Assume I have $y = f(x) \ne \mathrm{constant} $ and $(x_1 , \ldots ,x_N)$ a sequence N random input points, is there a set of criteria or a theorem that tells me that the output sequence $(y_1, \ldots, y_N)$ is a random sequence as well. Can it be extended to multivariable and vector functions $f(\vec{x})$ and $\vec{f}(\vec{x})$?
Given a random sequence of input points does it always produce a random output, excluding $f(x)$ = constant?
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sequences-and-series
stochastic-processes
random
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0I don't see any problem considering the constant function outputting a random variable ; it gives something with probability $1$ and something else with probability $0$... but it is still random, if you define "random" properly. – 2012-08-11
2 Answers
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The answer is yes. It doesnt matter if you have a deteministic or stochastic function $f$. If your function is of deterministic type then you have
$f_Y(y)=f_X(g^{-1}(y))|\frac{d g(y)^{-1}}{d y}|$
where $g$ is the function that you applied $f_X$ is the previous density with respect to $X$ and $f_Y$ is the density afther you apply the function.
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You need to define random in this context. Let $f(n)= n \pmod 2 \text{ for } n \in [0,63]$. If you feed it random input, each input value corresponds to $6$ bits. The output has only one bit per value, so by some definitions it is less random. It can be even worse if you don't cover the whole domain of $f$. If all your inputs are odd, the output will be constant $1$, despite being "random".