Are there distributions that satisfy this equation for more functions? For random variable X following any distribution, the following is true for any function $\phi(x)=ax+b$, where $a$ and $b$ are constants: $$\mathbb{E}[\phi(X)] = \phi(\mathbb{E}[X])$$
Are there distributions that allow this to be true for more types of $\phi$ besides linear functions?
I know I should be looking at $$E[\phi(X)] = \int_{-\infty}^\infty \phi(x)f(x)dx$$
but I was wondering if there were any particular examples that come to mind.
Thank you in advance!
 A: Recall that a function $\phi$ is odd if and only if $\phi(-x)=-\phi(x)$ for every $x$. Let $X$ denote an integrable random variable $X$ such that $X$ and $-X$ coincide in distribution, and $\phi$ an odd function. Then $\mathbb E(\phi(X))=0$ and $\mathbb E(X)=0$ hence $\phi(\mathbb E(X))=\phi(0)=0$. More generally:

Assume that there exists $a$ and $b$ such that $\phi(a-x)=b-\phi(x)$ for every $x$ and such that the distributions of $X$ and $a-X$ coincide. Then, if $X$ and $\phi(X)$ are both integrable,
  $$
\mathbb E(\phi(X))=\tfrac12b=\phi(\mathbb E(X)). 
$$

When the distribution of $X$ has a density $f$, $X$ and $-X$ coincide in distribution if and only if $f$ is even, in the sense that $f(-x)=f(x)$ for every $x$. In particular, $\mathbb E(X)=0$ when $X$ is integrable. Likewise, $X$ and $a-X$ coincide in distribution if and only if $f(a-x)=f(x)$ for every $x$. In particular, $\mathbb E(X)=\frac12a$ when $X$ is integrable.
In the other direction, if $\phi$ is such that $\mathbb E(\phi(X))=\phi(\mathbb E(X))$ for every integrable random variable $X$, then $\phi$ is an affine function. To show this, consider a random variable $X$ taking two values $x$ and $y$ with probability $t$ and $1-t$ respectively and observe that $\phi(\mathbb E(X))=\phi(tx+(1-t)y)$ and that $\mathbb E(\phi(X))=t\phi(x)+(1-t)\phi(y)$, hence $\phi$ must be both convex and concave, that is, affine.
