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I am working through some inference maths as described in a paper and have a doubt about a certain step. At one point, the authors have to compute an expectation of the following expression

$$ \sum_{i=1}^{N} \frac{w_i}{\sigma^2} (y_i - \beta x_i)^T(y_i - \beta x_i) $$

Now the expectation has to be taken wrt to $Q(w_i)$. For me, this should be simply:

$$ \frac{1}{\sigma^2}\sum_{i=1}^{N} <w_i> (y_i - \beta x_i)^T(y_i - \beta x_i) $$

where $<w_i>$ is the expectation operator applied. However, the authors write this as: $$ \frac{1}{\sigma^2}\sum_{i=1}^{N} <w_i> y_ix_i $$

I was wondering if there was some trick that I missed.

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As provided, the answer to the question is that you are correct. The expression$$\frac{1}{\sigma^2}\sum_{i=1}^{N} <w_i> y_ix_i$$ is certainly inappropriate as definitely not equal to the expectation of$$\sum_{i=1}^{N} \frac{w_i}{\sigma^2} (y_i - \beta x_i)^T(y_i - \beta x_i)$$It would be nice though if you could provide an excerpt of the paper you refer to, so that we can check this is indeed what the author(s) meant.

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  • $\begingroup$ Thanks for the answer. I will attempt to make the question a bit more elaborate now. I just wanted to check if I did not miss anything. I might post another question in a while to keep things a bit less confusing. $\endgroup$
    – Luca
    Commented Nov 6, 2015 at 10:32
  • $\begingroup$ I created a new thread showing my work and the expression from the paper here (stats.stackexchange.com/questions/180469/…) $\endgroup$
    – Luca
    Commented Nov 6, 2015 at 11:14
  • $\begingroup$ Cannot delete it as it has answers :/ $\endgroup$
    – Luca
    Commented Nov 6, 2015 at 13:02
  • $\begingroup$ than what hope do mortals like me have ;-) $\endgroup$
    – Luca
    Commented Nov 6, 2015 at 13:27

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