Expectation of a composite function. 
My thoughts on the solution:
I noticed that the ln function "stretches" the intervals of X like: (-1; 0) to (-inf; 0), and (0; 1) to (0; ln2).
Intuitively, it becomes clear that due to the long interval of negative values of ln(1+x), the expectation will also be negative. But I haven't figured out how to strictly prove it.
In general, it is probably worth considering the integral: ∫1−1ln(x+1)f(x)dx, where f(x) is probability density function.
Thanks!
 A: Hint: based on what you know,
$$
f(\mathbb{E}[X]) = \ln(1+\mathbb{E}[X])=\ln 1 = 0
$$
and you are trying to say something about the sign of $\mathbb{E}[f(X)]$, which by the above is equivalent to comparing $\mathbb{E}[f(X)]$ and $f(\mathbb{E}[X])$. Is there some inequality between $f(\mathbb{E}[X])$ and $\mathbb{E}[f(X)]$ you can think of using? What property of $f$ does it rely on?
Further hint: (click to reveal)

 Use Jensen's inequality. Here, the properties you are using are $f(0)=0$, and concavity of $f$.

A: Note that $\ln(1+X)<X$ for all $X\ne0$ and equality holds for $X=0$. Hence $Y\le X$, so $EY\le EX$ (since expectation of non-negative random variable $X-Y$ is non-negative). Also, since expectation of a non-negative random variable is zero iff that random variable is zero with probability $1$, so $EY=0$ if an only if $\ln(1+X)=X$ with probability $1$, which happens if and only if $X$ is degenerate at zero. $EY$ is negative otherwise.
My method of defining sign of the expectation is applicable for every function $f(X)$ such that $f(X)\le X$ for all $X\in (-1,1)$ and the inequality is strict except for $X=0$, and similarly also applicable for functions such that $f(X)\ge X$ in this interval and inequality is strict except for $X=0$.
