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In Bishop on page 219, the equality below is given. Unfortunately, I fail to see how one finds the mean $\mu_{a}$ of this distribution. I know the definition of expectation operator, but how does one obtain the most right equality in the bottom equation? Can someone enlighten me? Thanks!

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$\int p(a)a\ da = $ (first equality in OP post)

$=\int (\int \delta(a - w^T\phi)q(w)dw)\ ada = $ (change odrer of integration)

$=\int (\int \delta(a - w^T\phi)ada)\ q(w)dw = $ ((4.146) in Bishop)

$=\int w^T\phi\ q(w)dw $

UPD. Added comments and fix

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    $\begingroup$ This makes no sense: how do you justify making "$a$" disappear at the end, to be replaced by "$w^T\phi$"? Are you sure you put the equalities in the order intended? Regardless, a little explanation would be extremely helpful. $\endgroup$ – whuber Sep 11 '17 at 14:27
  • $\begingroup$ Updated my post $\endgroup$ – Nik Sep 11 '17 at 14:42

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