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

[luost26]

$ER = (ERE^{-1})E \sim \mathcal{IG}_{SO(3)}(ERE^{-1}, \sigma^2)$ 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 $Z = ERE^{-1}E = ER$ is still $IG(R, \sigma^2)$.

[longlongman]

Because $ER \not = RE \not = (RE)^{T}$, 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 $Z = ER = ERE^{-1}E \sim \mathcal{IG}(R, \sigma^2)$ holds, then how can we sample a random value from $\mathcal{IG}(ERE^{-1}, \sigma^2)$.

[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.