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Assume that $n(t)$ is a White Gaussian Noise (WGN) process with $E[n(t)]=0$, $E[n(t)^2]=\sigma^2$ and $x(t)$ a deterministic function defined in $[0,T]$. How can I compute from first principles the variance of $g(T)$ defined as

$$g(T)=\int_0^Tx(t)n(t)dt.$$

Any references to elementary textbooks on stochastic processes are also welcome.

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is the e(t) in the definition of g(T) the n(t) you introduced before? –  tibL Apr 19 '12 at 17:20
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You need the autocorrelation function of the process, not just the variance. –  Dilip Sarwate Apr 19 '12 at 17:25
    
Yes, the noise is white and e(t)=n(t). The text is now correct. –  Arrigo Benedetti Apr 19 '12 at 21:57
    
If the autocorrelation function is $$E[n(t)n(s)] = R_n(t-s)=\begin{cases}\sigma^2,&t=s,\\0,&t\neq s,\end{cases}$$ then the integral expression in Nate Eldredge's answer gives $\operatorname{var}(g(T))=0$. If the autocorrelation function is $\sigma^2\delta(t-s)$ (note the difference) then see my comment on that answer as well as this question. –  Dilip Sarwate Apr 20 '12 at 11:06
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Assuming that $e(t)$ is supposed to be $n(t)$ and that you know the covariances $E[n(s) n(t)]$, then note that $$g(T)^2 = \int_0^T \int_0^T x(s) x(t) n(s) n(t)\,ds\,dt$$ and so by Fubini's theorem $$E[g(T)^2] = \int_0^T \int_0^T x(s) x(t) E[n(s) n(t)]\,ds\,dt.$$

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@ArrigoBenedetti This calculation occurs in engineering applications very often where $n(t)$ is assumed to be a white Gaussian noise process with autocorrelation function $\sigma^2\delta(t-s)$ and so $E[g(T)] = 0$ while the variance becomes $$\text{var}(g(T)=\sigma^2\int_0^T x^2(t) \mathrm dt.$$ See for example Appendix A of this document. –  Dilip Sarwate Apr 19 '12 at 19:11
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These heuristics are nice, but it may be important to mention that they're not too much more than this. In particular, the integrals, as shown, don't exist. –  cardinal Apr 20 '12 at 4:20
    
@cardinal Would you please comment on or respond to this question of mine on this issue? Thanks. –  Dilip Sarwate Apr 20 '12 at 11:10
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The application of Fubini's theorem is illegal here, as mentioned by @cardinal. –  Did Jan 15 '13 at 6:55
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