Hi ! I was wondering if instead of the given formula,
$$P[y_n|x_n;w;r_n] = N(y_n|r_n^Twx_n, \sigma^2)$$
would work ? Since multiplying \(r_n^T\) by \(w\) would return \(w_k\)
Hi,
yes, this formula would also work if you are a bit careful with the shapes. You need to assume that W is a matrix in \(R^{d \times K}\). And then
$$P[y_n|x_n; W; r_n] = N(y_n| r_n^T W^T x_n,\sigma^2)$$
However, it is a bit more convenient for derivations to use \(r_{nk}\) as a power here.
Lab 6 Ex3.1
Hi ! I was wondering if instead of the given formula,
$$P[y_n|x_n;w;r_n] = N(y_n|r_n^Twx_n, \sigma^2)$$
would work ? Since multiplying \(r_n^T\) by \(w\) would return \(w_k\)
2
Hi,
yes, this formula would also work if you are a bit careful with the shapes. You need to assume that W is a matrix in \(R^{d \times K}\). And then
$$P[y_n|x_n; W; r_n] = N(y_n| r_n^T W^T x_n,\sigma^2)$$
However, it is a bit more convenient for derivations to use \(r_{nk}\) as a power here.
2
Add comment