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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.

Note:

Both RHS and LHS depends on the whole random process.

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    What is an AEGN? Something Something Gaussian Noise?2011-10-27
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    Additive white Gaussian noise.2011-10-27
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    E = white? $ $ $ $2011-10-27
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    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.2011-10-27
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    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.2011-10-27
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    @DidierPiau, I didn't get your comment. It is the notation used in Likelihood Function. How does the observation $ x(t) $ pdf depends on the parameter $ A $. The RHS is the pdf of the observation $ x(t) $ given $ A $. Does it make sense? Thanks.2011-10-27
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    Drazick: No it does not (make sense). Writing the LHS as $p(x(t),A)$ is saying that the LHS depends on ($A$ and on) the real valued random variable $x(t)$. But the RHS depends on the whole process $(x(s))_s$. If you want the LHS to depend on $(x(s))_s$, you should write something like $p(x,A)$ where $x=(x(s))_s$.2011-10-27
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    I see. Well, I meant the whole process on LHS. that why I used $ x(t) $ instead of $ x(t_0) $ for that matter. I will edit it.2011-10-28

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