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Let us consider a random variable $Z$ with a probability density function $f$ with respect to the Haar measure on $\mathrm{SO}(3)$. Next, we consider two fixed normal vectors $u,v$ in $\mathbb{R}^3$. Define a real random variable $$ Y = \langle Zu, v \rangle = \| Zu \| \| v \| \cos \theta = \cos \theta\,, $$

where $\theta$ is the angle between $Zu$ and $v$. Then $Y$ lines almost surely in $[-1,1]$. I would like to compute

$$ \mathbb{P} ( Y \in [a,b] ). $$

In other words, I would like to know what is the probability that the cosine of an angle between $v$ and $u$ rotated by $Z$ lies in some interval. However, I do not know how to transfer integration from $\mathrm{SO}(3)$ to $\mathbb{R}$.

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  • $\begingroup$ $Zu$ should be uniformly distributed on the unit $2$-sphere for symmetry reasons. Then the scalar product with $v$ projects onto a single coordinate, so you get the uniform distribution on $[-1,1]$ by a fun coincidence (mathoverflow.net/questions/33129/…) $\endgroup$ Commented May 23 at 14:51
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    $\begingroup$ the OP has not specified what $f$ is, so there is no symmetry one can use; $f$ need not be isotropic on the unit sphere. $\endgroup$ Commented May 23 at 15:19
  • $\begingroup$ Ah sorry, yes, I misread the question. $\endgroup$ Commented May 23 at 16:31

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Use the Euler angle parameterisation of the rotation matrix, $$Z=\begin{pmatrix} R(\alpha)&0\\ 0&1\end{pmatrix} \begin{pmatrix} 1&0\\ 0&R(\theta)\end{pmatrix} \begin{pmatrix} R(\alpha')&0\\ 0&1\end{pmatrix},$$ $$R(\theta)=\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix},$$ then the Haar measure$^\ast$ is $$\mu(\alpha,\alpha',\theta)=\sin\theta d\alpha d\alpha' d\theta.$$ The probability distribution function of $Z$ is given as $f(\alpha,\alpha',\theta)$. The two unit vectors $u,v$ have components $$u=(\sin \gamma\cos\phi,\sin\gamma\sin\phi,\cos\gamma),$$ $$v=(\sin \gamma'\cos\phi',\sin\gamma'\sin\phi',\cos\gamma').$$ The inner product $Y$ is given by $$Y=\sin \gamma \bigl\{\sin (\alpha'+\phi) [\cos \gamma' \sin \theta-\sin \gamma' \cos \theta \sin (\alpha-\phi')]+\cos (\alpha'+\phi)\sin \gamma' \cos (\alpha-\phi') \bigr\}$$ $$\qquad+\cos \gamma \bigl\{\sin \gamma' \sin \theta \sin (\alpha-\phi')+\cos \gamma' \cos \theta\bigr\}.$$ Moments of $Y$ can then be computed by integrating $$\mathbb{E}[Y^p]=\int_0^{2\pi}d\alpha \int_0^{2\pi}d\alpha' \int_0^{\pi}\sin\theta d\theta \,f(\alpha,\alpha',\theta)Y^p.$$ Further calculations would require knowledge of the function $f$.


$^\ast$ To calculate the Haar measure, first compute the metric tensor $$g_{mn}=-\operatorname{tr}Z^\top(\partial Z/\partial\phi_m)Z^\top(\partial Z/\partial\phi_n),\;\;\{\phi_1,\phi_2,\phi_3)=\{\alpha,\alpha',\theta\},$$ $$\Rightarrow g=\begin{pmatrix} 2&2\cos\theta&0\\ 2\cos\theta&2&0\\ 0&0&2\end{pmatrix},$$ and then the measure is $\sqrt{\det g}\propto\sin\theta.$
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  • $\begingroup$ Thank you for your comment. Could you please provide some theory on this that I could read on? $\endgroup$
    – user528399
    Commented May 23 at 17:41
  • $\begingroup$ for the Euler angle parameterization, see here; I wrote out the Haar measure calculation, which I copied from here (appendix B); the computation of $Y$ is just tedious algebra (I used Mathematica). $\endgroup$ Commented May 23 at 17:47

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