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I know that Gauss-Markov process expected value is always zero but in normal distribution, mean can vary and Gauss-Markov process is a subtype of Gaussian process.

The reason behind this question is my homework. I have to find $E[y(x)]$ and $R_y[t_1,t_2]$ assuming $x(t)$ is Gaussian process. The transformation is linear: $y(t)=t^{2}x(t)+e^{-x(t)}$. If I assume $E[x(t)]=0$ the solutions are very easy, for example mean: $E[y(t)]=E[t^{2}x(t)]+E[\lambda e^{-\lambda x(t)}]=t^{2}E[x(t)]+\frac{1}{\lambda}=0+1=1$ I am unsure if I understand Gaussian process behaviour properly.

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A Gaussian process need not have zero mean; indeed, $E[X(t)]$ can be a function of $t$. Also, even if $E[X(t)] = 0$ for all $t$, it is not true that $$E[y(t)] = E[t^2X(t) + e^{-X(t)}] = 0 + 1 = 1.$$ Note that $E[e^{-X(t)}]$ will give you the value of the moment generating function of the Gaussian random variable $X(t)$ at $-1$. – Dilip Sarwate Feb 13 '12 at 15:59
Thanks, especially for pointing out moment generating function. – Maciej Kucia Feb 13 '12 at 17:29
I would understand the term "Gaussian process" to mean a process $X$ in which for every finite set of times $t_1,\ldots,t_n$ and every set of real coefficients $c_1,\ldots,c_n$, the linear combination $c_1 X(t_1)+\cdots+c_n X(t_n)$ has a Gaussian distribution. That doesn't require zero mean. But maybe a definition posited in some particular book doesn't do it that way. – Michael Hardy Feb 13 '12 at 17:35
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This is a matter of definition. Some textbooks or authors will take "Gaussian process" to include the requirement that the process have mean zero; others do not. You should check the definition in your book, or ask your instructor.

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Thanks for help. I will ask my instructor, unfortunately he has a winter break right now. For now the best I can do is to find answers assuming $E[x(t)]=m(t)$ and after that checking solution for $m(t)=0$. – Maciej Kucia Feb 13 '12 at 17:33

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