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I try to understand a statement in this paper:

http://papers.nips.cc/paper/4305-spike-and-slab-variational-inference-for-multi-task-and-multiple-kernel-learning.pdf

In particular, I am talking about the reparameterisation from (1c) to (3).

Given is a random variable $w_{qm}$, that is distributed according to:

(1c) $w_{qm} \sim \pi \mathcal{N}(w_{qm}|0,\sigma_w²)+(1-\pi)\delta_0(w_{qm})$

Specifically, with probability $\pi$, each $w_{qm}$ is zero, and with probability, it is drawn from a Gaussian.

Further, assume a Gaussian random variable $\tilde{w}_{qm}\sim \mathcal{N}(\tilde{w}_{qm}|0,\sigma_w²)$ and a Bernoulli random variable $s_{qm}\sim\pi^{s_{qm}}(1-\pi)^{1-s_{qm}}$. The new random variable $\tilde{w}_{qm}s_{qm}$ is assumed to follow (1c), resulting in

(3) $p(\tilde{w}_{qm},s_{qm})=\mathcal{N}(w_{qm}|0,\sigma_w²)[\pi^{s_{qm}}(1-\pi)^{1-s_{qm}}]$

I don't really see how the authors did it ((1c)→(3)). I am grateful for any help.

Thanx so much in advance!

Krn

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  • $\begingroup$ You will need to add the material in question. No one is going to read the paper for you. $\endgroup$ Commented Dec 14, 2014 at 2:08
  • $\begingroup$ Thanx gung. I'll do wasn't sure what's best. $\endgroup$
    – Krn
    Commented Dec 14, 2014 at 2:09

1 Answer 1

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It looks like (3) is just a typo. The authors meant to write $$ p(\tilde{w}_{qm},s_{qm}) = \mathcal{N}(\tilde{w}_{qm}|0,\sigma_w²)[\pi^{s_{qm}}(1-\pi)^{1-s_{qm}}] $$ This is evident from later equations in the paper, such as (6).

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    $\begingroup$ Thanx Tom for taking the time! However, I guess I still don't get the core of my problem. Shouldn't I be able to transform (1c) to (3), or are the authors just assuming that's the same. I can't find the argument justifying that the two equations are the same. $\endgroup$
    – Krn
    Commented Dec 14, 2014 at 12:05
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    $\begingroup$ The equations are not the same. They correspond to two different ways of constructing $w_{qm}$, leading to the same distribution over $w_{qm}$. $\endgroup$
    – Tom Minka
    Commented Dec 14, 2014 at 12:08
  • $\begingroup$ @TomMinka, Although it is a very old answer and comment, I wonder could you please elaborate a little more on this? How it is "leading to the same distribution over $w_{qm}$"? $\endgroup$
    – user85361
    Commented Jul 25, 2022 at 16:46
  • $\begingroup$ By defining $w_{qm} = \tilde{w}_{qm} s_{qm}$ and working out its marginal distribution from the joint distribution $p(\tilde{w}_{qm}, s_{qm})$. $\endgroup$
    – Tom Minka
    Commented Nov 29, 2022 at 18:04

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