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Let $Y\sim \sum^N_{i=1}\omega_iN(m_i,h^2 V)$.

The text I'm reading states that $Var(Y)=(1+h^2)V$, when $m_i=\theta_i$, where $\theta_i$ are draws taken from $P(\theta|D)$, and $V=Var(\theta|D)$

I get $Var(Y)=\sum_im_i^2\omega_i - E(Y)^2+h^2V$, but I don't see how it's equal to $V+h^2V$...

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The formula below, which I think is called the law of total variance, can be useful:

$$ \text{Var}(A) = \text{E} \Big( \text{Var} (A \,|\, B) \Big) + \text{Var} \Big( \text{E} (A \,|\, B) \Big) $$

Applying this formula when

$$ X \,|\, \theta \sim \text{N} (\theta, h^2V) \\ \text{Var}(\theta) = V $$

gives

\begin{align} \text{Var} (X) & = \text{E} \Big( \text{Var} (X \,|\, \theta) \Big) + \text{Var} \Big( \text{E} (X \,|\, \theta) \Big) \\ & = \text{E} (h^2 V) + \text{Var}(\theta) \\ & = h^2 V + V \\ & = (h^2 + 1) V \end{align}

For the rest, I guess that $$ \text{Var}(Y) = \sum_i \omega_i^2 \, \text{Var}(X_i)=(h^2+1) \, V \, \sum_i \omega_i^2 = (h^2+1) \, V $$

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  • $\begingroup$ Ocram, a sum of r.v. is not the same as a mixture... $\endgroup$ – An old man in the sea. Aug 17 '18 at 14:02
  • $\begingroup$ your question is unclear $\endgroup$ – ocram Aug 17 '18 at 14:06
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    $\begingroup$ In your case, the mixture is indeed a sum: to the discrete distribution supported on the $m_i$ with probabilities $\omega_i$ you have added a Normal distribution of mean $0$ and variance $\sigma^2.$ $\endgroup$ – whuber Aug 17 '18 at 15:25
  • $\begingroup$ @Anoldmaninthesea: Does it answer your question, then? $\endgroup$ – ocram Aug 18 '18 at 14:23
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Well, we can deduce that $Var(E(Y|i))=\sum_i\omega_im_i^2-E^2(Y)$.

And, $Var(E(Y|i))=Var(\theta_i)=V$

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