This answer says that any casual sequence of the kind $y_n = y_{n-1} + y_{n-2} + y_{n-3} + \ldots $ will stay constant-0 because $y_0$ is a sum of zeroes, so is $y_1$ and the rest of the sequence. I can also put it the way $Y(z) = z^{-1}Y(z)+z^{-2}Y(z)+z^{-3}Y(z) + \ldots$, which either means that $z^{-1}+z^{-2}+\ldots+z^{-n} = 1$, which is unlikely, or simply $Y(z) = 0$, which is more plausible. But, Fibonacci is also of this kind. Does it mean that it is non-casual or what?
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Fibonacci is defined as $$F_1 = 1, F_2 = 1\\F_{n+1} = F_n + F_{n-1}$$
and never hits zero since it is obviously increasing. If you would take $$G_1=0,G_2=0\\G_{n+1} = G_n + G_{n-1},$$
then of course the sequence will be constantly zero.
EDIT:
According to your link, a sequence is causal if $x_n=0$ for $n<0$, however, this means that the definition of causal sequences only applies to sequences where $x_n$ is defined both for positive and negative values of $n$. In the case of the Fibonacci sequence, however, $F_n$ is defined only for $n\geq 0$, so it is, by definition, not causal.
5xum
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It seems there is a way to make the Fibonacci causal, http://math.stackexchange.com/questions/279868 – Val Jul 31 '14 at 11:38
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@val There is a simpler way of proving that: Any causal sequence with a recursive formula $a_n=\alpha_1a_{n-1} +\dots + \alpha_k a_{n-k}$ is obviously constantly equal to $0$. This means Fibonacci is not causal. – 5xum Jul 31 '14 at 11:40
A sequence is called causal **iff** $x_n = 0$ for all $n <0$.– Val Jul 31 '14 at 11:08