Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I just found Conjugate Prior which I think answer my question! – user54626 Jan 25 '13 at 5:17
up vote 0 down vote accepted

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, $$ \mathbb E(\theta\mid s_1,\ldots,s_n)=\tau^2\sum_{k=1}^n\frac{s_k}{\sigma_k^2},\qquad\frac1{\tau^2}=\frac1{\sigma^2}+\sum\limits_{k=1}^n\frac1{\sigma_k^2}. $$

share|cite|improve this answer
Thanks Did. I believe your argument is valid. – user54626 Jan 28 '13 at 20:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.