Let $Y=h(X),$ Find $E\{Y\}$ Problem:
Let $X: (\Omega, \mathscr{A}) \rightarrow (\mathbb{R},B)$ be a random variable with the uniform distribution $P^X=\frac{1}{2\pi}\mathbb{1}_{\{(0,2\pi)\}}$ on the interval $(0,2\pi),$ and $h:(\mathbb{R},B) \rightarrow (\mathbb{R},B)$ be given by $h(x)=\sin(x).$
Let $Y=h(X)$
Find $E\{Y\}$
Attempt
I want to use the expectation rule so:
$$E\{Y\}=E\{h(X)\}=\int h(x)P^X(dx)=\int \sin(x) \frac{1}{2\pi}\mathbb{1}_{\{(0,2\pi)\}} (dx)=$$
$$\frac{1}{2\pi} \int_0^{2\pi} \sin(x)dx=0$$
I have tried a few things and this is my "best" attempt.. but it still just doesn't seem right.
 A: You solved this correctly. The question can be phrased as this: Choose an angle from $0$ to $2\pi.$ What is the average value of its sine? Or, even better - choose a point uniformly from the unit circle. What is the average value of the $y$ coordinate?
Indeed, we select $\theta \in [0, 2\pi)$ uniformly, and so the density is $f_{\Theta} (\theta) = \frac{1}{2\pi}$ when in our interval.
Now by the law of the unconscious statistician, $E(\sin \Theta) = \int_0 ^{2\pi} \frac{1}{2\pi} \cdot \sin \theta \ d\theta = 0$. Done!
A: Well I think I did find this incomplete.

*

*What's $\Omega$? $\mathbb R$? [Edit: Ah ok it's not necessarily $\mathbb R$. I was figuring it out as I was typing this.]


*Also I think it's $\mathbb{1}_{(0,2\pi)}$ not $\mathbb{1}_{\{(0,2\pi)\}}$. $\mathbb{1}_{\{(0,2\pi)\}}$ sounds like the interval $(0,2\pi)$ is just a part of a collection of sets instead of a subset of $\mathbb R$ like $\Omega = \{\{3\}, (0,2\pi), \{4,7\}, [10,16] \cup \{28.7\}\}$ and then we have like a uniform discrete distribution on those 5 elements of $\Omega$ or something.


*Also I'm not sure about $\int_{\Omega} h(x)P^X(dx)$. I think there should be some $\omega$ s.t. we have something like $\int_{\Omega} h(X(\omega))dP^X(\omega)$ or $\int_{\Omega} h(X(\omega))P^X(d\omega)$
Ah in this case I think $\Omega$ need not be $\mathbb R$... Wait lemme check Rosenthal...



Ah yeah ok I believe you're missing also the probability measure for the probability space. So let's say it's $\mathbb P$ s.t. the probability space is $(\Omega, \mathscr A, \mathbb P)$. So I think it's supposed to be...
$$E\{Y\}=E\{h(X)\}=\int_{\Omega} h(X(\omega))\mathbb P(d\omega)$$
$$=\int_{\mathbb R} h(x)P^X(dx)$$
$$=\int_{\mathbb R} \sin(x) \left[ \left(\frac{1}{2\pi}\mathbb{1}_{(0,2\pi)}\right) \circ (dx) \right]$$
$$=\int_{\mathbb R} \sin(x) \left[\left(\frac{1}{2\pi} \mathbb{1}_{(0,2\pi)}\right) \circ (x)\right] dx$$
$$=\int_{\mathbb R} \sin(x) \frac{1}{2\pi} \left[\left(\mathbb{1}_{(0,2\pi)}\right) \circ (x)\right] dx$$
$$=\int_{\mathbb R} \sin(x) \frac{1}{2\pi} \mathbb{1}_{(0,2\pi)}(x) dx$$
$$=\frac{1}{2\pi} \int_{\mathbb R} \sin(x) \mathbb{1}_{(0,2\pi)}(x) dx$$
$$=\frac{1}{2\pi} \int_{\mathbb R \cap (0,2\pi)} \sin(x) dx \ \text{(I guess no need for this line though like...}$$
$$\text{...(0,2π) is already understood as a subset of the ambient space)}$$
$$=\frac{1}{2\pi} \int_{(0,2\pi)} \sin(x) dx$$
$$=\frac{1}{2\pi} \int_0^{2\pi} \sin(x)dx=0$$
