$sin(1/x)$: expectation vs infimum

Question

I want to show that $$ \underset{Q \in \mathcal{P}^0}{\inf} \ \{\mathbb{E}_{x \sim Q} \sin(1/x)\} > \inf_{x \in \mathbb{R}}\{ \sin(1/x)\} = -1$$ holds where $\mathcal{P}^0$ is the set of all probability distributions supported on $\mathbb{R}$.

Is this possible? I don't have a good intuition of "inf" over the set of all probability distributions.

Answer 1

Take any probability distribution $X$ on $\mathbb{R}$, let its associated density function be $f_X$.

Then, $$\mathbb{E}(sin(1/X))=\displaystyle\int_{-\infty}^{+\infty}sin(1/x) f_X(x) dx$$

So, it immediately follows that

$$|\mathbb{E}(sin(1/X))|=\left|\displaystyle\int_{-\infty}^{+\infty}sin(1/x) f_X(x) dx\right|\leq \displaystyle\int_{-\infty}^{+\infty}|sin(1/x) f_X(x)|dx\leq \displaystyle\int_{-\infty}^{+\infty} f_X(x) dx=1.$$

Answer 2

A probability distribution $\mathcal{Q}$ just gives 'weight' to different regions of the real line (hence the names probability mass/density functions). Then $\mathbb{E}_\mathcal{Q}[X]$ is just a weighted average of $X$ over the real line. Now think of every possible way you may assign weights over the real line (where every weight is positive and all the weight sums to 1) and then use these weightings to calculate an average over $X$. A stronger statement that you can then prove immediately is "all weighted averages of a set $S$ are lower-bounded by the minimum of $S$", and your original statement is a subcase of this.