# Completeness of $X$ with normal distribution $N(\theta,\theta^2)$ with equal mean and standard deviation: Integral of scale of a function

Suppose that $$f:\mathbb{R} \to \mathbb{R}$$ is a Borel function such that, for every $$a > 0$$, we always have:

$$\begin{equation*} \int_{\mathbb{R}}f(ax)\exp \left(-\frac{1}{2}(x-1)^2\right)dx =0 \end{equation*}$$

Does it follow that $$f = 0$$ almost surely ? Does the answer change if we limit the attention only to $$f \in L^{\infty}$$ ?

The background of the question is that, suppose $$X$$ obeys the distribution $$N(\theta, \theta^2)$$ for some $$\theta> 0$$ and we wish to determine if $$X$$ is a (boundly) completes statistic for $$\theta$$. Expanding the definition out with a change of variable will give the statement of the problem above. I am looking for an analysis solution .

Intuitively, the exponential centers at $$1$$ and the scale is not centered at $$1$$. A counterexample would be surprising but I cannot prove it.

Noting that the integral equation can be written as

$$\mathbb E \left (f(aZ+a)\right )=0, \, Z \sim \mathcal N(0,1),$$

a counterexample can be designed as follows:

$$\color{blue}{f(x)=\frac{\max(x,0)}{|x|\Phi(1)}+\frac{\min(x,0)}{|x|(1-\Phi(1))}=\begin{cases} \frac{1}{\Phi(1)} & x>0\\ \frac{-1}{1-\Phi(1)} & x<0 \end{cases}}$$

for $$x\neq0$$ where $$\Phi(x)=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{1}{2}x^2\right)\text{d}x$$ denotes the cdf of $$Z$$, and $$f(0)$$ can be set to any number.

PS: From the section "Relation to sufficient statistics" in the Wikipedia page on complete statistics, it can be indirectly proven that there is no complete statistic for $$\mathcal N(\theta,\theta^2)$$.

• This is so simple! Thank you. Your idea works. Your choice of constants actually does not work. But I will accept your answer.
– 温泽海
Commented Apr 17 at 14:03
• @温泽海 You are welcome! The first equation is correct. I just added a missing $-$ in the second equivalent equation for $x<0$.
– Amir
Commented Apr 17 at 14:26
• @温泽海 you may also check my answer to the following related question: math.stackexchange.com/q/4892966/1231520
– Amir
Commented Apr 17 at 14:31