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Assume $\theta$~$N(0,\sigma^2)$, and we have $n$ realization of signals $s_i$, where $s_i$~$N(\theta,\sigma_i^2)$.

Now the question is: what is $E[\theta|s_1,s_2,\dots,s_n]$?

Thanks in advance.

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I just found Conjugate Prior which I think answer my question! en.wikipedia.org/wiki/Conjugate_prior –  user54626 Jan 25 '13 at 5:17

1 Answer 1

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Since the random vector $(\theta,s_1,\ldots,s_n)$ is centered gaussian, $\mathbb E(\theta\mid s_1,\ldots,s_n)=\sum\limits_ka_ks_k$ for some deterministic vector $(a_1,\ldots,a_n)$. For every $i$, $\mathbb E(\theta s_i)=\sum\limits_ka_k\mathbb E(s_ks_i)$ yields the relation $$ \sigma^2=a_i(\sigma^2+\sigma_i^2)+\sum\limits_{k\ne i}a_k\sigma^2=a_i\sigma_i^2+\sigma^2\sum_{k=1}^na_k. $$ Hence, $a_i\sigma_i^2$ does not depend on $i$. Assume that $a_k\sigma_k^2=\tau^2$ for every $k$, then $$ \sigma^2=\tau^2+\sigma^2\tau^2\sum_{k=1}^n\frac1{\sigma_k^2}, $$ which yields the value of $\tau^2$. Finally, $$ \mathbb E(\theta\mid s_1,\ldots,s_n)=\tau^2\sum_{k=1}^n\frac{s_k}{\sigma_k^2},\qquad\frac1{\tau^2}=\frac1{\sigma^2}+\sum\limits_{k=1}^n\frac1{\sigma_k^2}. $$

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Thanks Did. I believe your argument is valid. –  user54626 Jan 28 '13 at 20:27

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