What is $\mathbb E[X\mid \sin(X)]$ when $X$ is uniform on $[0,\pi]$?

Question

Suppose that $X$ is uniformly distributed on $[0,\pi]$. Find $\mathbb{E}[X \mid\sin (X)]$.

My attempt:

$$\mathbb{E}[X\mid \sin (X)] = \int_{0}^{\pi} \dfrac{1}{\pi} f_{X\mid \sin (X)}(x) \, \mathrm{d}x$$

How would I calculate $f(X\mid \sin (X))$?

Answer 1

It might help to take a look at the graph of sin(x), $0\leq x \leq \pi$.
You see that once you know the value of sin($x$) (let's call this value $y_0$), there are only two possible values left for $x$.
Moreover, the function sin(x) is symmetric in x = $\pi/2$. So the two possible values for $x$ are distributed symmetrically around x = $\frac{\pi}{2}$.
Let's call these values $x_0$ = arcsin($y_0$) and $x_1 = \pi - x_0$.

Since the initial distribution for $X$ was uniform, there is no preference for $x_0$ or $x_1$. The probability density function now becomes some sort of probability mass function (for discrete random variables).
P($X = x_0|\text{sin}(X) = y_0$) = 1/2
P($X = x_1|\text{sin}(X) = y_0$) = 1/2
And now we can easily calculate that $\mathbb{E}[X|\text{sin}(X) = y_0]$ = $\frac{\pi}{2}$.

Note that this argument works for all values of $y_0$ between 0 and 1.
We can therefore conclude that
$\mathbb{E}[X|\text{sin}(X)]$ = $\frac{\pi}{2}$.