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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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    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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