Find all solutions of $f$ such that $ \operatorname{Cov}(f(x),x)=c $ where $x \sim N(\mu,\sigma^2)$ and $c$ does not depend on $(\mu,\sigma^2)$

Question

Find all solutions of $f$ such that $$ \operatorname{Cov}(f(x),x)=c$$ where $x \sim N(\mu,\sigma^2)$ and $c$ does not depend on $(\mu,\sigma^2)$

I think the only solution is $f$ is constant (almost surely). I am trying to first show that $c=0$, $E[f(x)]=0$ and $E[f(x)x]=0$ and then that $E[f(x)]=0$ implies $f(x)=0$.

I can more or less show that $E[f(x)]=0$ implies $f(x)=0$ by \begin{align*} E[f(x)]=(2 \pi \sigma^2)^{-1/2}\int f(x)e^{-(x-\mu)^2/2\sigma^2}dx \end{align*} Then changing to complex coordinates and analytic continuity I think this can be viewed as a Fourier transform so if $E[f(x)]=0$ then $f$ is zero.... although this is obviously not rigorous (and not 100\% sure correct).

To show that $c=0$, $E[f(x)]=0$ and $E[f(x)x]=0$, I was thinking of using choices of $(\mu,\sigma^2)$ and Cauchy-Schwarz.

Answer 1

Note that the functions under consideration must be integrable (finite $L^1$-norm). Denote the set of all solutions $f$ by $\mathcal{F}$. To start, I decided to restrict attention to smooth functions $f \in C^{\infty}(\mathbb{R})$. For such $f$ it is easy to estimate that off of a $\delta$-ball about $\mu$ such that $\delta/\sigma \to \infty$ as $\sigma \to 0$, \begin{eqnarray*} \left|(2\pi\sigma^2)^{-1/2} \int_{B(\mu,\delta)^C}(x-\mu)f(x)e^{-(x-\mu)^2/2\sigma^2} dx \right| & \leq & (2\pi\sigma^2)^{-1/2} \|f\|_{L^1(\mathbb{R})} \sup_{\mathbb{R}\backslash B(\mu,\delta)} |x-\mu|e^{-(x-\mu)^2/2\sigma^2} \\ & \leq & (2\pi)^{-1/2} \|f\|_{L^1(\mathbb{R})}{\delta\over\sigma} e^{-\delta^2/2\sigma^2} \\ & = & \epsilon(\sigma) \to 0. \end{eqnarray*} and note that if we take $\delta = \sqrt[3]{\sigma}$ for example, the $\epsilon(\sigma)$ goes to zero faster than any power of $\sigma$. This leads to \begin{eqnarray*} |\operatorname{Cov}(f(x),x)| & = & (2\pi\sigma^2)^{-1/2} \left|\int_R (x-\mu)f(x) e^{-(x-\mu)^2/2\sigma^2} dx \right| \\ & \leq & (2\pi\sigma^2)^{-1/2} \left| \int_{\mu-\delta}^{\mu+\delta} f(x)(x-\mu)dx \right| + (2\pi)^{-1/2}\|f\|_{L^1(\mathbb{R})}{\delta \over \sigma} e^{-\delta^2/2\sigma^2} \\ & = & (2\pi\sigma^2)^{-1/2} \left( \max|f| \delta^2 + \epsilon(\sigma) \right) \to 0 \end{eqnarray*} where $\max |f|$ refers to the largest magnitude of $f$ on $B(\mu,\delta)$. So the constant $c$ must be zero for any $f \in C^{\infty}$.

We estimate the function on the $\delta$-ball about $\mu$ by $f(x) = f(\mu) + f'(\mu)(x-\mu) + h(x)$ where $h(x) = O(|x-\mu|^2)$. Now we can write \begin{eqnarray*} \operatorname{Cov}(f(x),x) &=& (2\pi \sigma^2)^{-1/2} \int_R (x-\mu)f(x)e^{-(x-\mu)^2/2\sigma^2}dx \\ & = & (2\pi)^{-1/2} f'(\mu) \int_{\mu-\delta}^{\mu+\delta} (x-\mu)^2e^{-(x-\mu)^2/2\sigma^2} {dx \over \sigma} \\ && + (2\pi)^{-1/2}\int_{\mu-\delta}^{\mu+\delta}(x-\mu)h(x) e^{-(x-\mu)^2/2\sigma^2}{dx \over \sigma} + \epsilon(\sigma) \\ & = & (2\pi)^{-1/2} f'(\mu)\sigma^2 \int_{-\delta/\sigma}^{\delta/\sigma} w^2e^{-w^2/2}dw + \text{const}\cdot \sigma^3 + \epsilon(\sigma) \\ & \to & f'(\mu) \sigma^2 + \text{l.o.t.} \end{eqnarray*}

Now if $f \in \mathcal{F} \cap C^{\infty}(\mathbb{R})$ and $f$ is nonconstant, then for some $a \in \mathbb{R}$ we have $f'(a) \neq 0$. Taking $\mu = a$ and $\sigma$ small, we find $$ \operatorname{Cov}(f,x) = \sigma^2 f'(a) + o(\sigma^4), $$ which is nonzero for $\sigma > 0$ small enough. This contradicts the assumption $f \in \mathcal{F}$, so we conclude that all smooth functions that are in $\mathcal{F}$ must be constant.

By the continuity of the covariance operator $\operatorname{Cov}(\cdot, x)$ we can extend this result from smooth functions to $L^1$ functions.

Note that we need $\sigma \to 0$ to get the first estimate and $\delta \to 0$ to get the linear estimate on $f$. For the covariance to go to zero with $\sigma$ we also need $\delta^2/\sigma \to 0$. If we choose $\delta(\sigma) = \sigma^{1/3}$, we can satisfy all these conditions.