Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the following data: $$ x(t) = A + \omega(t) $$ where $ \omega(t) $ is an AWGN with zero mean, what would be likelihood function $p(x(t);A)$?

I know it could be proven to be: $$ p(x;A) = C \exp\left(- \frac{\int (x(t) - A))^2 dt}{2\sigma^2}\right) $$

Yet I don't know the reason and the formal proof.


Both RHS and LHS depends on the whole random process.

share|cite|improve this question
What is an AEGN? Something Something Gaussian Noise? –  Chris Taylor Oct 27 '11 at 7:42
Additive white Gaussian noise. –  Emre Oct 27 '11 at 9:27
E = white? $ $ $ $ –  Did Oct 27 '11 at 9:35
Apparently, E is a typo since on some keyboards W and E are close. Could you please check if my correction fits the sense of the question. –  Ilya Oct 27 '11 at 10:48
Drazick: the LHS of the last equation in your post depends on $x(t)$ while the RHS depends on $(x(s))_{0\leqslant s\leqslant t}$. You might wish to explain. –  Did Oct 27 '11 at 10:54

Your Answer


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

Browse other questions tagged or ask your own question.