Vector of normal distributes random variables

If I have $n$ random variables $X^n=(X_t^{(n)})_{t\ge 0}$, all $X^i$ normal distributed and they are independent. Now I define new processes:

$$Z_t:=X^{(1)}_t+\dots+X^{(n)}_t$$

Since $X^{(i)}$ are independent, $Z_t$ is normal distributed too. My question is: Is $Z=(Z_t)_{t\ge 0}$ a Gaussian process, i.e. is for $t_1<\dots<t_n$ the random vector $(Z_{t_1},\dots,Z_{t_n})$ multivariat normal distributed? If so why?

hulik

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Please use $X^{(i)}$ instead of $X^i$. You may need to compute the variance of the processes at some time –  Dilip Sarwate Mar 7 '12 at 13:30
Hint: linear transformations of Gaussian processes yield Gaussian processes. –  Dilip Sarwate Mar 7 '12 at 13:33
@ Dilip Sarwate: I edited my post. Please could you give some further explanations. All I know is, that $Z_t$ is normal distributed, but how should I use here linear transformation. Thank you for your help –  user20869 Mar 8 '12 at 13:45
See for example here or here –  Dilip Sarwate Mar 8 '12 at 14:19

I write $m$ instead of $n$ for the number of time points, because $n$ is already used for the number of random processes.
Since each $X^{(i)}$ is a Gaussian process, the random vector $$(X_{t_1}^{(i)},\dots X_{t_m}^{(i)})$$ has a multivariate normal distribution for each $i=1,\dots,n$. Putting $n$ independent vectors of this kind together makes a normally distributed vector of length $mn$: $$(X_{t_1}^{(1)},\dots X_{t_m}^{(1)}, X_{t_1}^{(2)},\dots X_{t_m}^{(2)}, \dots X_{t_1}^{(n)},\dots X_{t_m}^{(n)})$$ Pushforward by a linear map that sums the entries corresponding to the same time is also a normal distribution. Thus, $(Z_{t_1} ,\dots Z_{t_m} )$ is normally distributed.