I know the sum of independent Gaussian distributed random variables also has a Gaussian distribution. say we have a sequence of Gaussian random variables not necessarily independent $ X_0, X_1, X_2, \cdots, X_{n-1}, X_n$ and say $ Z_n = X_n - X_{n-1} $ if we know for all $n \geq 1$ & the sequence $$ Z_1,Z_2, \cdots, Z_n $$ are all independent with gaussian distribution.

now let $$ \{Y_n\} = \{(a_1X_1 - a_0 X_0), \cdots, (a_nX_n - a_{n-1} X_{n-1}) \} $$ where $a_n \in [0,\infty) $ how can be sure $Y_n$ is a sequence of Gaussian random variables too ?

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    $\begingroup$ Can you say something about the relationship between $X_0$ and the $Z_i$? Are they independent too? $\endgroup$ – Dinesh Jan 6 '11 at 17:28
  • $\begingroup$ lets say they were independent $\endgroup$ – almost_sure Jan 6 '11 at 17:31
  • $\begingroup$ Why do you expect this to be true? $\endgroup$ – Rasmus Jan 6 '11 at 17:47

Verify that $$ Y_1 = a_1 Z_1 + (a_1 - a_0 )X_0, $$ $$ Y_2 = a_2 Z_2 + (a_2 - a_1 )Z_1 + (a_2 - a_1 )X_0, $$ etc. Following the comments above, if the $Z_i$ are independent of $X_0$, then each $Y_i$ is a linear combination of independent Gaussian random variables, hence Gaussian.

EDIT: Let's explain why an additional assumption (such as the one we used, namely that the $Z_i$ are independent of $X_0$) is necessary here. Suppose that the following assertion is true: ($\star$) There exist normal random variables $X$ and $Y$, and non-zero constants $a$, $b$, $a'$, and $b'$, such that $aX+bY$ is normal but $a'X+b'Y$ is not. Then, letting $X_0 = -bY$ and $X_n=aX$ for all $n \geq 1$, we have $Z_1 = aX+bY$ and $Z_n = 0$ for all $n \geq 2$. Hence, the $X_i$ are normal and the $Z_i$ are independent normal (using the common convention that constants are normal with variance $0$). However, $(a'/a)X_1-(b'/b)X_0 = a'X + b'Y$, and is hence not normal. So, we only need to explain why we should expect that ($\star$) is true (though it might be difficult to give an example where it is satisfied). For this purpose, consider the fact that $X$ and $Y$ are jointly normal if and only if ANY linear combination of $X$ and $Y$ is univariate normal...

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