How to derive the closed form for M, S, and C in squash_sin(m, s, max_action=None) about pilco

[mdbenito]

Consider two gaussians X_i and X_j and their sines Y_i = sin X_i and Y_j = sin X_j. You want to compute the covariance E[(Y_i - EY_i) (Y_j - EY_j]. After expanding the product and simplifying a bit, one of the terms is E[Y_i Y_j]. This is the product of two sines so you can use the formula expressing it as a a difference of cosines. Use linearity and after some arithmetic you have entry ij in the covariance matrix, already looking similar to the code. The final part is to vectorise that. For this note what the convention in Tensorflow or numpy is when a column vector is added to a row vector and you are almost done.

Hope that helps!

[thaipduong]

Never mind. I figured it out. Thanks.

[hejia-zhang]

Hi @thaipduong, could you give me some hints on it? I am also trying to figure this out while have no clues yet.

[thaipduong]

@hjzh4 , I found my notes on this here: https://drive.google.com/file/d/1MqZMkRM6vkLJIOOhe8gXFddpMflLJvB2/view?usp=sharing
It's the same as what @mdbenito mentioned above. Hope it helps.

[hejia-zhang]

@thaipduong @mdbenito , thanks a lot guys! I just figured it out!
could you also take a look at issue #15 I think there is something missed in the formula for computing C