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A point $(X,Y)$ is uniformly distributed on the unit square $[0,1]^2$. Let $\theta$ be the angle between the x-axis and the line segment that connects $(0,0)$ to the point $(X,Y)$. Find the expected value $E[\theta]$.
The question also gives the hint that ${d\cos\theta\over d\theta}= {-\sin\theta}$, ${d\tan\theta\over d\theta}= {1\over \cos^2\theta}$.

Here is what I did:
Let: $Z=\tan\theta={Y\over X}$.
Here I assume X and Y are independent since coordinate in x-axis doesn't depend on y-axis.
Since $f_X(x)=1$, $f_Y(y)=1$, $f_{X,Y}(x,y)=1$. $0\le x\le1$, $0\le y\le1$. I get:
$$f_Z(z)={z\over2}, 0\lt z\lt1$$ $$f_Z(z)=1-{1\over2z}, z\gt1$$ $$E[\theta]=E[\arctan Z]=\int_{0}^{1}\arctan(z)\times{z\over2}dz+\int_{1}^{\infty}\arctan(z)\times(1-{1\over2z})dz$$ But I don't know how to calculate $\int_{1}^{\infty}\arctan(z)\times(1-{1\over2z})dz$.
Am I doing correctly on this question? I also noticed that I didn't use the hint the question provides.

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The problem is symmetric with respect to the line $\theta = \pi/4$. That is, for any point $(x,y)$ at $\pi/4 + \Delta$ there is a corresponding point at $\pi/4 - \Delta$. Hence: ${\cal E}(\theta) = \pi/4$.

enter image description here

Alternatively:

$${\cal E}(\theta) = \int\limits_{x=0}^1 \int\limits_{y=0}^1 \arctan (y/x)\ dx\ dy = \pi/4$$

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  • $\begingroup$ but the point is only inside the unit square.I think the domain of $\theta$ is from 0 to $\pi\over2$ $\endgroup$ – clement Oct 14 '18 at 22:58
  • $\begingroup$ both ways are better than mine. thank you! $\endgroup$ – clement Oct 15 '18 at 0:05
  • $\begingroup$ So please accept or at least up-vote my answer. $\endgroup$ – David G. Stork Oct 15 '18 at 0:06

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