\theta$ in (say) a right neighborhood of $0$ (including $0$), where $|f(x)|=O(1/(1+|x|^k))$ for each $k>0$ (note that the latter condition is not satisfied by the function $f=I_{(0,\infty)}-c$ in the example in the question). Then one can differentiate the integral in $(1)$ in $\theta$ to the right of $0$ to get \begin{equation} \left. \int_{\mathbb R}\frac{\partial^k g_\theta(x)}{\partial\theta^k} \right|_{\theta=0}\,f(x)\,dx=0, \tag{2} \end{equation} for all $k=0,1,\dots$. Expanding $g_\theta(x)/g_0(x)=e^{(\theta x)-(\theta x)^2/2}$ into the Maclaurin series, we have \begin{equation*} \left. \frac{\partial^k g_\theta(x)}{\partial\theta^k} \right|_{\theta=0}=a_k x^k\,e^{-1/2}, \end{equation*} where \begin{equation*} a_k:=\left(-\frac{1}{2}\right)^k k! \sum _{n=\left\lceil k/2\right\rceil }^k \frac1{n!}\,\binom{n}{k-n} (-2)^n. \end{equation*} The sequence $(a_k)$ is the sequence A001464 at http://oeis.org/A001464.

It appears that $a_k\ne0$ for all $k=0,1,\dots$ except $k=2$. If that is indeed so, then, by $(2)$, we would have \begin{equation} \int_{\mathbb R}x^kf(x)\,dx=0 \tag{3} \end{equation} for all $k=0,1,\dots$ except $k=2$. Letting $y:=x^4$, we see that $x^2=\sqrt y$ can be approximated arbitrarily closely by polynomials in $y=x^4$. So, one would have $(3)$ for all $k=0,1,\dots$, including $k=2$, if $|f(x)|$ decreases fast enough as $|x|\to\infty$. This would imply that $f=0$ almost everywhere.

  • $\begingroup$ Yes, this makes more sense. I forgot the extra $x$ when taking the derivatives. (Also, my assumption on $f$ doesn't work since we can take $\theta=0$.) $\endgroup$ Feb 28 '17 at 19:15

$g_\theta(x):=e^{-(\theta x-1)^2/2}=\exp(-\frac{1}{2\frac{1}{\theta^2}}[x-\frac{1}{\theta}]^2)$ is the kernel of $N(\frac{1}{\theta},\frac{1}{\theta^2})$ with density $\frac{1}{\sqrt {2\pi \frac{1}{\theta^2}}}\exp\left(-\frac{1}{2\frac{1}{\theta^2}}[x-\frac{1}{\theta}]^2\right)$

which is a curved exponential family. Therefore the guess is that such a family is not complete(See counter example in Example 1 of https://www.stat.tamu.edu/~suhasini/teaching613/chapter2.pdf), the couter-example can be constructed by finding a function $f,h$ such that their Laplace transforms are not the same but $\int_{\mathbb R}g_\theta(x)f(x)\,dx=\int_{\mathbb R}g_\theta(x)h(x)\,dx$. Then the uniqueness theorem of Laplace transform applies. i.e. $f-h\neq0$ yet $\int_{\mathbb R}g_\theta(x)[f-h](x)\,dx=0$

  • $\begingroup$ This non-complete family from the notes you quote is a (joint) density on $\mathbb R^2$, so is not very directly related to the OP's setting. $\endgroup$ Feb 28 '17 at 5:52
  • $\begingroup$ @ChristianRemling Yes, you're correct. What I was thinking is that if it cannot be complete on an n-fold product of the same domain then it cannot be complete on each of its component. Thank you for the input! $\endgroup$
    – Henry.L
    Feb 28 '17 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.