Further questions about left/right-multiply in issue #14

[longlongman]

Thanks for your quick reply #14 (comment).

It is still confusing to me. I also take the column vectors as the base vectors. In this case, $RE$ is more intuitive to me (we first get $E$ around $I$ and transform it with $R$).

Moreover, $ER=(ER{E}^{-1})E\sim {\mathcal{IG}}_{SO(3)}(ER{E}^{-1},{\sigma }^2)$, it seems that $RE$ and $ER$ do different things, because $ER{E}^{-1}=R$ is not always true.

[luost26]

$ER=(ER{E}^{-1})E\sim {\mathcal{IG}}_{SO(3)}(ER{E}^{-1},{\sigma }^2)$ is not correct.

Let's assume left-multiply applies. $ER\sim IG(R,{\sigma }^2)$ is a shorthand for another random variable $Y\sim IG(R,{\sigma }^2)$, where $Y=ER$ and $E\sim IG(I,{\sigma }^2)$.

Given $E\sim IG(I,{\sigma }^2)$, the distribution of $Z=ER{E}^{-1}E=ER$ is still $IG(R,{\sigma }^2)$.

[longlongman]

Because $ER\ne RE\ne (RE{)}^T$, I am not sure whether it is safe to say $Z=ER\sim \mathcal{IG}(R,{\sigma }^2)$ when we only know $Y=RE\sim \mathcal{IG}(R,{\sigma }^2)$.

Moreover, if $Z=ER=ER{E}^{-1}E\sim \mathcal{IG}(R,{\sigma }^2)$ holds, then how can we sample a random value from $\mathcal{IG}(ER{E}^{-1},{\sigma }^2)$.

[luost26]

Thank you for your time. I've checked the code and found $R$ is a column-vector rotation matrix (so $RE$ applies instead of $ER$ where $R$ is row-vector rot. mat.). I think $RE$ is correct and it is a mistake in the code.
Though numerically this leads to minor biases, I do think fixing it might improve the performance.

[longlongman]

Thank you for the great open-source project. I have learned a lot from your code.